<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 mergesort(vector<int>& nums, int l, int r){
        int n = r - l;
        if(n < 2) return;
        if(n == 2){
            if(nums[l] > nums[r-1]) swap(nums[l], nums[r-1]);
            return;
        }
        n = n / 2 + l;
        mergesort(nums, l, n);
        mergesort(nums, n, r);
        vector<int> right(nums.begin() + l, nums.begin() + n);
        int i = 0;
        while(i < right.size() && n < r){
            if(nums[n] > right[i]) {
                nums[l] = right[i];
                l++;
                i++;
            }
            else{
                nums[l] = nums[n];
                l++;
                n++;
            }
        }
        while(i < right.size()) nums[l++] = right[i++];
    }
public:
    vector<int> sortArray(vector<int>& nums) {
        mergesort(nums, 0, nums.size());
        return nums;
    }
};
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