Minimum Elements to Add to Form a Given Sum
<p>You are given an integer array <code>nums</code> and two integers <code>limit</code> and <code>goal</code>. The array <code>nums</code> has an interesting property that <code>abs(nums[i]) &lt;= limit</code>.</p>

<p>Return <em>the minimum number of elements you need to add to make the sum of the array equal to </em><code>goal</code>. The array must maintain its property that <code>abs(nums[i]) &lt;= limit</code>.</p>

<p>Note that <code>abs(x)</code> equals <code>x</code> if <code>x &gt;= 0</code>, and <code>-x</code> otherwise.</p>

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

<pre>
<strong>Input:</strong> nums = [1,-1,1], limit = 3, goal = -4
<strong>Output:</strong> 2
<strong>Explanation:</strong> You can add -2 and -3, then the sum of the array will be 1 - 1 + 1 - 2 - 3 = -4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,-10,9,1], limit = 100, goal = 0
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= limit &lt;= 10<sup>6</sup></code></li>
	<li><code>-limit &lt;= nums[i] &lt;= limit</code></li>
	<li><code>-10<sup>9</sup> &lt;= goal &lt;= 10<sup>9</sup></code></li>
</ul>
Greedy; Array
int minElements(int* nums, int numsSize, int limit, int goal) {
    long long sum = 0;
    for (int i = 0; i < numsSize; i++) {
        sum += nums[i];
    }
    long long diff = labs(goal - sum);
    return (diff + limit - 1) / limit;
}