<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 l, int r){
        int m = l+(r-l)/2;
        
        int i=l, j=m+1;
        
        vector<int> temp;
        while(i<=m || j<=r){
            int val1 = (i<=m ? nums[i]:INT_MAX);
            int val2 = (j<=r ? nums[j]:INT_MAX);
            
            if(val1 <= val2){
                temp.push_back(val1);
                i++;
            }else{
                temp.push_back(val2);
                j++;
            }
        }
        
        for(int i=0;i<r-l+1;i++) nums[i+l] = temp[i];
    }
    
    void mergeSort(vector<int>& nums, int l, int r){
        if(l>=r) return;
        
        int m = l+(r-l)/2;
        
        mergeSort(nums, l, m);
        mergeSort(nums, m+1, r);
        merge(nums, l, r);
    }
    
    vector<int> sortArray(vector<int>& nums) {
        mergeSort(nums, 0, nums.size()-1);
        
        return nums;
    }
};
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