<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> &n, int st, int en, int m){
        vector<int> l,r;
        for(int i=st;i<=en;i++){
            if(i<=m){
                l.push_back(n[i]);
            }
            else r.push_back(n[i]);
        }
        int i=0, j=0,k=st;
        while(i<l.size()&&j<r.size()){
            if(l[i]<=r[j])
                n[k] = l[i++];
            else n[k] = r[j++];
            ++k;
        }
        
        while(i<l.size())
            n[k++] = l[i++];
        
        while(j<r.size())
            n[k++] = r[j++];
            ++k;
        
    }
    
    void mergesort(vector<int> &n, int st, int en){
        int mid = (st+en)>>1;
        if(st<en){
            mergesort(n,st,mid);
            mergesort(n,mid+1,en);
            
            merge(n,st,en,mid);
        }
    }
    
    vector<int> sortArray(vector<int>& nums) {
        mergesort(nums,0,nums.size()-1);
        return nums;
    }
};
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