<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(int s,int e,vector<int>& nums){
        int mid = s+(e-s)/2;
        int len1 = mid - s +1;
        int len2 = e -mid;

        int *first = new int[len1];
        int *second = new int[len2];

        int k = s;

        for(int i =0;i<len1;i++){
            first[i]=nums[k++];
        }

        k = mid+1;

        for(int i =0;i<len2;i++){
            second[i]=nums[k++];
        }

        int index1 = 0, index2 = 0 ;
        k =s;
        while(index1<len1&&index2<len2){
            if(first[index1]<=second[index2]){
                nums[k++]=first[index1++];
            }
            else{
                nums[k++]=second[index2++];
            }

        }
        while(index1<len1){
            nums[k++]=first[index1++];
        }
        while(index2<len2){
            nums[k++]=second[index2++];
        }

        delete[] first;
        delete[] second;
    }
    void solve(int s,int e,vector<int>& nums){
        if(s>=e){
            return;
        }

        int mid = s+(e-s)/2;

        solve(s,mid,nums);

        solve(mid+1,e,nums);

        merge(s,e,nums);
    }
public:
    vector<int> sortArray(vector<int>& nums) {
       int s =0;
       int e = nums.size()-1;

       solve(s,e,nums);

       return nums;
    }
};
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