<p>Given an array of integers <code>nums</code>, sort the array in ascending order and return it.</p>

<p>You must solve the problem <strong>without using any built-in</strong> functions in <code>O(nlog(n))</code> time complexity and with the smallest space complexity possible.</p>

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

<pre>
<strong>Input:</strong> nums = [5,2,3,1]
<strong>Output:</strong> [1,2,3,5]
<strong>Explanation:</strong> After sorting the array, the positions of some numbers are not changed (for example, 2 and 3), while the positions of other numbers are changed (for example, 1 and 5).
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,1,1,2,0,0]
<strong>Output:</strong> [0,0,1,1,2,5]
<strong>Explanation:</strong> Note that the values of nums are not necessairly unique.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-5 * 10<sup>4</sup> &lt;= nums[i] &lt;= 5 * 10<sup>4</sup></code></li>
</ul>

class Solution {
private:
    void mergeTwoSortedArrays(vector < int > &sortedArray, span < int > &arrayA, span < int > &arrayB) {
        int pointerA = 0;
        int pointerB = 0;
        while (pointerA < arrayA.size() && pointerB < arrayB.size()) {
            sortedArray.push_back(min(arrayA[pointerA], arrayB[pointerB]));
            if (arrayA[pointerA] < arrayB[pointerB]) {
                ++pointerA;
            } else {
                ++pointerB;
            }
        }
        while (pointerA < arrayA.size()) {
            sortedArray.push_back(arrayA[pointerA]);
            ++pointerA;
        }
        while (pointerB < arrayB.size()) {
            sortedArray.push_back(arrayB[pointerB]);
            ++pointerB;
        }
    }

    void mergeSort(span < int > &nums) {
        if (nums.size() == 1) {
            return;
        }
        int mid = nums.size() / 2;
        span<int> firstHalf(nums.begin(), mid);
        span<int> secondHalf(nums.begin() + mid, nums.size() - mid);
        mergeSort(firstHalf);
        mergeSort(secondHalf);
        vector < int > sortedArray;
        mergeTwoSortedArrays(sortedArray, firstHalf, secondHalf);
        for (int i = 0; i < nums.size(); ++i) {
            nums[i] = sortedArray[i];
        }
    }
public:
    vector<int> sortArray(vector<int>& nums) {
        span<int> numsSpan(nums.begin(), nums.size());
        mergeSort(numsSpan);
        return nums;
    }
};
/*

[2, 5, 1, 3]

puttingElementIndex
leftArrayIndex
rightArrayIndex

*/
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