<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(int lo, int hi, vector<int>&arr) {
        if(lo>=hi) return ;
        int n=hi-lo+10;  ;
        vector<int>temp(n,0) ;
        int mid=(lo+hi)>>1; 
        int i=lo,j=mid+1;
        int k=0;  
        while(i<=mid && j<=hi) {
            if(arr[i]<=arr[j]) {
                temp[k++]=arr[i++] ; 
            }
            else {
                temp[k++]=arr[j++] ; 
            }
        }
        while(i<=mid) temp[k++] = arr[i++] ; 
        while(j<=hi)  temp[k++]=arr[j++] ; 
        for(int k=0; k<(hi-lo+1) ; k++) arr[k+lo]=temp[k] ; 
    }
    void mergesort(int lo, int hi, vector<int>&arr) {
        if(lo>=hi) return ; 
        int mid=(lo+hi)>>1; 
        mergesort(lo,mid,arr) ; 
        mergesort(mid+1,hi,arr) ; 
        merge(lo,hi,arr) ; 
    }
    vector<int> sortArray(vector<int>& arr) {
          int n=arr.size() ;    
          mergesort(0,n-1,arr) ; 
          return arr; 
    }
};
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