<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 mid, int r) {
        vector<int> temp;
        int left = l, right = mid + 1;
        while(left <= mid) {
            while(right <= r && nums[right] <= nums[left]) {
                temp.push_back(nums[right]);
                ++right;
            }
            temp.push_back(nums[left]);
            ++left;
        }
        while(right <= r) {
            temp.push_back(nums[right]);
            ++right;
        }
        for(int i = l; i <=r; ++i) nums[i] = temp[i - l];
    }

    void mergeSort(vector<int>& nums, int l, int r) {
        if(l < r) {
            int mid = l + (r - l) / 2;
            mergeSort(nums, l, mid);
            mergeSort(nums, mid + 1, r);
            merge(nums, l, mid, r);
        }
    }

    vector<int> sortArray(vector<int>& nums) {
        mergeSort(nums, 0, nums.size() - 1);
        return nums;
    }
};
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