Largest Perimeter Triangle
<p>Given an integer array <code>nums</code>, return <em>the largest perimeter of a triangle with a non-zero area, formed from three of these lengths</em>. If it is impossible to form any triangle of a non-zero area, return <code>0</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [2,1,2]
<strong>Output:</strong> 5
<strong>Explanation:</strong> You can form a triangle with three side lengths: 1, 2, and 2.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,1,10]
<strong>Output:</strong> 0
<strong>Explanation:</strong> 
You cannot use the side lengths 1, 1, and 2 to form a triangle.
You cannot use the side lengths 1, 1, and 10 to form a triangle.
You cannot use the side lengths 1, 2, and 10 to form a triangle.
As we cannot use any three side lengths to form a triangle of non-zero area, we return 0.
</pre>

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

<ul>
	<li><code>3 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
</ul>
Greedy; Array; Math; Sorting
int cmp(const void* a, const void* b) {
    return *(int*) b - *(int*) a;
}

int largestPerimeter(int* nums, int numsSize) {
    qsort(nums, numsSize, sizeof(int), cmp);
    for (int i = 2; i < numsSize; i++) {
        if (nums[i - 2] < nums[i - 1] + nums[i]) {
            return nums[i - 2] + nums[i - 1] + nums[i];
        }
    }
    return 0;
}