<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> &nums, int low, int mid, int high){
        int l=low,h=mid+1;
        vector<int> arr;
        while(l<=mid && h<=high){
            if(nums[l]<=nums[h]) {arr.push_back(nums[l]);l++;}
            else {arr.push_back(nums[h]);h++;}
        }
        while(l<=mid) {arr.push_back(nums[l]);l++;}
        while(h<=high) {arr.push_back(nums[h]);h++;}
        int k=0;
        for(int i=low;i<=high;i++){
            nums[i]=arr[k++];
        }
        return;
    }
    void merge_sort(vector<int>& nums,int low,int high){
        if(low<high) {
            int mid=(low+high)/2;
            merge_sort(nums,low,mid);
            merge_sort(nums,mid+1,high);
            Merge(nums,low,mid,high);
        }
    }
    vector<int> sortArray(vector<int>& nums) {
        merge_sort(nums,0,nums.size()-1);
        return nums;
    }
};
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