<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 mSort(vector<int> &nums,int l,int r){
        if(r-l <= 1){
            return;
        }
        if(r-l == 2){
            int a,b;
            a = min(nums[l],nums[l+1]);
            b = max(nums[l],nums[l+1]);
            nums[l] = a;
            nums[l+1] = b;
            return;
        }
        if(r-l == 3){
            int a,b,c;
            a = min({nums[l],nums[l+1],nums[l+2]});
            b = max({nums[l],nums[l+1],nums[l+2]});
            c = nums[l] + nums[l+1] + nums[l+2] - a - b;
            nums[l] = a;
            nums[l+1] = c;
            nums[l+2] = b;
            return;
        }
        int m = (l+r)/2;
        vector<int> ans;
        mSort(nums,l,m);
        mSort(nums,m,r);
        int i=l;
        int j=m;
        while(i<m && j<r){
            if(nums[i] < nums[j]){
                ans.push_back(nums[i]);
                i++;
            }
            else{
                ans.push_back(nums[j]);
                j++;
            }
        }
        while(i<m){
            ans.push_back(nums[i]);
            i++;
        }
        while(j<r){
            ans.push_back(nums[j]);
            j++;
        }
        for(int k=l;k<r;k++){
            nums[k] = ans[k-l];
        }

        return;
    }

    vector<int> sortArray(vector<int>& nums) {
        mSort(nums,0,nums.size());

        return nums;
    }
};
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