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1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
aarch64
-O0
ARM64 gcc 15.2.0
cmp: sub sp, sp, #16 str x0, [sp, 8] str x1, [sp] ldr x0, [sp, 8] ldr w1, [x0] ldr x0, [sp] ldr w0, [x0] sub w0, w1, w0 add sp, sp, 16 ret twoSum: stp x29, x30, [sp, -64]! mov x29, sp str x0, [sp, 24] str w1, [sp, 20] str w2, [sp, 16] ldr w0, [sp, 20] add w0, w0, 1 sxtw x0, w0 lsl x0, x0, 3 bl malloc str x0, [sp, 48] str wzr, [sp, 60] b .L4 .L5: ldrsw x0, [sp, 60] lsl x0, x0, 2 ldr x1, [sp, 24] add x1, x1, x0 ldrsw x0, [sp, 60] lsl x0, x0, 3 ldr x2, [sp, 48] add x0, x2, x0 ldr w1, [x1] str w1, [x0] ldrsw x0, [sp, 60] lsl x0, x0, 3 ldr x1, [sp, 48] add x0, x1, x0 ldr w1, [sp, 60] str w1, [x0, 4] ldr w0, [sp, 60] add w0, w0, 1 str w0, [sp, 60] .L4: ldr w1, [sp, 60] ldr w0, [sp, 20] cmp w1, w0 blt .L5 ldrsw x1, [sp, 20] adrp x0, cmp add x3, x0, :lo12:cmp mov x2, 8 ldr x0, [sp, 48] bl qsort str wzr, [sp, 60] ldr w0, [sp, 20] sub w0, w0, #1 str w0, [sp, 56] b .L6 .L10: ldrsw x0, [sp, 60] lsl x0, x0, 3 ldr x1, [sp, 48] add x0, x1, x0 ldr w1, [x0] ldrsw x0, [sp, 56] lsl x0, x0, 3 ldr x2, [sp, 48] add x0, x2, x0 ldr w0, [x0] add w0, w1, w0 str w0, [sp, 44] ldr w1, [sp, 44] ldr w0, [sp, 16] cmp w1, w0 bne .L7 mov x0, 8 bl malloc str x0, [sp, 32] ldrsw x0, [sp, 60] lsl x0, x0, 3 ldr x1, [sp, 48] add x0, x1, x0 ldr w1, [x0, 4] ldr x0, [sp, 32] str w1, [x0] ldrsw x0, [sp, 56] lsl x0, x0, 3 ldr x1, [sp, 48] add x1, x1, x0 ldr x0, [sp, 32] add x0, x0, 4 ldr w1, [x1, 4] str w1, [x0] ldr x0, [sp, 48] bl free ldr x0, [sp, 32] b .L8 .L7: ldr w1, [sp, 44] ldr w0, [sp, 16] cmp w1, w0 bge .L9 ldr w0, [sp, 60] add w0, w0, 1 str w0, [sp, 60] b .L6 .L9: ldr w0, [sp, 56] sub w0, w0, #1 str w0, [sp, 56] .L6: ldr w1, [sp, 60] ldr w0, [sp, 56] cmp w1, w0 blt .L10 ldr x0, [sp, 48] bl free mov x0, 0 .L8: ldp x29, x30, [sp], 64 ret
1
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
aarch64
-O1
ARM64 gcc 15.2.0
cmp: ldr w2, [x0] ldr w0, [x1] sub w0, w2, w0 ret twoSum: stp x29, x30, [sp, -48]! mov x29, sp stp x19, x20, [sp, 16] stp x21, x22, [sp, 32] mov x22, x0 mov w20, w1 mov w21, w2 add w0, w1, 1 sbfiz x0, x0, 3, 32 bl malloc mov x19, x0 cmp w20, 0 ble .L3 mov x3, x0 sxtw x0, w20 mov x1, 0 .L4: ldr w2, [x22, x1, lsl 2] stp w2, w1, [x3], 8 add x1, x1, 1 cmp x1, x0 bne .L4 adrp x3, cmp add x3, x3, :lo12:cmp mov x2, 8 mov x0, x19 bl qsort sub w0, w20, #1 cmp w0, 0 ble .L5 mov w2, 0 b .L10 .L14: mov x0, 8 bl malloc mov x21, x0 ldr w0, [x22, 4] ldr w1, [x20, 4] stp w0, w1, [x21] mov x0, x19 bl free b .L2 .L8: sub w0, w0, #1 .L9: cmp w2, w0 bge .L5 .L10: sbfiz x1, x2, 3, 32 add x22, x19, x1 sbfiz x3, x0, 3, 32 add x20, x19, x3 ldr w1, [x19, x1] ldr w3, [x19, x3] add w1, w1, w3 cmp w1, w21 beq .L14 bge .L8 add w2, w2, 1 b .L9 .L3: adrp x3, cmp add x3, x3, :lo12:cmp mov x2, 8 sxtw x1, w20 bl qsort .L5: mov x0, x19 bl free mov x21, 0 .L2: mov x0, x21 ldp x19, x20, [sp, 16] ldp x21, x22, [sp, 32] ldp x29, x30, [sp], 48 ret
2
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
aarch64
-O2
ARM64 gcc 15.2.0
cmp: ldr w2, [x0] ldr w0, [x1] sub w0, w2, w0 ret twoSum: stp x29, x30, [sp, -48]! mov x29, sp stp x21, x22, [sp, 32] mov x22, x0 add w0, w1, 1 stp x19, x20, [sp, 16] mov w20, w1 mov w21, w2 sbfiz x0, x0, 3, 32 bl malloc sxtw x4, w20 mov x19, x0 cmp w20, 0 ble .L4 mov x1, 0 .L5: ldr w2, [x22, x1, lsl 2] stp w2, w1, [x0], 8 add x1, x1, 1 cmp x1, x4 bne .L5 mov x0, x19 adrp x3, cmp mov x2, 8 add x3, x3, :lo12:cmp bl qsort subs w1, w20, #1 beq .L7 mov w3, 0 b .L12 .L21: add w3, w3, 1 cmp w1, w3 ble .L7 .L12: ubfiz x0, x3, 3, 32 ubfiz x2, x1, 3, 32 add x22, x19, x0 add x20, x19, x2 ldr w0, [x19, x0] ldr w2, [x19, x2] add w0, w0, w2 cmp w0, w21 beq .L20 blt .L21 sub w1, w1, #1 cmp w1, w3 bgt .L12 .L7: mov x0, x19 mov x21, 0 bl free mov x0, x21 ldp x19, x20, [sp, 16] ldp x21, x22, [sp, 32] ldp x29, x30, [sp], 48 ret .L20: mov x0, 8 bl malloc mov x21, x0 ldr w1, [x20, 4] ldr w0, [x22, 4] stp w0, w1, [x21] mov x0, x19 bl free mov x0, x21 ldp x19, x20, [sp, 16] ldp x21, x22, [sp, 32] ldp x29, x30, [sp], 48 ret .L4: mov x1, x4 adrp x3, cmp mov x2, 8 add x3, x3, :lo12:cmp bl qsort b .L7
3
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
aarch64
-O3
ARM64 gcc 15.2.0
cmp: ldr w2, [x0] ldr w0, [x1] sub w0, w2, w0 ret twoSum: stp x29, x30, [sp, -64]! mov x29, sp str x23, [sp, 48] mov x23, x0 add w0, w1, 1 stp x21, x22, [sp, 32] mov w22, w1 mov w21, w2 sbfiz x0, x0, 3, 32 stp x19, x20, [sp, 16] bl malloc sxtw x1, w22 mov x19, x0 cmp w22, 0 ble .L4 sub w20, w22, #1 cmp w20, 2 bls .L16 adrp x2, .LC0 lsr w3, w22, 2 movi v29.4s, 0x4 mov x0, x23 ldr q31, [x2, #:lo12:.LC0] add x3, x23, w3, uxtw 4 mov x2, x19 .L6: ldr q30, [x0], 16 st2 {v30.4s - v31.4s}, [x2], 32 add v31.4s, v31.4s, v29.4s cmp x3, x0 bne .L6 and w0, w22, -4 tst x22, 3 beq .L7 .L5: ubfiz x2, x0, 3, 32 add x3, x19, w0, uxtw 3 ldr w5, [x23, w0, uxtw 2] add w4, w0, 1 str w5, [x19, x2] str w0, [x3, 4] cmp w22, w4 ble .L8 add x23, x23, w0, uxtw 2 add x3, x2, 8 add x5, x19, x3 add w0, w0, 2 ldr w6, [x23, 4] str w6, [x19, x3] str w4, [x5, 4] cmp w22, w0 ble .L7 add x2, x2, 16 ldr w4, [x23, 8] add x3, x19, x2 str w4, [x19, x2] str w0, [x3, 4] .L7: mov x0, x19 adrp x3, cmp mov x2, 8 add x3, x3, :lo12:cmp bl qsort mov w3, w20 .L15: mov w4, 0 b .L14 .L27: add w4, w4, 1 cmp w3, w4 ble .L9 .L14: ubfiz x2, x4, 3, 32 ubfiz x1, x3, 3, 32 add x20, x19, x1 add x22, x19, x2 ldr w0, [x19, x2] ldr w1, [x19, x1] add w0, w0, w1 cmp w0, w21 beq .L26 blt .L27 sub w3, w3, #1 cmp w3, w4 bgt .L14 .L9: mov x0, x19 mov x21, 0 bl free ldr x23, [sp, 48] mov x0, x21 ldp x19, x20, [sp, 16] ldp x21, x22, [sp, 32] ldp x29, x30, [sp], 64 ret .L8: mov x0, x19 adrp x3, cmp mov x2, 8 add x3, x3, :lo12:cmp bl qsort mov w3, w20 cbnz w20, .L15 b .L9 .L26: mov x0, 8 bl malloc mov x21, x0 ldr w2, [x20, 4] ldr w1, [x22, 4] mov x0, x19 stp w1, w2, [x21] bl free ldr x23, [sp, 48] mov x0, x21 ldp x19, x20, [sp, 16] ldp x21, x22, [sp, 32] ldp x29, x30, [sp], 64 ret .L4: adrp x3, cmp mov x2, 8 add x3, x3, :lo12:cmp bl qsort b .L9 .L16: mov w0, 0 b .L5 .LC0: .word 0 .word 1 .word 2 .word 3
4
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
aarch64
-O0
armv8-a clang 21.1.0
cmp: sub sp, sp, #16 str x0, [sp, #8] str x1, [sp] ldr x8, [sp, #8] ldr w8, [x8] ldr x9, [sp] ldr w9, [x9] subs w0, w8, w9 add sp, sp, #16 ret twoSum: sub sp, sp, #80 stp x29, x30, [sp, #64] add x29, sp, #64 stur x0, [x29, #-16] stur w1, [x29, #-20] stur w2, [x29, #-24] ldur w8, [x29, #-20] add w9, w8, #1 mov w8, w9 sxtw x8, w8 lsl x0, x8, #3 bl malloc str x0, [sp, #8] str wzr, [sp, #28] b .LBB1_1 .LBB1_1: ldr w8, [sp, #28] ldur w9, [x29, #-20] subs w8, w8, w9 b.ge .LBB1_4 b .LBB1_2 .LBB1_2: ldur x8, [x29, #-16] ldrsw x9, [sp, #28] ldr w8, [x8, x9, lsl #2] ldr x9, [sp, #8] ldrsw x10, [sp, #28] lsl x10, x10, #3 str w8, [x9, x10] ldr w8, [sp, #28] ldr x9, [sp, #8] ldrsw x10, [sp, #28] add x9, x9, x10, lsl #3 str w8, [x9, #4] b .LBB1_3 .LBB1_3: ldr w8, [sp, #28] add w8, w8, #1 str w8, [sp, #28] b .LBB1_1 .LBB1_4: ldr x0, [sp, #8] ldursw x1, [x29, #-20] mov x2, #8 adrp x3, cmp add x3, x3, :lo12:cmp bl qsort str wzr, [sp, #28] ldur w8, [x29, #-20] subs w8, w8, #1 str w8, [sp, #24] b .LBB1_5 .LBB1_5: ldr w8, [sp, #28] ldr w9, [sp, #24] subs w8, w8, w9 b.ge .LBB1_13 b .LBB1_6 .LBB1_6: ldr x8, [sp, #8] ldrsw x9, [sp, #28] lsl x9, x9, #3 ldr w8, [x8, x9] ldr x9, [sp, #8] ldrsw x10, [sp, #24] lsl x10, x10, #3 ldr w9, [x9, x10] add w8, w8, w9 str w8, [sp, #20] ldr w8, [sp, #20] ldur w9, [x29, #-24] subs w8, w8, w9 b.ne .LBB1_8 b .LBB1_7 .LBB1_7: mov x0, #8 bl malloc str x0, [sp, #32] ldr x8, [sp, #8] ldrsw x9, [sp, #28] add x8, x8, x9, lsl #3 ldr w8, [x8, #4] ldr x9, [sp, #32] str w8, [x9] ldr x8, [sp, #8] ldrsw x9, [sp, #24] add x8, x8, x9, lsl #3 ldr w8, [x8, #4] ldr x9, [sp, #32] str w8, [x9, #4] ldr x0, [sp, #8] bl free ldr x8, [sp, #32] stur x8, [x29, #-8] b .LBB1_14 .LBB1_8: ldr w8, [sp, #20] ldur w9, [x29, #-24] subs w8, w8, w9 b.ge .LBB1_10 b .LBB1_9 .LBB1_9: ldr w8, [sp, #28] add w8, w8, #1 str w8, [sp, #28] b .LBB1_11 .LBB1_10: ldr w8, [sp, #24] subs w8, w8, #1 str w8, [sp, #24] b .LBB1_11 .LBB1_11: b .LBB1_12 .LBB1_12: b .LBB1_5 .LBB1_13: ldr x0, [sp, #8] bl free stur xzr, [x29, #-8] b .LBB1_14 .LBB1_14: ldur x0, [x29, #-8] ldp x29, x30, [sp, #64] add sp, sp, #80 ret
5
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
aarch64
-O1
armv8-a clang 21.1.0
cmp: ldr w8, [x0] ldr w9, [x1] sub w0, w8, w9 ret twoSum: stp x29, x30, [sp, #-48]! stp x22, x21, [sp, #16] stp x20, x19, [sp, #32] mov x29, sp mov w8, w1 mov x22, x0 mov w20, w2 sbfiz x8, x8, #3, #32 mov w21, w1 add x0, x8, #8 bl malloc cmp w21, #1 mov x19, x0 b.lt .LBB1_3 mov x8, xzr mov w9, w21 add x10, x19, #4 .LBB1_2: ldr w11, [x22, x8, lsl #2] stp w11, w8, [x10, #-4] add x8, x8, #1 add x10, x10, #8 cmp x9, x8 b.ne .LBB1_2 .LBB1_3: sxtw x1, w21 adrp x3, cmp add x3, x3, :lo12:cmp mov x0, x19 mov w2, #8 bl qsort subs w8, w21, #1 b.le .LBB1_7 mov w9, wzr .LBB1_5: ubfiz x10, x9, #3, #32 sbfiz x11, x8, #3, #32 sxtw x21, w8 mov w22, w9 ldr w10, [x19, x10] ldr w11, [x19, x11] add w10, w11, w10 cmp w10, w20 b.eq .LBB1_8 cset w8, ge cinc w9, w22, lt sub w8, w21, w8 cmp w9, w8 b.lt .LBB1_5 .LBB1_7: mov x20, xzr b .LBB1_9 .LBB1_8: mov w0, #8 bl malloc add x8, x19, x22, lsl #3 add x9, x19, x21, lsl #3 mov x20, x0 ldr w8, [x8, #4] ldr w9, [x9, #4] stp w8, w9, [x0] .LBB1_9: mov x0, x19 bl free mov x0, x20 ldp x20, x19, [sp, #32] ldp x22, x21, [sp, #16] ldp x29, x30, [sp], #48 ret
6
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
aarch64
-O2
armv8-a clang 21.1.0
cmp: ldr w8, [x0] ldr w9, [x1] sub w0, w8, w9 ret .LCPI1_0: .word 0 .word 1 .word 2 .word 3 twoSum: stp x29, x30, [sp, #-48]! stp x22, x21, [sp, #16] stp x20, x19, [sp, #32] mov x29, sp mov w21, w1 mov x22, x0 mov w20, w2 sbfiz x8, x21, #3, #32 add x0, x8, #8 bl malloc cmp w21, #0 mov x19, x0 b.le .LBB1_3 cmp w21, #8 mov w1, w21 b.hs .LBB1_4 mov x8, xzr b .LBB1_7 .LBB1_3: sxtw x1, w21 adrp x3, cmp add x3, x3, :lo12:cmp mov x0, x19 mov w2, #8 bl qsort b .LBB1_13 .LBB1_4: movi v0.4s, #4 movi v1.4s, #8 adrp x9, .LCPI1_0 and x8, x1, #0x7ffffff8 ldr q3, [x9, :lo12:.LCPI1_0] add x9, x19, #32 add x10, x22, #16 mov x11, x8 .LBB1_5: add v5.4s, v3.4s, v0.4s sub x12, x9, #32 subs x11, x11, #8 ldp q2, q4, [x10, #-16] add x10, x10, #32 st2 { v2.4s, v3.4s }, [x12] add v3.4s, v3.4s, v1.4s st2 { v4.4s, v5.4s }, [x9] add x9, x9, #64 b.ne .LBB1_5 cmp x8, x1 b.eq .LBB1_9 .LBB1_7: add x9, x19, x8, lsl #3 add x9, x9, #4 .LBB1_8: ldr w10, [x22, x8, lsl #2] stp w10, w8, [x9, #-4] add x8, x8, #1 add x9, x9, #8 cmp x1, x8 b.ne .LBB1_8 .LBB1_9: adrp x3, cmp add x3, x3, :lo12:cmp mov x0, x19 mov w2, #8 bl qsort subs w8, w21, #1 b.eq .LBB1_13 mov w9, wzr .LBB1_11: ubfiz x10, x9, #3, #32 sbfiz x11, x8, #3, #32 sxtw x21, w8 mov w22, w9 ldr w10, [x19, x10] ldr w11, [x19, x11] add w10, w11, w10 cmp w10, w20 b.eq .LBB1_15 cset w8, ge cinc w9, w22, lt sub w8, w21, w8 cmp w9, w8 b.lt .LBB1_11 .LBB1_13: mov x20, xzr .LBB1_14: mov x0, x19 bl free mov x0, x20 ldp x20, x19, [sp, #32] ldp x22, x21, [sp, #16] ldp x29, x30, [sp], #48 ret .LBB1_15: mov w0, #8 bl malloc add x8, x19, x22, lsl #3 add x9, x19, x21, lsl #3 mov x20, x0 ldr w8, [x8, #4] ldr w9, [x9, #4] stp w8, w9, [x0] b .LBB1_14
7
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
aarch64
-O3
armv8-a clang 21.1.0
cmp: ldr w8, [x0] ldr w9, [x1] sub w0, w8, w9 ret .LCPI1_0: .word 0 .word 1 .word 2 .word 3 twoSum: stp x29, x30, [sp, #-48]! stp x22, x21, [sp, #16] stp x20, x19, [sp, #32] mov x29, sp mov w21, w1 mov x22, x0 mov w20, w2 sbfiz x8, x21, #3, #32 add x0, x8, #8 bl malloc cmp w21, #0 mov x19, x0 b.le .LBB1_3 cmp w21, #8 mov w1, w21 b.hs .LBB1_4 mov x8, xzr b .LBB1_7 .LBB1_3: sxtw x1, w21 adrp x3, cmp add x3, x3, :lo12:cmp mov x0, x19 mov w2, #8 bl qsort b .LBB1_10 .LBB1_4: movi v0.4s, #4 movi v1.4s, #8 adrp x9, .LCPI1_0 and x8, x1, #0x7ffffff8 ldr q3, [x9, :lo12:.LCPI1_0] add x9, x19, #32 add x10, x22, #16 mov x11, x8 .LBB1_5: add v5.4s, v3.4s, v0.4s sub x12, x9, #32 subs x11, x11, #8 ldp q2, q4, [x10, #-16] add x10, x10, #32 st2 { v2.4s, v3.4s }, [x12] add v3.4s, v3.4s, v1.4s st2 { v4.4s, v5.4s }, [x9] add x9, x9, #64 b.ne .LBB1_5 cmp x8, x1 b.eq .LBB1_9 .LBB1_7: add x9, x19, x8, lsl #3 add x9, x9, #4 .LBB1_8: ldr w10, [x22, x8, lsl #2] stp w10, w8, [x9, #-4] add x8, x8, #1 add x9, x9, #8 cmp x1, x8 b.ne .LBB1_8 .LBB1_9: adrp x3, cmp add x3, x3, :lo12:cmp mov x0, x19 mov w2, #8 bl qsort subs w8, w21, #1 b.ne .LBB1_12 .LBB1_10: mov x20, xzr .LBB1_11: mov x0, x19 bl free mov x0, x20 ldp x20, x19, [sp, #32] ldp x22, x21, [sp, #16] ldp x29, x30, [sp], #48 ret .LBB1_12: mov x21, xzr b .LBB1_14 .LBB1_13: add x21, x21, #1 cmp w21, w8 b.ge .LBB1_10 .LBB1_14: lsl x9, x21, #3 sbfiz x10, x8, #3, #32 ldr w9, [x19, x9] ldr w10, [x19, x10] add w9, w10, w9 cmp w9, w20 b.eq .LBB1_17 b.lt .LBB1_13 sub w8, w8, #1 cmp w21, w8 b.lt .LBB1_14 b .LBB1_10 .LBB1_17: mov w0, #8 sxtw x22, w8 bl malloc add x8, x19, x21, lsl #3 add x9, x19, x22, lsl #3 mov x20, x0 ldr w8, [x8, #4] ldr w9, [x9, #4] stp w8, w9, [x0] b .LBB1_11
8
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
mips64
-O0
mips64 clang 21.1.0
cmp: .Lfunc_begin0 = .Ltmp0 daddiu $sp, $sp, -32 sd $ra, 24($sp) sd $fp, 16($sp) move $fp, $sp sd $4, 8($fp) sd $5, 0($fp) ld $1, 8($fp) lw $1, 0($1) ld $2, 0($fp) lw $2, 0($2) subu $1, $1, $2 move $2, $1 move $sp, $fp ld $fp, 16($sp) ld $ra, 24($sp) daddiu $sp, $sp, 32 jr $ra nop twoSum: .Lfunc_begin1 = .Ltmp3 daddiu $sp, $sp, -96 sd $ra, 88($sp) sd $fp, 80($sp) sd $gp, 72($sp) move $fp, $sp lui $1, %hi(%neg(%gp_rel(twoSum))) daddu $1, $1, $25 daddiu $gp, $1, %lo(%neg(%gp_rel(twoSum))) sd $gp, 8($fp) move $1, $6 move $2, $5 sd $4, 56($fp) sw $2, 52($fp) sw $1, 48($fp) lw $1, 52($fp) dsll $1, $1, 3 daddiu $4, $1, 8 ld $25, %call16(malloc)($gp) jalr $25 nop sd $2, 16($fp) sw $zero, 36($fp) b .LBB1_1 nop .LBB1_1: lw $1, 36($fp) lw $2, 52($fp) slt $1, $1, $2 beqz $1, .LBB1_5 nop b .LBB1_3 nop .LBB1_3: ld $1, 56($fp) lw $3, 36($fp) dsll $2, $3, 2 daddu $1, $1, $2 lw $1, 0($1) ld $2, 16($fp) dsll $3, $3, 3 daddu $2, $2, $3 sw $1, 0($2) lw $1, 36($fp) ld $2, 16($fp) sll $3, $1, 0 dsll $3, $3, 3 daddu $2, $2, $3 sw $1, 4($2) b .LBB1_4 nop .LBB1_4: lw $1, 36($fp) addiu $1, $1, 1 sw $1, 36($fp) b .LBB1_1 nop .LBB1_5: ld $gp, 8($fp) ld $4, 16($fp) lw $5, 52($fp) ld $7, %got_disp(cmp)($gp) ld $25, %call16(qsort)($gp) daddiu $6, $zero, 8 jalr $25 nop sw $zero, 36($fp) lw $1, 52($fp) addiu $1, $1, -1 sw $1, 32($fp) b .LBB1_6 nop .LBB1_6: lw $1, 36($fp) lw $2, 32($fp) slt $1, $1, $2 beqz $1, .LBB1_17 nop b .LBB1_8 nop .LBB1_8: ld $2, 16($fp) lw $1, 36($fp) dsll $1, $1, 3 daddu $1, $2, $1 lw $1, 0($1) lw $3, 32($fp) dsll $3, $3, 3 daddu $2, $2, $3 lw $2, 0($2) addu $1, $1, $2 sw $1, 28($fp) lw $1, 28($fp) lw $2, 48($fp) bne $1, $2, .LBB1_11 nop b .LBB1_10 nop .LBB1_10: ld $gp, 8($fp) ld $25, %call16(malloc)($gp) daddiu $4, $zero, 8 ld $gp, 8($fp) jalr $25 nop sd $2, 40($fp) ld $1, 16($fp) lw $2, 36($fp) dsll $2, $2, 3 daddu $1, $1, $2 lw $1, 4($1) ld $2, 40($fp) sw $1, 0($2) ld $1, 16($fp) lw $2, 32($fp) dsll $2, $2, 3 daddu $1, $1, $2 lw $1, 4($1) ld $2, 40($fp) sw $1, 4($2) ld $4, 16($fp) ld $25, %call16(free)($gp) jalr $25 nop ld $1, 40($fp) sd $1, 64($fp) b .LBB1_18 nop .LBB1_11: lw $1, 28($fp) lw $2, 48($fp) slt $1, $1, $2 beqz $1, .LBB1_14 nop b .LBB1_13 nop .LBB1_13: lw $1, 36($fp) addiu $1, $1, 1 sw $1, 36($fp) b .LBB1_15 nop .LBB1_14: lw $1, 32($fp) addiu $1, $1, -1 sw $1, 32($fp) b .LBB1_15 nop .LBB1_15: b .LBB1_16 nop .LBB1_16: b .LBB1_6 nop .LBB1_17: ld $gp, 8($fp) ld $4, 16($fp) ld $25, %call16(free)($gp) jalr $25 nop daddiu $1, $zero, 0 sd $zero, 64($fp) b .LBB1_18 nop .LBB1_18: ld $2, 64($fp) move $sp, $fp ld $gp, 72($sp) ld $fp, 80($sp) ld $ra, 88($sp) daddiu $sp, $sp, 96 jr $ra nop
9
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
mips64
-O1
mips64 clang 21.1.0
cmp: .Lfunc_begin0 = .Ltmp0 daddiu $sp, $sp, -16 sd $ra, 8($sp) sd $fp, 0($sp) move $fp, $sp lw $1, 0($5) lw $2, 0($4) subu $2, $2, $1 move $sp, $fp ld $fp, 0($sp) ld $ra, 8($sp) jr $ra daddiu $sp, $sp, 16 twoSum: .Lfunc_begin1 = .Ltmp3 daddiu $sp, $sp, -64 sd $ra, 56($sp) sd $fp, 48($sp) sd $gp, 40($sp) sd $20, 32($sp) sd $19, 24($sp) sd $18, 16($sp) sd $17, 8($sp) sd $16, 0($sp) move $fp, $sp lui $1, %hi(%neg(%gp_rel(twoSum))) daddu $1, $1, $25 daddiu $gp, $1, %lo(%neg(%gp_rel(twoSum))) move $17, $6 move $18, $5 move $19, $4 dsll $1, $5, 3 ld $25, %call16(malloc)($gp) jalr $25 daddiu $4, $1, 8 move $16, $2 blez $18, .LBB1_3 sll $5, $18, 0 daddiu $2, $16, 4 dsll $1, $5, 3 daddu $3, $2, $1 addiu $4, $zero, 0 .LBB1_2: lw $1, 0($19) sw $4, 0($2) sw $1, -4($2) daddiu $19, $19, 4 daddiu $2, $2, 8 bne $2, $3, .LBB1_2 addiu $4, $4, 1 .LBB1_3: ld $7, %got_disp(cmp)($gp) ld $25, %call16(qsort)($gp) move $4, $16 jalr $25 daddiu $6, $zero, 8 slti $1, $18, 2 bnez $1, .LBB1_9 daddiu $19, $zero, 0 addiu $2, $18, -1 addiu $3, $zero, 0 .LBB1_5: sll $1, $3, 0 dsll $1, $1, 3 daddu $18, $16, $1 lw $1, 0($18) sll $4, $2, 0 dsll $4, $4, 3 daddu $20, $16, $4 lw $4, 0($20) addu $4, $4, $1 beq $4, $17, .LBB1_8 nop slt $1, $4, $17 addu $3, $3, $1 xori $1, $1, 1 subu $2, $2, $1 slt $1, $3, $2 bnez $1, .LBB1_5 nop b .LBB1_9 nop .LBB1_8: ld $25, %call16(malloc)($gp) jalr $25 daddiu $4, $zero, 8 move $19, $2 lw $1, 4($18) sw $1, 0($2) lw $1, 4($20) sw $1, 4($2) .LBB1_9: ld $25, %call16(free)($gp) jalr $25 move $4, $16 move $2, $19 move $sp, $fp ld $16, 0($sp) ld $17, 8($sp) ld $18, 16($sp) ld $19, 24($sp) ld $20, 32($sp) ld $gp, 40($sp) ld $fp, 48($sp) ld $ra, 56($sp) jr $ra daddiu $sp, $sp, 64
10
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
mips64
-O2
mips64 clang 21.1.0
cmp: .Lfunc_begin0 = .Ltmp0 daddiu $sp, $sp, -16 sd $ra, 8($sp) sd $fp, 0($sp) move $fp, $sp lw $1, 0($5) lw $2, 0($4) subu $2, $2, $1 move $sp, $fp ld $fp, 0($sp) ld $ra, 8($sp) jr $ra daddiu $sp, $sp, 16 twoSum: .Lfunc_begin1 = .Ltmp3 daddiu $sp, $sp, -64 sd $ra, 56($sp) sd $fp, 48($sp) sd $gp, 40($sp) sd $20, 32($sp) sd $19, 24($sp) sd $18, 16($sp) sd $17, 8($sp) sd $16, 0($sp) move $fp, $sp lui $1, %hi(%neg(%gp_rel(twoSum))) daddu $1, $1, $25 daddiu $gp, $1, %lo(%neg(%gp_rel(twoSum))) move $17, $6 move $18, $5 move $19, $4 dsll $1, $5, 3 ld $25, %call16(malloc)($gp) jalr $25 daddiu $4, $1, 8 blez $18, .LBB1_8 move $16, $2 daddiu $2, $16, 4 sll $5, $18, 0 dsll $1, $5, 3 daddu $3, $2, $1 addiu $4, $zero, 0 .LBB1_2: lw $1, 0($19) sw $4, 0($2) sw $1, -4($2) daddiu $19, $19, 4 daddiu $2, $2, 8 bne $2, $3, .LBB1_2 addiu $4, $4, 1 ld $7, %got_disp(cmp)($gp) ld $25, %call16(qsort)($gp) move $4, $16 jalr $25 daddiu $6, $zero, 8 addiu $1, $zero, 1 beq $18, $1, .LBB1_10 daddiu $19, $zero, 0 addiu $2, $18, -1 addiu $3, $zero, 0 .LBB1_5: sll $1, $3, 0 dsll $1, $1, 3 daddu $18, $16, $1 lw $1, 0($18) sll $4, $2, 0 dsll $4, $4, 3 daddu $20, $16, $4 lw $4, 0($20) addu $4, $4, $1 beq $4, $17, .LBB1_9 nop slt $1, $4, $17 addu $3, $3, $1 xori $1, $1, 1 subu $2, $2, $1 slt $1, $3, $2 bnez $1, .LBB1_5 nop b .LBB1_10 nop .LBB1_8: sll $5, $18, 0 ld $7, %got_disp(cmp)($gp) ld $25, %call16(qsort)($gp) move $4, $16 jalr $25 daddiu $6, $zero, 8 b .LBB1_10 daddiu $19, $zero, 0 .LBB1_9: ld $25, %call16(malloc)($gp) jalr $25 daddiu $4, $zero, 8 move $19, $2 lw $1, 4($18) sw $1, 0($2) lw $1, 4($20) sw $1, 4($2) .LBB1_10: ld $25, %call16(free)($gp) jalr $25 move $4, $16 move $2, $19 move $sp, $fp ld $16, 0($sp) ld $17, 8($sp) ld $18, 16($sp) ld $19, 24($sp) ld $20, 32($sp) ld $gp, 40($sp) ld $fp, 48($sp) ld $ra, 56($sp) jr $ra daddiu $sp, $sp, 64
11
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
mips64
-O3
mips64 clang 21.1.0
cmp: .Lfunc_begin0 = .Ltmp0 daddiu $sp, $sp, -16 sd $ra, 8($sp) sd $fp, 0($sp) move $fp, $sp lw $1, 0($5) lw $2, 0($4) subu $2, $2, $1 move $sp, $fp ld $fp, 0($sp) ld $ra, 8($sp) jr $ra daddiu $sp, $sp, 16 twoSum: .Lfunc_begin1 = .Ltmp3 daddiu $sp, $sp, -64 sd $ra, 56($sp) sd $fp, 48($sp) sd $gp, 40($sp) sd $20, 32($sp) sd $19, 24($sp) sd $18, 16($sp) sd $17, 8($sp) sd $16, 0($sp) move $fp, $sp lui $1, %hi(%neg(%gp_rel(twoSum))) move $19, $4 move $17, $6 move $18, $5 daddu $1, $1, $25 daddiu $gp, $1, %lo(%neg(%gp_rel(twoSum))) dsll $1, $5, 3 ld $25, %call16(malloc)($gp) jalr $25 daddiu $4, $1, 8 blez $18, .LBB1_8 move $16, $2 sll $5, $18, 0 daddiu $2, $16, 4 addiu $4, $zero, 0 dsll $1, $5, 3 daddu $3, $2, $1 .LBB1_2: lw $1, 0($19) sw $4, 0($2) daddiu $19, $19, 4 sw $1, -4($2) daddiu $2, $2, 8 bne $2, $3, .LBB1_2 addiu $4, $4, 1 ld $7, %got_disp(cmp)($gp) ld $25, %call16(qsort)($gp) move $4, $16 jalr $25 daddiu $6, $zero, 8 addiu $1, $zero, 1 beq $18, $1, .LBB1_10 daddiu $19, $zero, 0 addiu $2, $18, -1 addiu $3, $zero, 0 .LBB1_5: sll $1, $3, 0 sll $4, $2, 0 dsll $1, $1, 3 dsll $4, $4, 3 daddu $18, $16, $1 daddu $20, $16, $4 lw $1, 0($18) lw $4, 0($20) addu $4, $4, $1 beq $4, $17, .LBB1_9 nop slt $1, $4, $17 addu $3, $3, $1 xori $1, $1, 1 subu $2, $2, $1 slt $1, $3, $2 bnez $1, .LBB1_5 nop b .LBB1_10 nop .LBB1_8: ld $7, %got_disp(cmp)($gp) ld $25, %call16(qsort)($gp) sll $5, $18, 0 move $4, $16 jalr $25 daddiu $6, $zero, 8 b .LBB1_10 daddiu $19, $zero, 0 .LBB1_9: ld $25, %call16(malloc)($gp) jalr $25 daddiu $4, $zero, 8 lw $1, 4($18) move $19, $2 sw $1, 0($2) lw $1, 4($20) sw $1, 4($2) .LBB1_10: ld $25, %call16(free)($gp) jalr $25 move $4, $16 move $2, $19 move $sp, $fp ld $16, 0($sp) ld $17, 8($sp) ld $18, 16($sp) ld $19, 24($sp) ld $20, 32($sp) ld $gp, 40($sp) ld $fp, 48($sp) ld $ra, 56($sp) jr $ra daddiu $sp, $sp, 64
12
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
mips64
-O0
mips64 gcc 15.2.0
cmp: daddiu $sp,$sp,-32 sd $fp,24($sp) move $fp,$sp sd $4,0($fp) sd $5,8($fp) ld $2,0($fp) lw $3,0($2) ld $2,8($fp) lw $2,0($2) subu $2,$3,$2 move $sp,$fp ld $fp,24($sp) daddiu $sp,$sp,32 jr $31 nop twoSum: daddiu $sp,$sp,-80 sd $31,72($sp) sd $fp,64($sp) sd $28,56($sp) move $fp,$sp lui $28,%hi(%neg(%gp_rel(twoSum))) daddu $28,$28,$25 daddiu $28,$28,%lo(%neg(%gp_rel(twoSum))) sd $4,32($fp) move $3,$5 move $2,$6 sll $3,$3,0 sw $3,40($fp) sll $2,$2,0 sw $2,44($fp) lw $2,40($fp) addiu $2,$2,1 dsll $2,$2,3 move $4,$2 ld $2,%call16(malloc)($28) mtlo $2 mflo $25 jalr $25 nop sd $2,8($fp) sw $0,0($fp) b .L4 nop .L5: lw $2,0($fp) dsll $2,$2,2 ld $3,32($fp) daddu $3,$3,$2 lw $2,0($fp) dsll $2,$2,3 ld $4,8($fp) daddu $2,$4,$2 lw $3,0($3) sw $3,0($2) lw $2,0($fp) dsll $2,$2,3 ld $3,8($fp) daddu $2,$3,$2 lw $3,0($fp) sw $3,4($2) lw $2,0($fp) addiu $2,$2,1 sw $2,0($fp) .L4: lw $3,0($fp) lw $2,40($fp) slt $2,$3,$2 bne $2,$0,.L5 nop lw $2,40($fp) ld $7,%got_disp(cmp)($28) li $6,8 # 0x8 move $5,$2 ld $4,8($fp) ld $2,%call16(qsort)($28) mtlo $2 mflo $25 jalr $25 nop sw $0,0($fp) lw $2,40($fp) addiu $2,$2,-1 sw $2,4($fp) b .L6 nop .L10: lw $2,0($fp) dsll $2,$2,3 ld $3,8($fp) daddu $2,$3,$2 lw $3,0($2) lw $2,4($fp) dsll $2,$2,3 ld $4,8($fp) daddu $2,$4,$2 lw $2,0($2) addu $2,$3,$2 sw $2,16($fp) lw $3,16($fp) lw $2,44($fp) bne $3,$2,.L7 nop li $4,8 # 0x8 ld $2,%call16(malloc)($28) mtlo $2 mflo $25 jalr $25 nop sd $2,24($fp) lw $2,0($fp) dsll $2,$2,3 ld $3,8($fp) daddu $2,$3,$2 lw $3,4($2) ld $2,24($fp) sw $3,0($2) lw $2,4($fp) dsll $2,$2,3 ld $3,8($fp) daddu $3,$3,$2 ld $2,24($fp) daddiu $2,$2,4 lw $3,4($3) sw $3,0($2) ld $4,8($fp) ld $2,%call16(free)($28) mtlo $2 mflo $25 jalr $25 nop ld $2,24($fp) b .L8 nop .L7: lw $3,16($fp) lw $2,44($fp) slt $2,$3,$2 beq $2,$0,.L9 nop lw $2,0($fp) addiu $2,$2,1 sw $2,0($fp) b .L6 nop .L9: lw $2,4($fp) addiu $2,$2,-1 sw $2,4($fp) .L6: lw $3,0($fp) lw $2,4($fp) slt $2,$3,$2 bne $2,$0,.L10 nop ld $4,8($fp) ld $2,%call16(free)($28) mtlo $2 mflo $25 jalr $25 nop move $2,$0 .L8: move $sp,$fp ld $31,72($sp) ld $fp,64($sp) ld $28,56($sp) daddiu $sp,$sp,80 jr $31 nop
13
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
mips64
-O1
mips64 gcc 15.2.0
cmp: lw $3,0($4) lw $2,0($5) jr $31 subu $2,$3,$2 twoSum: daddiu $sp,$sp,-64 sd $31,56($sp) sd $28,48($sp) sd $20,40($sp) sd $19,32($sp) sd $18,24($sp) sd $17,16($sp) sd $16,8($sp) lui $28,%hi(%neg(%gp_rel(twoSum))) daddu $28,$28,$25 daddiu $28,$28,%lo(%neg(%gp_rel(twoSum))) move $16,$4 move $17,$5 move $20,$6 addiu $4,$5,1 ld $25,%call16(malloc)($28) 1: jalr $25 dsll $4,$4,3 blez $17,.L4 move $19,$2 move $4,$16 move $5,$0 .L5: lw $3,0($4) sw $3,0($2) sw $5,4($2) move $18,$5 addiu $5,$5,1 daddiu $4,$4,4 bne $17,$5,.L5 daddiu $2,$2,8 ld $7,%got_disp(cmp)($28) li $6,8 # 0x8 ld $25,%call16(qsort)($28) 1: jalr $25 move $4,$19 blez $18,.L6 move $2,$0 b .L11 move $5,$20 .L15: ld $25,%call16(malloc)($28) 1: jalr $25 li $4,8 # 0x8 move $18,$2 lw $2,4($17) sw $2,0($18) lw $2,4($16) sw $2,4($18) ld $25,%call16(free)($28) 1: jalr $25 move $4,$19 b .L16 move $2,$18 .L9: .L10: slt $3,$2,$18 beq $3,$0,.L17 ld $25,%call16(free)($28) .L11: dsll $17,$2,3 daddu $17,$19,$17 dsll $16,$18,3 daddu $16,$19,$16 lw $3,0($17) lw $4,0($16) addu $3,$3,$4 beq $3,$5,.L15 slt $3,$3,$20 beql $3,$0,.L9 addiu $18,$18,-1 b .L10 addiu $2,$2,1 .L4: ld $7,%got_disp(cmp)($28) li $6,8 # 0x8 move $5,$17 ld $25,%call16(qsort)($28) 1: jalr $25 move $4,$2 .L6: ld $25,%call16(free)($28) .L17: 1: jalr $25 move $4,$19 move $18,$0 move $2,$18 .L16: ld $31,56($sp) ld $28,48($sp) ld $20,40($sp) ld $19,32($sp) ld $18,24($sp) ld $17,16($sp) ld $16,8($sp) jr $31 daddiu $sp,$sp,64
14
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
mips64
-O2
mips64 gcc 15.2.0
cmp: lw $3,0($4) lw $2,0($5) jr $31 subu $2,$3,$2 twoSum: daddiu $sp,$sp,-80 sd $28,64($sp) lui $28,%hi(%neg(%gp_rel(twoSum))) daddu $28,$28,$25 daddiu $28,$28,%lo(%neg(%gp_rel(twoSum))) ld $25,%call16(malloc)($28) addiu $2,$5,1 sd $20,56($sp) sd $19,48($sp) sd $17,32($sp) sd $16,24($sp) sd $31,72($sp) sd $18,40($sp) move $16,$4 dsll $4,$2,3 move $17,$5 1: jalr $25 move $20,$6 blez $17,.L5 move $19,$2 move $6,$2 move $3,$0 .L6: lwu $2,0($16) dsll $7,$3,32 dsrl $7,$7,32 dsll $2,$2,32 move $18,$3 or $2,$2,$7 addiu $3,$3,1 sd $2,0($6) daddiu $16,$16,4 bne $17,$3,.L6 daddiu $6,$6,8 ld $25,%call16(qsort)($28) ld $7,%got_disp(cmp)($28) li $6,8 # 0x8 move $5,$17 1: jalr $25 move $4,$19 beq $18,$0,.L23 ld $25,%call16(free)($28) b .L13 move $3,$0 .L22: slt $2,$3,$18 beq $2,$0,.L23 ld $25,%call16(free)($28) .L13: sll $16,$18,0 .L24: dsll $17,$3,32 dsll $16,$16,32 dsrl $17,$17,32 dsrl $16,$16,32 dsll $17,$17,3 dsll $16,$16,3 daddu $17,$19,$17 daddu $16,$19,$16 lw $4,0($16) lw $2,0($17) addu $2,$2,$4 beq $2,$20,.L21 slt $4,$2,$20 bnel $4,$0,.L22 addiu $3,$3,1 addiu $18,$18,-1 slt $2,$3,$18 bnel $2,$0,.L24 sll $16,$18,0 ld $25,%call16(free)($28) .L23: 1: jalr $25 move $4,$19 ld $31,72($sp) ld $28,64($sp) ld $20,56($sp) ld $19,48($sp) ld $18,40($sp) ld $17,32($sp) ld $16,24($sp) move $2,$0 jr $31 daddiu $sp,$sp,80 .L21: ld $25,%call16(malloc)($28) 1: jalr $25 li $4,8 # 0x8 lwu $3,4($17) lwu $4,4($16) ld $25,%call16(free)($28) dsll $3,$3,32 or $3,$3,$4 sd $3,0($2) move $4,$19 1: jalr $25 sd $2,0($sp) ld $31,72($sp) ld $2,0($sp) ld $28,64($sp) ld $20,56($sp) ld $19,48($sp) ld $18,40($sp) ld $17,32($sp) ld $16,24($sp) jr $31 daddiu $sp,$sp,80 .L5: ld $25,%call16(qsort)($28) ld $7,%got_disp(cmp)($28) li $6,8 # 0x8 move $5,$17 1: jalr $25 move $4,$2 b .L23 ld $25,%call16(free)($28)
15
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
mips64
-O3
mips64 gcc 15.2.0
cmp: lw $3,0($4) lw $2,0($5) jr $31 subu $2,$3,$2 twoSum: daddiu $sp,$sp,-80 sd $28,64($sp) lui $28,%hi(%neg(%gp_rel(twoSum))) daddu $28,$28,$25 daddiu $28,$28,%lo(%neg(%gp_rel(twoSum))) ld $25,%call16(malloc)($28) addiu $2,$5,1 sd $20,56($sp) sd $19,48($sp) sd $17,32($sp) sd $16,24($sp) sd $31,72($sp) sd $18,40($sp) move $16,$4 dsll $4,$2,3 move $17,$5 1: jalr $25 move $20,$6 blez $17,.L5 move $19,$2 move $6,$2 move $3,$0 .L6: lwu $2,0($16) dsll $7,$3,32 dsrl $7,$7,32 dsll $2,$2,32 move $18,$3 or $2,$2,$7 addiu $3,$3,1 sd $2,0($6) daddiu $16,$16,4 bne $17,$3,.L6 daddiu $6,$6,8 ld $25,%call16(qsort)($28) ld $7,%got_disp(cmp)($28) li $6,8 # 0x8 move $5,$17 1: jalr $25 move $4,$19 beq $18,$0,.L23 ld $25,%call16(free)($28) b .L13 move $3,$0 .L22: slt $2,$3,$18 beq $2,$0,.L23 ld $25,%call16(free)($28) .L13: sll $16,$18,0 .L24: dsll $17,$3,32 dsll $16,$16,32 dsrl $17,$17,32 dsrl $16,$16,32 dsll $17,$17,3 dsll $16,$16,3 daddu $17,$19,$17 daddu $16,$19,$16 lw $4,0($16) lw $2,0($17) addu $2,$2,$4 beq $2,$20,.L21 slt $4,$2,$20 bnel $4,$0,.L22 addiu $3,$3,1 addiu $18,$18,-1 slt $2,$3,$18 bnel $2,$0,.L24 sll $16,$18,0 ld $25,%call16(free)($28) .L23: 1: jalr $25 move $4,$19 ld $31,72($sp) ld $28,64($sp) ld $20,56($sp) ld $19,48($sp) ld $18,40($sp) ld $17,32($sp) ld $16,24($sp) move $2,$0 jr $31 daddiu $sp,$sp,80 .L21: ld $25,%call16(malloc)($28) 1: jalr $25 li $4,8 # 0x8 lwu $3,4($17) lwu $4,4($16) ld $25,%call16(free)($28) dsll $3,$3,32 or $3,$3,$4 sd $3,0($2) move $4,$19 1: jalr $25 sd $2,0($sp) ld $31,72($sp) ld $2,0($sp) ld $28,64($sp) ld $20,56($sp) ld $19,48($sp) ld $18,40($sp) ld $17,32($sp) ld $16,24($sp) jr $31 daddiu $sp,$sp,80 .L5: ld $25,%call16(qsort)($28) ld $7,%got_disp(cmp)($28) li $6,8 # 0x8 move $5,$17 1: jalr $25 move $4,$2 b .L23 ld $25,%call16(free)($28)
16
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
riscv64
-O0
RISC-V 64 clang 21.1.0
cmp: addi sp, sp, -32 sd ra, 24(sp) sd s0, 16(sp) addi s0, sp, 32 sd a0, -24(s0) sd a1, -32(s0) ld a0, -24(s0) lw a0, 0(a0) ld a1, -32(s0) lw a1, 0(a1) subw a0, a0, a1 ld ra, 24(sp) ld s0, 16(sp) addi sp, sp, 32 ret twoSum: addi sp, sp, -80 sd ra, 72(sp) sd s0, 64(sp) addi s0, sp, 80 sd a0, -32(s0) sw a1, -36(s0) sw a2, -40(s0) lw a0, -36(s0) slli a0, a0, 3 addi a0, a0, 8 call malloc sd a0, -72(s0) li a0, 0 sw a0, -52(s0) j .LBB1_1 .LBB1_1: lw a0, -52(s0) lw a1, -36(s0) bge a0, a1, .LBB1_4 j .LBB1_2 .LBB1_2: ld a0, -32(s0) lw a2, -52(s0) slli a1, a2, 2 add a0, a0, a1 lw a0, 0(a0) ld a1, -72(s0) slli a2, a2, 3 add a1, a1, a2 sw a0, 0(a1) lw a0, -52(s0) ld a1, -72(s0) slli a2, a0, 3 add a1, a1, a2 sw a0, 4(a1) j .LBB1_3 .LBB1_3: lw a0, -52(s0) addiw a0, a0, 1 sw a0, -52(s0) j .LBB1_1 .LBB1_4: ld a0, -72(s0) lw a1, -36(s0) .Lpcrel_hi0: auipc a2, %pcrel_hi(cmp) addi a3, a2, %pcrel_lo(.Lpcrel_hi0) li a2, 8 call qsort li a0, 0 sw a0, -52(s0) lw a0, -36(s0) addiw a0, a0, -1 sw a0, -56(s0) j .LBB1_5 .LBB1_5: lw a0, -52(s0) lw a1, -56(s0) bge a0, a1, .LBB1_13 j .LBB1_6 .LBB1_6: ld a1, -72(s0) lw a0, -52(s0) slli a0, a0, 3 add a0, a0, a1 lw a0, 0(a0) lw a2, -56(s0) slli a2, a2, 3 add a1, a1, a2 lw a1, 0(a1) addw a0, a0, a1 sw a0, -60(s0) lw a0, -60(s0) lw a1, -40(s0) bne a0, a1, .LBB1_8 j .LBB1_7 .LBB1_7: li a0, 8 call malloc sd a0, -48(s0) ld a0, -72(s0) lw a1, -52(s0) slli a1, a1, 3 add a0, a0, a1 lw a0, 4(a0) ld a1, -48(s0) sw a0, 0(a1) ld a0, -72(s0) lw a1, -56(s0) slli a1, a1, 3 add a0, a0, a1 lw a0, 4(a0) ld a1, -48(s0) sw a0, 4(a1) ld a0, -72(s0) call free ld a0, -48(s0) sd a0, -24(s0) j .LBB1_14 .LBB1_8: lw a0, -60(s0) lw a1, -40(s0) bge a0, a1, .LBB1_10 j .LBB1_9 .LBB1_9: lw a0, -52(s0) addiw a0, a0, 1 sw a0, -52(s0) j .LBB1_11 .LBB1_10: lw a0, -56(s0) addiw a0, a0, -1 sw a0, -56(s0) j .LBB1_11 .LBB1_11: j .LBB1_12 .LBB1_12: j .LBB1_5 .LBB1_13: ld a0, -72(s0) call free li a0, 0 sd a0, -24(s0) j .LBB1_14 .LBB1_14: ld a0, -24(s0) ld ra, 72(sp) ld s0, 64(sp) addi sp, sp, 80 ret
17
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
riscv64
-O1
RISC-V 64 clang 21.1.0
cmp: lw a0, 0(a0) lw a1, 0(a1) subw a0, a0, a1 ret twoSum: addi sp, sp, -48 sd ra, 40(sp) sd s0, 32(sp) sd s1, 24(sp) sd s2, 16(sp) sd s3, 8(sp) sd s4, 0(sp) mv s3, a2 mv s2, a1 mv s1, a0 slli s0, a1, 3 addi a0, s0, 8 call malloc mv s4, a0 blez s2, .LBB1_3 li a0, 0 addi a1, s4, 4 add a2, a1, s0 .LBB1_2: lw a3, 0(s1) sw a3, -4(a1) sw a0, 0(a1) addi a0, a0, 1 addi a1, a1, 8 addi s1, s1, 4 bne a1, a2, .LBB1_2 .LBB1_3: .Lpcrel_hi0: auipc a0, %pcrel_hi(cmp) addi a3, a0, %pcrel_lo(.Lpcrel_hi0) li a2, 8 mv a0, s4 mv a1, s2 call qsort li a0, 2 blt s2, a0, .LBB1_7 li a0, 0 addiw s2, s2, -1 .LBB1_5: slli s1, a0, 3 slli s0, s2, 3 add s1, s1, s4 add s0, s0, s4 lw a1, 0(s1) lw a2, 0(s0) addw a1, a1, a2 beq a1, s3, .LBB1_8 slt a1, a1, s3 xori a2, a1, 1 subw s2, s2, a2 addw a0, a0, a1 blt a0, s2, .LBB1_5 .LBB1_7: li s2, 0 j .LBB1_9 .LBB1_8: li a0, 8 call malloc mv s2, a0 lw a0, 4(s1) lw a1, 4(s0) sw a0, 0(s2) sw a1, 4(s2) .LBB1_9: mv a0, s4 call free mv a0, s2 ld ra, 40(sp) ld s0, 32(sp) ld s1, 24(sp) ld s2, 16(sp) ld s3, 8(sp) ld s4, 0(sp) addi sp, sp, 48 ret
18
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
riscv64
-O2
RISC-V 64 clang 21.1.0
cmp: lw a0, 0(a0) lw a1, 0(a1) subw a0, a0, a1 ret twoSum: addi sp, sp, -64 sd ra, 56(sp) sd s0, 48(sp) sd s1, 40(sp) sd s2, 32(sp) sd s3, 24(sp) sd s4, 16(sp) sd s5, 8(sp) mv s5, a2 mv s2, a1 mv s0, a0 slli s4, a1, 3 addi a0, s4, 8 call malloc mv s3, a0 blez s2, .LBB1_7 li a0, 0 addi a1, s3, 4 add a2, a1, s4 .LBB1_2: lw a3, 0(s0) sw a3, -4(a1) sw a0, 0(a1) addi a0, a0, 1 addi a1, a1, 8 addi s0, s0, 4 bne a1, a2, .LBB1_2 .Lpcrel_hi1: auipc a0, %pcrel_hi(cmp) addi a3, a0, %pcrel_lo(.Lpcrel_hi1) li a2, 8 mv a0, s3 mv a1, s2 call qsort addiw s2, s2, -1 beqz s2, .LBB1_10 li a0, 0 .LBB1_5: slli s1, a0, 3 slli s0, s2, 3 add s1, s1, s3 add s0, s0, s3 lw a1, 0(s1) lw a2, 0(s0) addw a1, a1, a2 beq a1, s5, .LBB1_9 slt a1, a1, s5 xori a2, a1, 1 subw s2, s2, a2 addw a0, a0, a1 blt a0, s2, .LBB1_5 j .LBB1_8 .LBB1_7: .Lpcrel_hi0: auipc a0, %pcrel_hi(cmp) addi a3, a0, %pcrel_lo(.Lpcrel_hi0) li a2, 8 mv a0, s3 mv a1, s2 call qsort .LBB1_8: li s2, 0 j .LBB1_10 .LBB1_9: li a0, 8 call malloc mv s2, a0 lw a0, 4(s1) lw a1, 4(s0) sw a0, 0(s2) sw a1, 4(s2) .LBB1_10: mv a0, s3 call free mv a0, s2 ld ra, 56(sp) ld s0, 48(sp) ld s1, 40(sp) ld s2, 32(sp) ld s3, 24(sp) ld s4, 16(sp) ld s5, 8(sp) addi sp, sp, 64 ret
19
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
riscv64
-O3
RISC-V 64 clang 21.1.0
cmp: lw a0, 0(a0) lw a1, 0(a1) subw a0, a0, a1 ret twoSum: addi sp, sp, -64 sd ra, 56(sp) sd s0, 48(sp) sd s1, 40(sp) sd s2, 32(sp) sd s3, 24(sp) sd s4, 16(sp) sd s5, 8(sp) mv s5, a2 mv s2, a1 mv s0, a0 slli s4, a1, 3 addi a0, s4, 8 call malloc mv s3, a0 blez s2, .LBB1_7 li a0, 0 addi a1, s3, 4 add a2, a1, s4 .LBB1_2: lw a3, 0(s0) sw a3, -4(a1) sw a0, 0(a1) addi a0, a0, 1 addi a1, a1, 8 addi s0, s0, 4 bne a1, a2, .LBB1_2 .Lpcrel_hi1: auipc a0, %pcrel_hi(cmp) addi a3, a0, %pcrel_lo(.Lpcrel_hi1) li a2, 8 mv a0, s3 mv a1, s2 call qsort addiw s2, s2, -1 beqz s2, .LBB1_10 li a0, 0 .LBB1_5: slli s1, a0, 3 slli s0, s2, 3 add s1, s1, s3 add s0, s0, s3 lw a1, 0(s1) lw a2, 0(s0) addw a1, a1, a2 beq a1, s5, .LBB1_9 slt a1, a1, s5 xori a2, a1, 1 subw s2, s2, a2 addw a0, a0, a1 blt a0, s2, .LBB1_5 j .LBB1_8 .LBB1_7: .Lpcrel_hi0: auipc a0, %pcrel_hi(cmp) addi a3, a0, %pcrel_lo(.Lpcrel_hi0) li a2, 8 mv a0, s3 mv a1, s2 call qsort .LBB1_8: li s2, 0 j .LBB1_10 .LBB1_9: li a0, 8 call malloc mv s2, a0 lw a0, 4(s1) lw a1, 4(s0) sw a0, 0(s2) sw a1, 4(s2) .LBB1_10: mv a0, s3 call free mv a0, s2 ld ra, 56(sp) ld s0, 48(sp) ld s1, 40(sp) ld s2, 32(sp) ld s3, 24(sp) ld s4, 16(sp) ld s5, 8(sp) addi sp, sp, 64 ret
20
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
riscv64
-O0
RISC-V 64 gcc 15.2.0
cmp: addi sp,sp,-32 sd ra,24(sp) sd s0,16(sp) addi s0,sp,32 sd a0,-24(s0) sd a1,-32(s0) ld a5,-24(s0) lw a4,0(a5) ld a5,-32(s0) lw a5,0(a5) subw a5,a4,a5 sext.w a5,a5 mv a0,a5 ld ra,24(sp) ld s0,16(sp) addi sp,sp,32 jr ra twoSum: addi sp,sp,-64 sd ra,56(sp) sd s0,48(sp) addi s0,sp,64 sd a0,-56(s0) mv a5,a1 mv a4,a2 sw a5,-60(s0) mv a5,a4 sw a5,-64(s0) lw a5,-60(s0) addiw a5,a5,1 sext.w a5,a5 slli a5,a5,3 mv a0,a5 call malloc mv a5,a0 sd a5,-32(s0) sw zero,-20(s0) j .L4 .L5: lw a5,-20(s0) slli a5,a5,2 ld a4,-56(s0) add a4,a4,a5 lw a5,-20(s0) slli a5,a5,3 ld a3,-32(s0) add a5,a3,a5 lw a4,0(a4) sw a4,0(a5) lw a5,-20(s0) slli a5,a5,3 ld a4,-32(s0) add a5,a4,a5 lw a4,-20(s0) sw a4,4(a5) lw a5,-20(s0) addiw a5,a5,1 sw a5,-20(s0) .L4: lw a5,-20(s0) mv a4,a5 lw a5,-60(s0) sext.w a4,a4 sext.w a5,a5 blt a4,a5,.L5 lw a4,-60(s0) lui a5,%hi(cmp) addi a3,a5,%lo(cmp) li a2,8 mv a1,a4 ld a0,-32(s0) call qsort sw zero,-20(s0) lw a5,-60(s0) addiw a5,a5,-1 sw a5,-24(s0) j .L6 .L10: lw a5,-20(s0) slli a5,a5,3 ld a4,-32(s0) add a5,a4,a5 lw a4,0(a5) lw a5,-24(s0) slli a5,a5,3 ld a3,-32(s0) add a5,a3,a5 lw a5,0(a5) addw a5,a4,a5 sw a5,-36(s0) lw a5,-36(s0) mv a4,a5 lw a5,-64(s0) sext.w a4,a4 sext.w a5,a5 bne a4,a5,.L7 li a0,8 call malloc mv a5,a0 sd a5,-48(s0) lw a5,-20(s0) slli a5,a5,3 ld a4,-32(s0) add a5,a4,a5 lw a4,4(a5) ld a5,-48(s0) sw a4,0(a5) lw a5,-24(s0) slli a5,a5,3 ld a4,-32(s0) add a4,a4,a5 ld a5,-48(s0) addi a5,a5,4 lw a4,4(a4) sw a4,0(a5) ld a0,-32(s0) call free ld a5,-48(s0) j .L8 .L7: lw a5,-36(s0) mv a4,a5 lw a5,-64(s0) sext.w a4,a4 sext.w a5,a5 bge a4,a5,.L9 lw a5,-20(s0) addiw a5,a5,1 sw a5,-20(s0) j .L6 .L9: lw a5,-24(s0) addiw a5,a5,-1 sw a5,-24(s0) .L6: lw a5,-20(s0) mv a4,a5 lw a5,-24(s0) sext.w a4,a4 sext.w a5,a5 blt a4,a5,.L10 ld a0,-32(s0) call free li a5,0 .L8: mv a0,a5 ld ra,56(sp) ld s0,48(sp) addi sp,sp,64 jr ra
21
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
riscv64
-O1
RISC-V 64 gcc 15.2.0
cmp: lw a0,0(a0) lw a5,0(a1) subw a0,a0,a5 ret twoSum: addi sp,sp,-48 sd ra,40(sp) sd s0,32(sp) sd s1,24(sp) sd s2,16(sp) sd s3,8(sp) sd s4,0(sp) mv s0,a0 mv s1,a1 mv s4,a2 addiw a0,a1,1 slli a0,a0,3 call malloc mv s3,a0 ble s1,zero,.L3 mv a3,s0 mv a4,a0 li a5,0 .L4: lw a2,0(a3) sw a2,0(a4) sw a5,4(a4) mv s2,a5 addiw a1,a5,1 mv a5,a1 addi a3,a3,4 addi a4,a4,8 bne s1,a1,.L4 lui a3,%hi(cmp) addi a3,a3,%lo(cmp) li a2,8 mv a0,s3 call qsort ble s2,zero,.L5 li a4,0 j .L10 .L14: li a0,8 call malloc mv s2,a0 lw a5,4(s1) sw a5,0(a0) lw a5,4(s0) sw a5,4(a0) mv a0,s3 call free j .L2 .L8: addiw s2,s2,-1 .L9: bge a4,s2,.L5 .L10: slli s1,a4,3 add s1,s3,s1 slli s0,s2,3 add s0,s3,s0 lw a3,0(s1) lw a5,0(s0) addw a5,a5,a3 beq a5,s4,.L14 bge a5,s4,.L8 addiw a4,a4,1 j .L9 .L3: lui a3,%hi(cmp) addi a3,a3,%lo(cmp) li a2,8 mv a1,s1 call qsort .L5: mv a0,s3 call free li s2,0 .L2: mv a0,s2 ld ra,40(sp) ld s0,32(sp) ld s1,24(sp) ld s2,16(sp) ld s3,8(sp) ld s4,0(sp) addi sp,sp,48 jr ra
22
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
riscv64
-O2
RISC-V 64 gcc 15.2.0
cmp: lw a0,0(a0) lw a5,0(a1) subw a0,a0,a5 ret twoSum: addi sp,sp,-48 addiw a5,a1,1 sd s2,16(sp) mv s2,a0 slli a0,a5,3 sd s0,32(sp) sd s1,24(sp) sd a1,0(sp) sd ra,40(sp) mv s1,a2 call malloc ld a1,0(sp) mv s0,a0 ble a1,zero,.L4 mv a0,s2 mv a4,s0 li a5,0 .L5: lw a3,0(a0) sw a5,4(a4) mv a6,a5 sw a3,0(a4) addiw a5,a5,1 addi a0,a0,4 addi a4,a4,8 bne a1,a5,.L5 lui a3,%hi(cmp) addi a3,a3,%lo(cmp) li a2,8 mv a0,s0 sd a6,0(sp) call qsort ld a6,0(sp) beq a6,zero,.L7 li a2,0 .L12: slli a4,a2,3 slli a5,a6,3 add a4,s0,a4 add a5,s0,a5 lw a1,0(a4) lw a3,0(a5) addw a3,a3,a1 beq a3,s1,.L20 bge a3,s1,.L10 addiw a2,a2,1 bgt a6,a2,.L12 .L7: mv a0,s0 call free ld ra,40(sp) ld s0,32(sp) li s1,0 ld s2,16(sp) mv a0,s1 ld s1,24(sp) addi sp,sp,48 jr ra .L10: addiw a6,a6,-1 bgt a6,a2,.L12 j .L7 .L20: li a0,8 sd a5,8(sp) sd a4,0(sp) call malloc ld a4,0(sp) ld a5,8(sp) mv s1,a0 lw a4,4(a4) lw a5,4(a5) mv a0,s0 sw a4,0(s1) sw a5,4(s1) call free ld ra,40(sp) ld s0,32(sp) ld s2,16(sp) mv a0,s1 ld s1,24(sp) addi sp,sp,48 jr ra .L4: lui a3,%hi(cmp) addi a3,a3,%lo(cmp) li a2,8 call qsort j .L7
23
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
riscv64
-O3
RISC-V 64 gcc 15.2.0
cmp: lw a0,0(a0) lw a5,0(a1) subw a0,a0,a5 ret twoSum: addi sp,sp,-48 addiw a5,a1,1 sd s2,16(sp) mv s2,a0 slli a0,a5,3 sd s0,32(sp) sd s1,24(sp) sd a1,0(sp) sd ra,40(sp) mv s1,a2 call malloc ld a1,0(sp) mv s0,a0 ble a1,zero,.L4 mv a0,s2 mv a4,s0 li a5,0 .L5: lw a3,0(a0) sw a5,4(a4) mv a6,a5 sw a3,0(a4) addiw a5,a5,1 addi a0,a0,4 addi a4,a4,8 bne a1,a5,.L5 lui a3,%hi(cmp) addi a3,a3,%lo(cmp) li a2,8 mv a0,s0 sd a6,0(sp) call qsort ld a6,0(sp) beq a6,zero,.L7 li a2,0 .L12: slli a4,a2,3 slli a5,a6,3 add a4,s0,a4 add a5,s0,a5 lw a1,0(a4) lw a3,0(a5) addw a3,a3,a1 beq a3,s1,.L20 bge a3,s1,.L10 addiw a2,a2,1 bgt a6,a2,.L12 .L7: mv a0,s0 call free ld ra,40(sp) ld s0,32(sp) li s1,0 ld s2,16(sp) mv a0,s1 ld s1,24(sp) addi sp,sp,48 jr ra .L10: addiw a6,a6,-1 bgt a6,a2,.L12 j .L7 .L20: li a0,8 sd a5,8(sp) sd a4,0(sp) call malloc ld a4,0(sp) ld a5,8(sp) mv s1,a0 lw a4,4(a4) lw a5,4(a5) mv a0,s0 sw a4,0(s1) sw a5,4(s1) call free ld ra,40(sp) ld s0,32(sp) ld s2,16(sp) mv a0,s1 ld s1,24(sp) addi sp,sp,48 jr ra .L4: lui a3,%hi(cmp) addi a3,a3,%lo(cmp) li a2,8 call qsort j .L7
24
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
x86-64
-O0
x86-64 clang 21.1.0
cmp: push rbp mov rbp, rsp mov qword ptr [rbp - 8], rdi mov qword ptr [rbp - 16], rsi mov rax, qword ptr [rbp - 8] mov eax, dword ptr [rax] mov rcx, qword ptr [rbp - 16] sub eax, dword ptr [rcx] pop rbp ret twoSum: push rbp mov rbp, rsp sub rsp, 64 mov qword ptr [rbp - 16], rdi mov dword ptr [rbp - 20], esi mov dword ptr [rbp - 24], edx mov eax, dword ptr [rbp - 20] add eax, 1 movsxd rdi, eax shl rdi, 3 call malloc@PLT mov qword ptr [rbp - 56], rax mov dword ptr [rbp - 36], 0 .LBB1_1: mov eax, dword ptr [rbp - 36] cmp eax, dword ptr [rbp - 20] jge .LBB1_4 mov rax, qword ptr [rbp - 16] movsxd rcx, dword ptr [rbp - 36] mov edx, dword ptr [rax + 4*rcx] mov rax, qword ptr [rbp - 56] movsxd rcx, dword ptr [rbp - 36] mov dword ptr [rax + 8*rcx], edx mov edx, dword ptr [rbp - 36] mov rax, qword ptr [rbp - 56] movsxd rcx, dword ptr [rbp - 36] mov dword ptr [rax + 8*rcx + 4], edx mov eax, dword ptr [rbp - 36] add eax, 1 mov dword ptr [rbp - 36], eax jmp .LBB1_1 .LBB1_4: mov rdi, qword ptr [rbp - 56] movsxd rsi, dword ptr [rbp - 20] mov edx, 8 lea rcx, [rip + cmp] call qsort@PLT mov dword ptr [rbp - 36], 0 mov eax, dword ptr [rbp - 20] sub eax, 1 mov dword ptr [rbp - 40], eax .LBB1_5: mov eax, dword ptr [rbp - 36] cmp eax, dword ptr [rbp - 40] jge .LBB1_13 mov rax, qword ptr [rbp - 56] movsxd rcx, dword ptr [rbp - 36] mov eax, dword ptr [rax + 8*rcx] mov rcx, qword ptr [rbp - 56] movsxd rdx, dword ptr [rbp - 40] add eax, dword ptr [rcx + 8*rdx] mov dword ptr [rbp - 44], eax mov eax, dword ptr [rbp - 44] cmp eax, dword ptr [rbp - 24] jne .LBB1_8 mov edi, 8 call malloc@PLT mov qword ptr [rbp - 32], rax mov rax, qword ptr [rbp - 56] movsxd rcx, dword ptr [rbp - 36] mov ecx, dword ptr [rax + 8*rcx + 4] mov rax, qword ptr [rbp - 32] mov dword ptr [rax], ecx mov rax, qword ptr [rbp - 56] movsxd rcx, dword ptr [rbp - 40] mov ecx, dword ptr [rax + 8*rcx + 4] mov rax, qword ptr [rbp - 32] mov dword ptr [rax + 4], ecx mov rdi, qword ptr [rbp - 56] call free@PLT mov rax, qword ptr [rbp - 32] mov qword ptr [rbp - 8], rax jmp .LBB1_14 .LBB1_8: mov eax, dword ptr [rbp - 44] cmp eax, dword ptr [rbp - 24] jge .LBB1_10 mov eax, dword ptr [rbp - 36] add eax, 1 mov dword ptr [rbp - 36], eax jmp .LBB1_11 .LBB1_10: mov eax, dword ptr [rbp - 40] add eax, -1 mov dword ptr [rbp - 40], eax .LBB1_11: jmp .LBB1_12 .LBB1_12: jmp .LBB1_5 .LBB1_13: mov rdi, qword ptr [rbp - 56] call free@PLT mov qword ptr [rbp - 8], 0 .LBB1_14: mov rax, qword ptr [rbp - 8] add rsp, 64 pop rbp ret
25
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
x86-64
-O1
x86-64 clang 21.1.0
cmp: mov eax, dword ptr [rdi] sub eax, dword ptr [rsi] ret twoSum: push rbp push r15 push r14 push r13 push r12 push rbx push rax mov ebp, edx mov r14d, esi mov r12, rdi movsxd r15, esi lea rdi, [8*r15 + 8] call malloc@PLT mov rbx, rax test r15d, r15d jle .LBB1_3 mov eax, r14d xor ecx, ecx .LBB1_2: mov edx, dword ptr [r12 + 4*rcx] mov dword ptr [rbx + 8*rcx], edx mov dword ptr [rbx + 8*rcx + 4], ecx inc rcx cmp rax, rcx jne .LBB1_2 .LBB1_3: lea rcx, [rip + cmp] mov edx, 8 mov rdi, rbx mov rsi, r15 call qsort@PLT xor r15d, r15d cmp r14d, 2 jl .LBB1_8 dec r14d xor eax, eax .LBB1_5: mov r12d, eax movsxd r13, r14d mov eax, dword ptr [rbx + 8*r13] add eax, dword ptr [rbx + 8*r12] cmp eax, ebp je .LBB1_6 setl al setge cl movzx ecx, cl sub r14d, ecx movzx eax, al add eax, r12d cmp eax, r14d jl .LBB1_5 jmp .LBB1_8 .LBB1_6: mov edi, 8 call malloc@PLT mov r15, rax mov eax, dword ptr [rbx + 8*r12 + 4] mov dword ptr [r15], eax mov eax, dword ptr [rbx + 8*r13 + 4] mov dword ptr [r15 + 4], eax .LBB1_8: mov rdi, rbx call free@PLT mov rax, r15 add rsp, 8 pop rbx pop r12 pop r13 pop r14 pop r15 pop rbp ret
26
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
x86-64
-O2
x86-64 clang 21.1.0
cmp: mov eax, dword ptr [rdi] sub eax, dword ptr [rsi] ret .LCPI1_1: .long 2 .long 2 .zero 4 .zero 4 .LCPI1_2: .long 4 .long 4 .zero 4 .zero 4 .LCPI1_3: .long 0 .long 1 .long 0 .long 0 twoSum: push rbp push r15 push r14 push r13 push r12 push rbx push rax mov ebp, edx mov r14d, esi mov r15, rdi movsxd r12, esi lea rdi, [8*r12 + 8] call malloc@PLT mov rbx, rax test r12d, r12d jle .LBB1_1 mov esi, r14d cmp r14d, 4 jae .LBB1_4 xor eax, eax jmp .LBB1_7 .LBB1_1: lea rcx, [rip + cmp] mov edx, 8 mov rdi, rbx mov rsi, r12 call qsort@PLT xor r15d, r15d jmp .LBB1_13 .LBB1_4: mov eax, esi and eax, 2147483644 movq xmm0, qword ptr [rip + .LCPI1_3] xor ecx, ecx movdqa xmm1, xmmword ptr [rip + .LCPI1_1] movdqa xmm2, xmmword ptr [rip + .LCPI1_2] .LBB1_5: movdqa xmm3, xmm0 paddd xmm3, xmm1 movq xmm4, qword ptr [r15 + 4*rcx] movq xmm5, qword ptr [r15 + 4*rcx + 8] punpckldq xmm5, xmm3 punpckldq xmm4, xmm0 movdqu xmmword ptr [rbx + 8*rcx], xmm4 movdqu xmmword ptr [rbx + 8*rcx + 16], xmm5 add rcx, 4 paddd xmm0, xmm2 cmp rax, rcx jne .LBB1_5 cmp eax, esi je .LBB1_8 .LBB1_7: mov ecx, dword ptr [r15 + 4*rax] mov dword ptr [rbx + 8*rax], ecx mov dword ptr [rbx + 8*rax + 4], eax inc rax cmp rsi, rax jne .LBB1_7 .LBB1_8: lea rcx, [rip + cmp] mov edx, 8 mov rdi, rbx call qsort@PLT xor r15d, r15d cmp r14d, 1 je .LBB1_13 dec r14d xor eax, eax .LBB1_10: mov r12d, eax movsxd r13, r14d mov eax, dword ptr [rbx + 8*r13] add eax, dword ptr [rbx + 8*r12] cmp eax, ebp je .LBB1_11 setl al setge cl movzx ecx, cl sub r14d, ecx movzx eax, al add eax, r12d cmp eax, r14d jl .LBB1_10 jmp .LBB1_13 .LBB1_11: mov edi, 8 call malloc@PLT mov r15, rax mov eax, dword ptr [rbx + 8*r12 + 4] mov dword ptr [r15], eax mov eax, dword ptr [rbx + 8*r13 + 4] mov dword ptr [r15 + 4], eax .LBB1_13: mov rdi, rbx call free@PLT mov rax, r15 add rsp, 8 pop rbx pop r12 pop r13 pop r14 pop r15 pop rbp ret
27
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
x86-64
-O3
x86-64 clang 21.1.0
cmp: mov eax, dword ptr [rdi] sub eax, dword ptr [rsi] ret .LCPI1_1: .long 2 .long 2 .zero 4 .zero 4 .LCPI1_2: .long 4 .long 4 .zero 4 .zero 4 .LCPI1_3: .long 0 .long 1 .long 0 .long 0 twoSum: push rbp push r15 push r14 push r13 push r12 push rbx push rax mov ebp, edx mov r14d, esi mov r15, rdi movsxd r12, esi lea rdi, [8*r12 + 8] call malloc@PLT mov rbx, rax test r12d, r12d jle .LBB1_1 mov esi, r14d cmp r14d, 4 jae .LBB1_4 xor eax, eax jmp .LBB1_7 .LBB1_1: lea rcx, [rip + cmp] mov edx, 8 mov rdi, rbx mov rsi, r12 call qsort@PLT xor r15d, r15d jmp .LBB1_13 .LBB1_4: mov eax, esi and eax, 2147483644 movq xmm0, qword ptr [rip + .LCPI1_3] xor ecx, ecx movdqa xmm1, xmmword ptr [rip + .LCPI1_1] movdqa xmm2, xmmword ptr [rip + .LCPI1_2] .LBB1_5: movdqa xmm3, xmm0 paddd xmm3, xmm1 movq xmm4, qword ptr [r15 + 4*rcx] movq xmm5, qword ptr [r15 + 4*rcx + 8] punpckldq xmm5, xmm3 punpckldq xmm4, xmm0 movdqu xmmword ptr [rbx + 8*rcx], xmm4 movdqu xmmword ptr [rbx + 8*rcx + 16], xmm5 add rcx, 4 paddd xmm0, xmm2 cmp rax, rcx jne .LBB1_5 cmp eax, esi je .LBB1_8 .LBB1_7: mov ecx, dword ptr [r15 + 4*rax] mov dword ptr [rbx + 8*rax], ecx mov dword ptr [rbx + 8*rax + 4], eax inc rax cmp rsi, rax jne .LBB1_7 .LBB1_8: lea rcx, [rip + cmp] mov edx, 8 mov rdi, rbx call qsort@PLT xor r15d, r15d cmp r14d, 1 je .LBB1_13 dec r14d xor eax, eax .LBB1_10: mov r12d, eax movsxd r13, r14d mov eax, dword ptr [rbx + 8*r13] add eax, dword ptr [rbx + 8*r12] cmp eax, ebp je .LBB1_11 setl al setge cl movzx ecx, cl sub r14d, ecx movzx eax, al add eax, r12d cmp eax, r14d jl .LBB1_10 jmp .LBB1_13 .LBB1_11: mov edi, 8 call malloc@PLT mov r15, rax mov eax, dword ptr [rbx + 8*r12 + 4] mov dword ptr [r15], eax mov eax, dword ptr [rbx + 8*r13 + 4] mov dword ptr [r15 + 4], eax .LBB1_13: mov rdi, rbx call free@PLT mov rax, r15 add rsp, 8 pop rbx pop r12 pop r13 pop r14 pop r15 pop rbp ret
28
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
x86-64
-O0
x86-64 gcc 15.2
cmp: push rbp mov rbp, rsp mov QWORD PTR [rbp-8], rdi mov QWORD PTR [rbp-16], rsi mov rax, QWORD PTR [rbp-8] mov edx, DWORD PTR [rax] mov rax, QWORD PTR [rbp-16] mov eax, DWORD PTR [rax] sub edx, eax mov eax, edx pop rbp ret twoSum: push rbp mov rbp, rsp sub rsp, 48 mov QWORD PTR [rbp-40], rdi mov DWORD PTR [rbp-44], esi mov DWORD PTR [rbp-48], edx mov eax, DWORD PTR [rbp-44] add eax, 1 cdqe sal rax, 3 mov rdi, rax call malloc mov QWORD PTR [rbp-16], rax mov DWORD PTR [rbp-4], 0 jmp .L4 .L5: mov eax, DWORD PTR [rbp-4] cdqe lea rdx, [0+rax*4] mov rax, QWORD PTR [rbp-40] add rax, rdx mov edx, DWORD PTR [rbp-4] movsx rdx, edx lea rcx, [0+rdx*8] mov rdx, QWORD PTR [rbp-16] add rdx, rcx mov eax, DWORD PTR [rax] mov DWORD PTR [rdx], eax mov eax, DWORD PTR [rbp-4] cdqe lea rdx, [0+rax*8] mov rax, QWORD PTR [rbp-16] add rdx, rax mov eax, DWORD PTR [rbp-4] mov DWORD PTR [rdx+4], eax add DWORD PTR [rbp-4], 1 .L4: mov eax, DWORD PTR [rbp-4] cmp eax, DWORD PTR [rbp-44] jl .L5 mov eax, DWORD PTR [rbp-44] movsx rsi, eax mov rax, QWORD PTR [rbp-16] mov ecx, OFFSET FLAT:cmp mov edx, 8 mov rdi, rax call qsort mov DWORD PTR [rbp-4], 0 mov eax, DWORD PTR [rbp-44] sub eax, 1 mov DWORD PTR [rbp-8], eax jmp .L6 .L10: mov eax, DWORD PTR [rbp-4] cdqe lea rdx, [0+rax*8] mov rax, QWORD PTR [rbp-16] add rax, rdx mov edx, DWORD PTR [rax] mov eax, DWORD PTR [rbp-8] cdqe lea rcx, [0+rax*8] mov rax, QWORD PTR [rbp-16] add rax, rcx mov eax, DWORD PTR [rax] add eax, edx mov DWORD PTR [rbp-20], eax mov eax, DWORD PTR [rbp-20] cmp eax, DWORD PTR [rbp-48] jne .L7 mov edi, 8 call malloc mov QWORD PTR [rbp-32], rax mov eax, DWORD PTR [rbp-4] cdqe lea rdx, [0+rax*8] mov rax, QWORD PTR [rbp-16] add rax, rdx mov edx, DWORD PTR [rax+4] mov rax, QWORD PTR [rbp-32] mov DWORD PTR [rax], edx mov eax, DWORD PTR [rbp-8] cdqe lea rdx, [0+rax*8] mov rax, QWORD PTR [rbp-16] add rax, rdx mov rdx, QWORD PTR [rbp-32] add rdx, 4 mov eax, DWORD PTR [rax+4] mov DWORD PTR [rdx], eax mov rax, QWORD PTR [rbp-16] mov rdi, rax call free mov rax, QWORD PTR [rbp-32] jmp .L8 .L7: mov eax, DWORD PTR [rbp-20] cmp eax, DWORD PTR [rbp-48] jge .L9 add DWORD PTR [rbp-4], 1 jmp .L6 .L9: sub DWORD PTR [rbp-8], 1 .L6: mov eax, DWORD PTR [rbp-4] cmp eax, DWORD PTR [rbp-8] jl .L10 mov rax, QWORD PTR [rbp-16] mov rdi, rax call free mov eax, 0 .L8: leave ret
29
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
x86-64
-O1
x86-64 gcc 15.2
cmp: mov eax, DWORD PTR [rdi] sub eax, DWORD PTR [rsi] ret twoSum: push r13 push r12 push rbp push rbx sub rsp, 8 mov r13, rdi mov r12d, esi mov ebp, edx lea edi, [rsi+1] movsx rdi, edi sal rdi, 3 call malloc mov rbx, rax test r12d, r12d jle .L3 movsx rdx, r12d mov esi, 0 .L4: mov eax, DWORD PTR [r13+0+rsi*4] mov DWORD PTR [rbx+rsi*8], eax mov DWORD PTR [rbx+4+rsi*8], esi add rsi, 1 cmp rsi, rdx jne .L4 mov ecx, OFFSET FLAT:cmp mov edx, 8 mov rdi, rbx call qsort lea edx, [r12-1] test edx, edx jle .L5 mov ecx, 0 jmp .L10 .L14: mov edi, 8 call malloc mov rbp, rax mov eax, DWORD PTR [r13+4] mov DWORD PTR [rbp+0], eax mov eax, DWORD PTR [r12+4] mov DWORD PTR [rbp+4], eax mov rdi, rbx call free jmp .L2 .L8: sub edx, 1 .L9: cmp ecx, edx jge .L5 .L10: movsx rax, ecx lea r13, [rbx+rax*8] movsx rax, edx lea r12, [rbx+rax*8] mov eax, DWORD PTR [r12] add eax, DWORD PTR [r13+0] cmp eax, ebp je .L14 jge .L8 add ecx, 1 jmp .L9 .L3: movsx rsi, r12d mov ecx, OFFSET FLAT:cmp mov edx, 8 mov rdi, rax call qsort .L5: mov rdi, rbx call free mov ebp, 0 .L2: mov rax, rbp add rsp, 8 pop rbx pop rbp pop r12 pop r13 ret
30
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
x86-64
-O2
x86-64 gcc 15.2
cmp: mov eax, DWORD PTR [rdi] sub eax, DWORD PTR [rsi] ret twoSum: push r14 mov r14d, esi push r13 mov r13, rdi lea edi, [rsi+1] push r12 movsx rdi, edi mov r12d, edx push rbp sal rdi, 3 push rbx call malloc movsx rsi, r14d mov rbx, rax test r14d, r14d jle .L4 lea ebp, [r14-1] cmp ebp, 2 jbe .L17 mov edx, r14d mov edi, 4 movdqa xmm1, XMMWORD PTR .LC0[rip] xor eax, eax shr edx, 2 movd xmm3, edi sal rdx, 4 pshufd xmm3, xmm3, 0 .L6: movdqu xmm0, XMMWORD PTR [r13+0+rax] movdqa xmm2, xmm0 punpckhdq xmm0, xmm1 punpckldq xmm2, xmm1 movups XMMWORD PTR [rbx+16+rax*2], xmm0 paddd xmm1, xmm3 movups XMMWORD PTR [rbx+rax*2], xmm2 add rax, 16 cmp rdx, rax jne .L6 mov eax, r14d and eax, -4 test r14b, 3 je .L26 .L5: cdqe .L8: movd xmm0, DWORD PTR [r13+0+rax*4] movd xmm4, eax punpckldq xmm0, xmm4 movq QWORD PTR [rbx+rax*8], xmm0 add rax, 1 cmp r14d, eax jg .L8 mov edx, 8 mov ecx, OFFSET FLAT:cmp mov rdi, rbx call qsort mov edx, ebp test ebp, ebp je .L10 .L16: xor ecx, ecx jmp .L15 .L28: add ecx, 1 cmp edx, ecx jle .L10 .L15: movsx rax, ecx lea rbp, [rbx+rax*8] movsx rax, edx lea r13, [rbx+rax*8] mov eax, DWORD PTR [r13+0] add eax, DWORD PTR [rbp+0] cmp eax, r12d je .L27 jl .L28 sub edx, 1 cmp edx, ecx jg .L15 .L10: mov rdi, rbx xor r12d, r12d call free pop rbx mov rax, r12 pop rbp pop r12 pop r13 pop r14 ret .L26: mov edx, 8 mov ecx, OFFSET FLAT:cmp mov rdi, rbx call qsort mov edx, ebp jmp .L16 .L27: mov edi, 8 call malloc movd xmm1, DWORD PTR [r13+4] movd xmm0, DWORD PTR [rbp+4] mov rdi, rbx mov r12, rax punpckldq xmm0, xmm1 movq QWORD PTR [rax], xmm0 call free pop rbx mov rax, r12 pop rbp pop r12 pop r13 pop r14 ret .L4: mov ecx, OFFSET FLAT:cmp mov edx, 8 mov rdi, rax call qsort jmp .L10 .L17: xor eax, eax jmp .L5 .LC0: .long 0 .long 1 .long 2 .long 3
31
1
Two Sum
Easy
/* 1. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1]. */ /** * Note: The returned array must be malloced, assume caller calls free(). */ typedef struct data_s { int val; int idx; } data_t; int cmp(const void *a, const void *b) { return ((data_t *)a)->val - ((data_t *)b)->val; } int* twoSum(int* nums, int numsSize, int target) { int *indices; int i, j, k; #if 0 for (i = 0; i < numsSize - 1; i ++) { for (j = i + 1; j < numsSize; j ++) { if (nums[i] + nums[j] == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = i; indices[1] = j; return indices; } } } #else data_t *array; array = malloc((numsSize + 1) * sizeof(data_t)); //assert(array); for (i = 0; i < numsSize; i ++) { array[i].val = nums[i]; array[i].idx = i; } qsort(array, numsSize, sizeof(data_t), cmp); i = 0; j = numsSize - 1; while (i < j) { k = array[i].val + array[j].val; if (k == target) { indices = malloc(2 * sizeof(int)); //assert(indices); indices[0] = array[i].idx; indices[1] = array[j].idx; free(array); return indices; } else if (k < target) { i ++; } else { j --; } } free(array); #endif return NULL; } /* Difficulty:Easy Total Accepted:586.7K Total Submissions:1.7M Companies LinkedIn Uber Airbnb Facebook Amazon Microsoft Apple Yahoo Dropbox Bloomberg Yelp Adobe Related Topics Array Hash Table Similar Questions 3Sum 4Sum Two Sum II - Input array is sorted Two Sum III - Data structure design Subarray Sum Equals K Two Sum IV - Input is a BST */
x86-64
-O3
x86-64 gcc 15.2
cmp: mov eax, DWORD PTR [rdi] sub eax, DWORD PTR [rsi] ret twoSum: push r14 mov r14d, esi push r13 mov r13, rdi lea edi, [rsi+1] push r12 movsx rdi, edi mov r12d, edx push rbp sal rdi, 3 push rbx call malloc movsx rsi, r14d mov rbx, rax test r14d, r14d jle .L4 lea ebp, [r14-1] cmp ebp, 2 jbe .L16 mov edx, r14d mov edi, 4 movdqa xmm1, XMMWORD PTR .LC0[rip] xor eax, eax shr edx, 2 movd xmm3, edi sal rdx, 4 pshufd xmm3, xmm3, 0 .L6: movdqu xmm0, XMMWORD PTR [r13+0+rax] movdqa xmm2, xmm0 punpckhdq xmm0, xmm1 punpckldq xmm2, xmm1 movups XMMWORD PTR [rbx+16+rax*2], xmm0 paddd xmm1, xmm3 movups XMMWORD PTR [rbx+rax*2], xmm2 add rax, 16 cmp rdx, rax jne .L6 test r14b, 3 je .L7 mov eax, r14d and eax, -4 .L5: movsx rdx, eax movd xmm4, eax lea ecx, [rax+1] movd xmm0, DWORD PTR [r13+0+rdx*4] punpckldq xmm0, xmm4 movq QWORD PTR [rbx+rdx*8], xmm0 cmp r14d, ecx jle .L8 movd xmm0, DWORD PTR [r13+4+rdx*4] movd xmm5, ecx add eax, 2 punpckldq xmm0, xmm5 movq QWORD PTR [rbx+8+rdx*8], xmm0 cmp r14d, eax jle .L7 movd xmm0, DWORD PTR [r13+8+rdx*4] movd xmm6, eax punpckldq xmm0, xmm6 movq QWORD PTR [rbx+16+rdx*8], xmm0 .L7: mov edx, 8 mov ecx, OFFSET FLAT:cmp mov rdi, rbx call qsort mov edx, ebp .L15: xor ecx, ecx jmp .L14 .L27: add ecx, 1 cmp edx, ecx jle .L9 .L14: movsx rax, ecx lea rbp, [rbx+rax*8] movsx rax, edx lea r13, [rbx+rax*8] mov eax, DWORD PTR [r13+0] add eax, DWORD PTR [rbp+0] cmp eax, r12d je .L26 jl .L27 sub edx, 1 cmp edx, ecx jg .L14 .L9: mov rdi, rbx xor r12d, r12d call free pop rbx mov rax, r12 pop rbp pop r12 pop r13 pop r14 ret .L8: mov edx, 8 mov ecx, OFFSET FLAT:cmp mov rdi, rbx call qsort mov edx, ebp test ebp, ebp jne .L15 jmp .L9 .L26: mov edi, 8 call malloc movd xmm1, DWORD PTR [r13+4] movd xmm0, DWORD PTR [rbp+4] mov rdi, rbx mov r12, rax punpckldq xmm0, xmm1 movq QWORD PTR [rax], xmm0 call free pop rbx mov rax, r12 pop rbp pop r12 pop r13 pop r14 ret .L4: mov ecx, OFFSET FLAT:cmp mov edx, 8 mov rdi, rax call qsort jmp .L9 .L16: xor eax, eax jmp .L5 .LC0: .long 0 .long 1 .long 2 .long 3
32
2
Add Two Numbers
Medium
/* 2. Add Two Numbers You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. You may assume the two numbers do not contain any leading zero, except the number 0 itself. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 */ /** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) { struct ListNode *head = NULL; struct ListNode *p, *q; int s = 0; while (l1 || l2 || s != 0) { p = malloc(sizeof(struct ListNode)); //assert(p); if (l1) { s += l1->val; l1 = l1->next; } if (l2) { s += l2->val; l2 = l2->next; } p->val = s % 10; p->next = NULL; s = s / 10; if (!head) { head = p; } if (q) { q->next = p; } q = p; } return head; } /* Difficulty:Medium Total Accepted:329.9K Total Submissions:1.2M Companies Amazon Microsoft Bloomberg Airbnb Adobe Related Topics Linked List Math Similar Questions Multiply Strings Add Binary Sum of Two Integers Add Strings Add Two Numbers II */
aarch64
-O0
ARM64 gcc 15.2.0
addTwoNumbers: stp x29, x30, [sp, -64]! mov x29, sp str x0, [sp, 24] str x1, [sp, 16] str xzr, [sp, 56] str wzr, [sp, 44] b .L2 .L7: mov x0, 16 bl malloc str x0, [sp, 32] ldr x0, [sp, 24] cmp x0, 0 beq .L3 ldr x0, [sp, 24] ldr w0, [x0] ldr w1, [sp, 44] add w0, w1, w0 str w0, [sp, 44] ldr x0, [sp, 24] ldr x0, [x0, 8] str x0, [sp, 24] .L3: ldr x0, [sp, 16] cmp x0, 0 beq .L4 ldr x0, [sp, 16] ldr w0, [x0] ldr w1, [sp, 44] add w0, w1, w0 str w0, [sp, 44] ldr x0, [sp, 16] ldr x0, [x0, 8] str x0, [sp, 16] .L4: ldr w1, [sp, 44] mov w0, 10 sdiv w2, w1, w0 mov w0, w2 lsl w0, w0, 2 add w0, w0, w2 lsl w0, w0, 1 sub w1, w1, w0 ldr x0, [sp, 32] str w1, [x0] ldr x0, [sp, 32] str xzr, [x0, 8] ldr w0, [sp, 44] mov w1, 26215 movk w1, 0x6666, lsl 16 smull x1, w0, w1 lsr x1, x1, 32 asr w1, w1, 2 asr w0, w0, 31 sub w0, w1, w0 str w0, [sp, 44] ldr x0, [sp, 56] cmp x0, 0 bne .L5 ldr x0, [sp, 32] str x0, [sp, 56] .L5: ldr x0, [sp, 48] cmp x0, 0 beq .L6 ldr x0, [sp, 48] ldr x1, [sp, 32] str x1, [x0, 8] .L6: ldr x0, [sp, 32] str x0, [sp, 48] .L2: ldr x0, [sp, 24] cmp x0, 0 bne .L7 ldr x0, [sp, 16] cmp x0, 0 bne .L7 ldr w0, [sp, 44] cmp w0, 0 bne .L7 ldr x0, [sp, 56] ldp x29, x30, [sp], 64 ret
33
2
Add Two Numbers
Medium
/* 2. Add Two Numbers You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. You may assume the two numbers do not contain any leading zero, except the number 0 itself. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 */ /** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) { struct ListNode *head = NULL; struct ListNode *p, *q; int s = 0; while (l1 || l2 || s != 0) { p = malloc(sizeof(struct ListNode)); //assert(p); if (l1) { s += l1->val; l1 = l1->next; } if (l2) { s += l2->val; l2 = l2->next; } p->val = s % 10; p->next = NULL; s = s / 10; if (!head) { head = p; } if (q) { q->next = p; } q = p; } return head; } /* Difficulty:Medium Total Accepted:329.9K Total Submissions:1.2M Companies Amazon Microsoft Bloomberg Airbnb Adobe Related Topics Linked List Math Similar Questions Multiply Strings Add Binary Sum of Two Integers Add Strings Add Two Numbers II */
aarch64
-O1
ARM64 gcc 15.2.0
addTwoNumbers: stp x29, x30, [sp, -80]! mov x29, sp stp x19, x20, [sp, 16] stp x21, x22, [sp, 32] mov x20, x0 orr x0, x0, x1 cbz x0, .L8 stp x23, x24, [sp, 48] stp x25, x26, [sp, 64] mov x21, x1 mov w19, 0 mov x22, 0 mov x26, 16 mov w25, 10 mov w24, 26215 movk w24, 0x6666, lsl 16 b .L7 .L6: orr x0, x21, x20 cmp x0, 0 ccmp w19, 0, 0, eq beq .L12 .L7: mov x23, x2 mov x0, x26 bl malloc mov x2, x0 cbz x20, .L3 ldr w0, [x20] add w19, w19, w0 ldr x20, [x20, 8] .L3: cbz x21, .L4 ldr w0, [x21] add w19, w19, w0 ldr x21, [x21, 8] .L4: sdiv w0, w19, w25 add w0, w0, w0, lsl 2 sub w0, w19, w0, lsl 1 str w0, [x2] str xzr, [x2, 8] smull x0, w19, w24 asr x0, x0, 34 sub w19, w0, w19, asr 31 cmp x22, 0 csel x22, x22, x2, ne cbz x23, .L6 str x2, [x23, 8] b .L6 .L12: ldp x23, x24, [sp, 48] ldp x25, x26, [sp, 64] .L1: mov x0, x22 ldp x19, x20, [sp, 16] ldp x21, x22, [sp, 32] ldp x29, x30, [sp], 80 ret .L8: mov x22, 0 b .L1
34
2
Add Two Numbers
Medium
/* 2. Add Two Numbers You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. You may assume the two numbers do not contain any leading zero, except the number 0 itself. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 */ /** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) { struct ListNode *head = NULL; struct ListNode *p, *q; int s = 0; while (l1 || l2 || s != 0) { p = malloc(sizeof(struct ListNode)); //assert(p); if (l1) { s += l1->val; l1 = l1->next; } if (l2) { s += l2->val; l2 = l2->next; } p->val = s % 10; p->next = NULL; s = s / 10; if (!head) { head = p; } if (q) { q->next = p; } q = p; } return head; } /* Difficulty:Medium Total Accepted:329.9K Total Submissions:1.2M Companies Amazon Microsoft Bloomberg Airbnb Adobe Related Topics Linked List Math Similar Questions Multiply Strings Add Binary Sum of Two Integers Add Strings Add Two Numbers II */
aarch64
-O2
ARM64 gcc 15.2.0
addTwoNumbers: stp x29, x30, [sp, -80]! mov x29, sp stp x19, x20, [sp, 16] mov x20, x1 mov w19, 0 stp x21, x22, [sp, 32] mov w21, 26215 mov w22, 10 stp x23, x24, [sp, 48] mov x23, x0 orr x0, x23, x20 str x25, [sp, 64] mov x24, 0 movk w21, 0x6666, lsl 16 cbnz x0, .L7 .L21: cbz w19, .L1 mov x23, 0 mov x20, 0 mov x0, 16 bl malloc .L4: sdiv w2, w19, w22 smull x1, w19, w21 str xzr, [x0, 8] cmp x24, 0 csel x24, x24, x0, ne asr x1, x1, 34 add w2, w2, w2, lsl 2 sub w2, w19, w2, lsl 1 str w2, [x0] sub w19, w1, w19, asr 31 cbz x25, .L6 str x0, [x25, 8] .L6: mov x25, x0 orr x0, x23, x20 cbz x0, .L21 .L7: mov x0, 16 bl malloc cbz x23, .L3 ldr w1, [x23] ldr x23, [x23, 8] add w19, w19, w1 cbz x20, .L4 .L3: ldr w1, [x20] ldr x20, [x20, 8] add w19, w19, w1 b .L4 .L1: ldr x25, [sp, 64] mov x0, x24 ldp x19, x20, [sp, 16] ldp x21, x22, [sp, 32] ldp x23, x24, [sp, 48] ldp x29, x30, [sp], 80 ret
35
2
Add Two Numbers
Medium
/* 2. Add Two Numbers You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. You may assume the two numbers do not contain any leading zero, except the number 0 itself. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 */ /** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) { struct ListNode *head = NULL; struct ListNode *p, *q; int s = 0; while (l1 || l2 || s != 0) { p = malloc(sizeof(struct ListNode)); //assert(p); if (l1) { s += l1->val; l1 = l1->next; } if (l2) { s += l2->val; l2 = l2->next; } p->val = s % 10; p->next = NULL; s = s / 10; if (!head) { head = p; } if (q) { q->next = p; } q = p; } return head; } /* Difficulty:Medium Total Accepted:329.9K Total Submissions:1.2M Companies Amazon Microsoft Bloomberg Airbnb Adobe Related Topics Linked List Math Similar Questions Multiply Strings Add Binary Sum of Two Integers Add Strings Add Two Numbers II */
aarch64
-O3
ARM64 gcc 15.2.0
addTwoNumbers: stp x29, x30, [sp, -80]! mov x29, sp stp x19, x20, [sp, 16] mov x20, x1 mov w19, 0 stp x21, x22, [sp, 32] mov w21, 26215 mov w22, 10 stp x23, x24, [sp, 48] mov x23, x0 orr x1, x23, x20 str x25, [sp, 64] mov x24, 0 movk w21, 0x6666, lsl 16 mov x0, 16 cbnz x1, .L7 .L21: cbz w19, .L1 mov x23, 0 mov x20, 0 bl malloc .L4: sdiv w2, w19, w22 smull x1, w19, w21 str xzr, [x0, 8] cmp x24, 0 csel x24, x24, x0, ne asr x1, x1, 34 add w2, w2, w2, lsl 2 sub w2, w19, w2, lsl 1 str w2, [x0] sub w19, w1, w19, asr 31 cbz x25, .L6 str x0, [x25, 8] .L6: mov x25, x0 orr x1, x23, x20 mov x0, 16 cbz x1, .L21 .L7: mov x0, 16 bl malloc cbz x23, .L3 ldr w1, [x23] ldr x23, [x23, 8] add w19, w19, w1 cbz x20, .L4 .L3: ldr w1, [x20] ldr x20, [x20, 8] add w19, w19, w1 b .L4 .L1: ldr x25, [sp, 64] mov x0, x24 ldp x19, x20, [sp, 16] ldp x21, x22, [sp, 32] ldp x23, x24, [sp, 48] ldp x29, x30, [sp], 80 ret
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LeetCode Assembly Dataset

441 LeetCode problems solved in C, compiled to assembly across 4 architectures, 2 compilers, and 4 optimization levels using GCC and Clang via the Godbolt Compiler Explorer API.

Dataset Summary

Stat Value
Total rows 14,112
Unique problems 441
Architectures x86-64, AArch64, MIPS64, RISC-V 64
Compilers GCC 15.2, Clang 21.1.0
Optimization levels -O0, -O1, -O2, -O3
Compilation success rate 100%
Difficulty split Easy: 98, Medium: 259, Hard: 84

Each problem has 32 assembly variants (4 architectures x 2 compilers x 4 optimization levels), making this useful for studying how the same algorithm compiles differently across ISAs, compilers, and optimization settings.

Key Insights

  • -O3 often produces more code than -O0, not less. Across all architectures, -O3 output is 37-73% larger on average due to aggressive inlining and loop unrolling.
  • -O1 consistently produces the most compact assembly across all four architectures.
  • Recursive tree problems see extreme -O3 blowup. The top 10 largest O3-vs-O0 ratios are all binary tree problems. Worst case: Longest Univalue Path goes from 90 lines (O0) to 3,809 lines (O3) -- a 42x increase.
  • Iterative and math problems shrink dramatically at -O3. Flip String to Monotone Increasing drops from 89 to 8 lines (91% reduction). Nim Game compiles down to just 4 lines.
  • AArch64 produces the most compact assembly at -O2 (avg 153 lines), followed by x86-64 (170), RISC-V (170), and MIPS64 (236). MIPS is 39% more verbose than AArch64.
  • Hard problems produce 2x more assembly than Easy problems at -O2 on x86-64 (247 vs 121 avg lines). Algorithmic complexity maps directly to code size.
  • The largest output is LFU Cache (1,121 lines at -O0 on x86-64), a complex doubly-linked list plus hash table implementation.
  • The smallest output is Nim Game at -O3 (4 lines on x86-64) -- the entire solution optimizes to a single bitwise AND.

Use Cases

  • Training or evaluating models on C-to-assembly translation
  • Studying how optimization levels affect generated code
  • Compiler behavior research

Schema

Column Type Description
id int Unique row identifier (0-14111)
problem_id int LeetCode problem number
problem_title string Problem name (e.g. "Two Sum")
difficulty string Easy, Medium, or Hard
c_source string Complete C source code
architecture string Target ISA: x86-64, aarch64, mips64, riscv64
optimization string Optimization flag: -O0, -O1, -O2, -O3
compiler string Compiler version used (e.g. x86-64 gcc 15.2, x86-64 clang 21.1.0)
assembly string Compiled assembly output

Usage

from datasets import load_dataset

ds = load_dataset("ronantakizawa/leetcode-assembly", split="train")

# Get all x86-64 GCC assembly at -O2
x86_gcc_O2 = ds.filter(lambda r: r["architecture"] == "x86-64" and r["compiler"].startswith("x86-64 gcc") and r["optimization"] == "-O2")

# Compare GCC vs Clang for the same problem
two_sum = ds.filter(lambda r: r["problem_id"] == 1 and r["architecture"] == "x86-64" and r["optimization"] == "-O2")
for row in two_sum:
    print(f"--- {row['compiler']} ---")
    print(row["assembly"][:200])
    print()

Sources

C solutions were collected from three open-source GitHub repositories of pure-C LeetCode solutions:

After deduplication, 441 unique problems remained. Solutions are self-contained C files using only the standard library (no C++ STL), producing clean and readable assembly output.

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