<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 {
    void Merge(vector<int> &nums ,int &start1 ,int &start2 , int &n){
        int m1 = start2 - start1;
        int m2 = n-(start2 - start1);
        vector<int> Arr;
        for(int i=0;i<m1;i++) Arr.push_back(nums[start1+i]);
        
        int i=0, j=0;
        int index = start1;
        while(i<m1 && j<m2){
            if(Arr[i] < nums[start2 + j]){
                nums[index] = Arr[i]; 
                i++;
            }
            else {
                nums[index] = nums[ start2 + j];
                j++;
            }
            index ++;
        }   
        while(i<m1){
            nums[index] = Arr[i]; 
            i++;
            index ++;
        }
    }
    void Merge_sort(vector<int> &nums,int &start ,int n){
        if(n<2)return;
        else{
            int start2 = start + n/2;
            Merge_sort(nums ,start , n/2);
            Merge_sort(nums ,start2 , (n+1)/2);
            Merge(nums,start ,start2 ,n);
        }
    }
public:
    Solution(){
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);
    }
    vector<int> sortArray(vector<int>& nums) {
        int start = 0;
        int n = nums.size();
        Merge_sort(nums,start,n);
        return nums;
    }   
};
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