<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 m, int r){
        vector<int> v={m-l+1, r-m, 0, 0, l}, la(v[0]+1, 1e6), ra(v[1]+1, 1e6);
        for(int i=0;i<v[0];i++) la[i] = nums[l+i];
        for(int i=0;i<v[1];i++) ra[i] = nums[m+1+i];
        while(v[4]<=r) nums[v[4]++] = (la[v[2]]<=ra[v[3]])?la[v[2]++]:ra[v[3]++]; }
    void merge_sort(vector<int>& nums, int begin, int end){
        if(begin >= end) return;
        int mid = begin + (end-begin)/2;
        merge_sort(nums, begin, mid);
        merge_sort(nums, mid+1, end);
        merge(nums, begin, mid, end);
    }
    vector<int> sortArray(vector<int>& nums) {
        merge_sort(nums, 0, nums.size()-1);
        return nums;
    }
};
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