<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:
    vector<int> sortArray(vector<int>& nums) {
        int n = nums.size();
        for(int p = 1 ; p <= n ; p *= 2){
            int s1 = 0, e1 = p-1, s2 = e1+1, e2 = s2+p-1;
            while(s1 < nums.size()){
                vector<int> left;
                for(int i = s1 ; i <= min(e1,n-1) ; i++)
                    left.push_back(nums[i]);
                vector<int> right;
                for(int i = s2 ; i <= min(e2,n-1) ; i++)
                    right.push_back(nums[i]);
                int i = 0, j = 0;
                for(int k = s1 ; k <= min(e2,n-1) ; k++){
                    if(i >= left.size()){
                        nums[k] = right[j++];
                    }
                    else if(j >= right.size()){
                        nums[k] = left[i++];
                    }
                    else if(left[i] > right[j]){
                        nums[k] = right[j++];
                    }
                    else{
                        nums[k] =left[i++];
                    }
                }
                s1 = e2+1;
                e1 = s1+p-1;
                s2 = e1+1;
                e2 = s2+p-1;
            }
        }
        return nums;
    }
};
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