<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(vector<int>&arr, int l, int m, int r){
        int mid=m;
        int s=l,e=r;
        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]=arr[k++];
        }
        k=mid+1;
        for(int i=0;i<len2;i++){
            second[i]=arr[k++];
        }

        //merge
        int index1=0,index2=0;
        k=s;
        //logic for inserting in sorted order
        while(index1<len1&&index2<len2){
            if(first[index1]<second[index2]){
                arr[k++]=first[index1++];
            }
            else{
                arr[k++]=second[index2++];
            }

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

         
    }
     void mergeSort(vector<int>&arr, int l, int r){
        int s=l,e=r;
        if(s>=e){
            return ;
        }
        //left part sort
        int mid=(s+e)/2;

        //sorting the left part of array
        mergeSort(arr,s,mid);
        //sorting the right part
        mergeSort(arr,mid+1,e);

        merge(arr,s,mid,e);

    }
    vector<int> sortArray(vector<int>& nums) {
        int n=nums.size();
        mergeSort(nums,0,n-1);
        return nums;
    }
};
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