<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?
class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        // i - Scan from the left side
        // j - Scan from the right side
        // k - The position to put in the sorted vector, starting from right (largest)
        int i = 0, j = nums.size() - 1, k = nums.size() - 1;
        vector<int> sorted_squares(nums.size(), 0);

        while (i <= j) {
            int squared_i = pow(nums[i], 2);
            int squared_j = pow(nums[j], 2);

            if (squared_i >= squared_j) {
                sorted_squares[k--] = squared_i;
                i++;
            } else {
                sorted_squares[k--] = squared_j;
                j--;
            }
        }

        return sorted_squares;
    }
};
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