<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 {
    vector<int> mergesort(int start,int end,vector<int>& nums){
        if(start==end){
            return{nums[start]};
        }
        vector<int> result;
        int mid=(start+end)/2;
        vector<int> left= mergesort(start,mid,nums);
        vector<int> right=mergesort(mid+1,end,nums);
        int i=0,j=0;
        while(i<=(mid-start) || j<=(end-mid-1)){
            int a=INT_MAX,b=INT_MAX;
            if(i<=mid-start){
                a=left[i];
            }
            if(j<=end-mid-1){
                b=right[j];
            }
            a<=b?i++:j++;
            result.push_back(min(a,b));
        }
        return result;
    }
public:
    vector<int> sortArray(vector<int>& nums) {
        return mergesort(0,nums.size()-1,nums);
    }
};
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