<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 {
private:
    void mergeSort(vector<int>& nums,int s,int e){
        if(s>=e){
            return;
        }
        int m=(s+e)/2;
        mergeSort(nums,s,m);
        mergeSort(nums,m+1,e);
        merge(nums,s,e);
    }
    void merge(vector<int>& nums,int s,int e){
        int mid=(s+e)/2;
        int l1=mid-s+1;
        int l2=e-mid;
        int *first=new int[l1];
        int *second=new int[l2];
        
        for (int i = 0; i < l1; i++) {
            first[i] = nums[s + i];
        }
        for (int i = 0; i < l2; i++) {
            second[i] = nums[mid + 1 + i];
        }
        int i = 0, j = 0;  // Pointers to traverse the temporary arrays
        int k = s;         // Pointer to update the main array

        // Merging the two halves back into nums[]
        while (i < l1 && j < l2) {
            if (first[i] <= second[j]) {
                nums[k++] = first[i++];
            } else {
                nums[k++] = second[j++];
            }
        }

        // Copying any remaining elements of first[]
        while (i < l1) {
            nums[k++] = first[i++];
        }

        // Copying any remaining elements of second[]
        while (j < l2) {
            nums[k++] = second[j++];
        }
       
    }
public:
    vector<int> sortArray(vector<int>& nums) {
        int n=nums.size();
        mergeSort(nums,0,n-1);
        vector<int> ans;
        for(int i=0;i<n;i++){
            ans.push_back(nums[i]);
        }
        return ans;
    }
};
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