Two Sum
<p>Given an array of integers <code>nums</code>&nbsp;and an integer <code>target</code>, return <em>indices of the two numbers such that they add up to <code>target</code></em>.</p>

<p>You may assume that each input would have <strong><em>exactly</em> one solution</strong>, and you may not use the <em>same</em> element twice.</p>

<p>You can return the answer in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,7,11,15], target = 9
<strong>Output:</strong> [0,1]
<strong>Explanation:</strong> Because nums[0] + nums[1] == 9, we return [0, 1].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,2,4], target = 6
<strong>Output:</strong> [1,2]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,3], target = 6
<strong>Output:</strong> [0,1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= target &lt;= 10<sup>9</sup></code></li>
	<li><strong>Only one valid answer exists.</strong></li>
</ul>

<p>&nbsp;</p>
<strong>Follow-up:&nbsp;</strong>Can you come up with an algorithm that is less than <code>O(n<sup>2</sup>)</code><font face="monospace">&nbsp;</font>time complexity?
Array; Hash Table
#include <stdlib.h>

int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
    int capacity = 1;
    while (capacity < numsSize * 2) capacity <<= 1;
    int* keys = malloc(capacity * sizeof(int));
    int* vals = malloc(capacity * sizeof(int));
    char* used = calloc(capacity, sizeof(char));
    if (!keys || !vals || !used) {
        free(keys);
        free(vals);
        free(used);
        *returnSize = 0;
        return NULL;
    }
    for (int i = 0; i < numsSize; ++i) {
        int x = nums[i];
        int y = target - x;
        unsigned int h = (unsigned int) y & (capacity - 1);
        while (used[h]) {
            if (keys[h] == y) {
                int* res = malloc(2 * sizeof(int));
                res[0] = vals[h];
                res[1] = i;
                *returnSize = 2;
                free(keys);
                free(vals);
                free(used);
                return res;
            }
            h = (h + 1) & (capacity - 1);
        }
        unsigned int h2 = (unsigned int) x & (capacity - 1);
        while (used[h2]) h2 = (h2 + 1) & (capacity - 1);
        used[h2] = 1;
        keys[h2] = x;
        vals[h2] = i;
    }
    *returnSize = 0;
    free(keys);
    free(vals);
    free(used);
    return NULL;
}
