<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) {
        return merge_sort(nums, 0, nums.size() - 1);
    }

    vector<int> merge_sort(vector<int>& nums, int start, int end)
    {
        if ((end - start) <= 5)
        {
            vector<int> res;
            for (int i = start; i <= end; i++)
                res.push_back(nums[i]);
            sort(res.begin(), res.end());
            return res;
        }

        int mid = start + (end - start) / 2;
        auto left = merge_sort(nums, start, mid);
        auto right = merge_sort(nums, mid + 1, end);

        int left_idx = 0;
        int right_idx = 0;
        vector<int> merge;

        while (left_idx < left.size() && right_idx < right.size())
        {
            if (left[left_idx] < right[right_idx])
            {
                merge.push_back(left[left_idx]);
                left_idx++;
            }
            else
            {
                merge.push_back(right[right_idx]);
                right_idx++;
            }
        }

        while (left_idx < left.size())
        {
            merge.push_back(left[left_idx]);
            left_idx++;
        }

        while (right_idx < right.size())
        {
            merge.push_back(right[right_idx]);
            right_idx++;
        }

        return merge;
    }
};
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