LeetCode 0001 Two Sum

Iterative Brute Force

Summary

Given an array of integers nums and an integer target, return the indices of the two numbers such that they add up to target.

Process

• Descend through nums, iterating from index 0.
• Within the first loop, descend through nums once more, this time iterating from index 1.
• If the difference of target and the inner value equals the outer value, return each index.
• If both loops complete without a successful comparison, return empty.

Metrics	Time Complexity	Space Complexity
Worst	O(n²)	O(1)
Average	𝚹(n²)	𝚹(1)
Best	Ω(1)	Ω(1)

int* two_sum(int* nums, int nums_size, int target, int* return_size);

int* two_sum(int* nums, int nums_size, int target, int* return_size) {
    *return_size = 2;

    int* solution = malloc(*return_size * sizeof(int));
    if (solution == NULL) {
        *return_size = 0;
        return NULL;
    }

    for (int i = 0; i < nums_size; i++) {
        for (int j = i + 1; j < nums_size; j++) {
            if (nums[i] == target - nums[j]) {
                solution[0] = i;
                solution[1] = j;
                return solution;
            }
        }
    }

    free(solution);
    *return_size = 0;
    return NULL;
}

int main(void) {
    int nums[] = { 3, 8, 2, -12, 24 };
    int nums_size = sizeof(nums) / sizeof(nums[0]);
    int target = 12;
    int return_size;

    int* solution = two_sum(nums, nums_size, target, &return_size);

    if (solution != NULL) {
        printf("[%d %d]\n", solution[0], solution[1]);
        free(solution);
    } else {
        printf("no matching combination\n");
    }

    return 0;
}