Squares of a Sorted Array
<p>Given an integer array <code>nums</code> sorted in <strong>non-decreasing</strong> order, return <em>an array of <strong>the squares of each number</strong> sorted in non-decreasing order</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [-4,-1,0,3,10]
<strong>Output:</strong> [0,1,9,16,100]
<strong>Explanation:</strong> After squaring, the array becomes [16,1,0,9,100].
After sorting, it becomes [0,1,9,16,100].
</pre>

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

<pre>
<strong>Input:</strong> nums = [-7,-3,2,3,11]
<strong>Output:</strong> [4,9,9,49,121]
</pre>

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

<ul>
	<li><code><span>1 &lt;= nums.length &lt;= </span>10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>nums</code> is sorted in <strong>non-decreasing</strong> order.</li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Squaring each element and sorting the new array is very trivial, could you find an <code>O(n)</code> solution using a different approach?
Array; Two Pointers; Sorting
/**
 * @param {number[]} nums
 * @return {number[]}
 */
var sortedSquares = function (nums) {
    const n = nums.length;
    const ans = Array(n).fill(0);
    for (let i = 0, j = n - 1, k = n - 1; i <= j; --k) {
        const [a, b] = [nums[i] * nums[i], nums[j] * nums[j]];
        if (a > b) {
            ans[k] = a;
            ++i;
        } else {
            ans[k] = b;
            --j;
        }
    }
    return ans;
};
