<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> &left, vector<int> &right, vector<int> &temp) {
        int n = left.size();
        int m = right.size();

        int i = 0, j = 0, k = 0;
        while (i < n && j < m) {
            if (left[i] <= right[j]) {
                temp[k++] = left[i++];
            } else {
                temp[k++] = right[j++];
            }
        }
        while (i < n) {
            temp[k++] = left[i++];
        }
        while (j < m) {
            temp[k++] = right[j++];
        }
    }

    vector<int> mergeSort(int i, int j, vector<int> &nums) {
        if (i > j) return {};
        if (i == j) return {nums[i]};

        int mid = (i + j) / 2;

        vector<int> left = mergeSort(i, mid, nums);
        vector<int> right = mergeSort(mid + 1, j, nums);
        vector<int> temp(left.size() + right.size());

        merge(left, right, temp);
        return temp;
    }

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

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