<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>

#define pq priority_queue
#define um unordered_map
#define ll long long
#define pb push_back
#define popb pop_back
#define mp make_pair

class Solution {
public:
    void mergeSort(vector<int>& nums, int l, int r){
        // printf("l: %d, r: %d\n", l, r);
        int mid = (r - l) / 2 + l;

        if(r - l < 2) return;
        mergeSort(nums, l, mid);
        mergeSort(nums, mid, r);

        vector<int> arr;
        for(int i = l; i < r; i++){
            arr.pb(nums[i]);
        }
        int len = arr.size();
        mid = len / 2;
        int idx1 = 0, idx2 = mid;
        while(l < r){
            if(idx1 == mid){
                nums[l++] = arr[idx2];
                idx2++;
            }else if(idx2 == len){
                nums[l++] = arr[idx1];
                idx1++;
            }else if(arr[idx1] < arr[idx2]){
                nums[l++] = arr[idx1];
                idx1++;
            }else{
                nums[l++] = arr[idx2];
                idx2++;
            }
        }
    }

    vector<int> sortArray(vector<int>& nums) {
        int len = nums.size();
        
        mergeSort(nums, 0, len);

        return nums;
    }
};
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