<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 {
public:

    void merge(vector<int>& nums, int l, int m, int h)
    {
        if(l >= h)
            return;

        vector<int> left(m-l+1);
        vector<int> right(h-m);

        //cout << "l = " << l << " m = " << m << " h = " << h << endl;
        //cout << "right size = " << right.size() << endl;

        for(int i = l; i<=m; i++)
        {
            left[i-l] = nums[i];
        }    
        
        for(int j = m+1; j<=h; j++)
        {
            right[j-m-1] = nums[j];
        }    
        vector<int> res(h-l+1);

        int i = 0, j = 0;
        int k = 0;
        while(i < m-l+1 && j < h-m)
        {
            if(left[i] < right[j])
            {
                
                res[k] = left[i];
                i++;
                k++;
            }
            else
            {
                res[k] = right[j];
                j++;
                k++;
            }
        }

        while(i < m-l+1)
        {
            res[k] = left[i];
            k++;
            i++;
        }
        while(j < h-m)
        {
            res[k] = right[j];
            k++;
            j++;
        }
        
        for(int i = l; i<=h; i++)
        {
           nums[i] = res[i-l];
        }

    }
    void mergeSort(vector<int>& nums, int l, int h)
    {
        if(l >= h)
            return;

        int m = l+(h-l)/2;
        mergeSort(nums, l, m);
        mergeSort(nums, m+1, h);

        merge(nums, l, m, h);
    }
    vector<int> sortArray(vector<int>& nums) {
        
        mergeSort(nums, 0, nums.size()-1);
        return nums;
    }


};
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