<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>

void merge(int* arr1,int n1,int* arr2,int n2,int* input,int size){
    int i=0, j=0, k=0;
    while(i<n1 && j<n2){
        if(arr1[i] < arr2[j]) input[k++] = arr1[i++];
        else input[k++] = arr2[j++];
    }
    while(i < n1) input[k++] = arr1[i++];
    while(j < n2) input[k++] = arr2[j++];
}


void mergeSort(int input[], int size){
    if(size == 1 || size == 0) return;
    int n1=size/2, n2=size-n1;
    int* arr1 = new int[n1];
    int* arr2 = new int[n2];
    for(int i=0;i<n1;i++) arr1[i] = input[i];
    int k=0;
    for(int j=n1;j<size;j++) arr2[k++] = input[j];
    
    mergeSort(arr1,n1);
    mergeSort(arr2,n2);
    merge(arr1,n1,arr2,n2,input,size);
 }


class Solution {
public:
    vector<int> sortArray(vector<int>& nums) {
        int n=(int)nums.size();
        int ans[n];
        for(int i=0;i<n;++i)ans[i]=nums[i];
        mergeSort(ans, n);
        vector<int> fin;
        for(auto &x:ans)fin.push_back(x);
        return fin;
    }
};
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