<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 {
    void selectionSort2(vector<int>& nums) {
        if (nums[0] > nums[1])
            swap(nums[0], nums[1]);
    }

    void merge(vector<int>& nums, vector<int>& left, vector<int>& right) {
        int i = 0;
        int l = 0, r = 0;
        const int m = left.size(), n = right.size();

        while (l < m && r < n) {
            if (left[l] < right[r])
                nums[i++] = left[l++];
            else
                nums[i++] = right[r++];
        }
        while (l < m) nums[i++] = left[l++];
        while (r < n) nums[i++] = right[r++];
    }

    vector<int> mergeSort(vector<int>& nums) {
        if (nums.size() > 2) {
            // Split the array into 2
            const int n = nums.size();
            const int half = n / 2;

            vector<int> left(nums.begin(), nums.begin() + half);
            vector<int> right(nums.begin() + half, nums.end());

            left = mergeSort(left);
            right = mergeSort(right);

            merge(nums, left, right);
        }
        else if (nums.size() > 1) {
            selectionSort2(nums);
        }

        return move(nums);
    }
public:
    vector<int> sortArray(vector<int>& nums) {
        // Merge Sort Recursive
        // TC: O(nlogn), SC: O(n)
        return mergeSort(nums);
    }

    Solution() {
        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
    }
};
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