<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:
    def sortArray(self, nums: List[int]) -> List[int]:
        if len(nums) <= 1:
            return nums
        n_len = len(nums)
        left = self.sortArray(nums[0:n_len//2])
        right = self.sortArray(nums[n_len//2:])
        return self.merge(left, right)
    
    def merge(self, left: List[int], right: List[int]) -> List[int]:
        lp, rp, index = 0, 0, 0
        output = []
        while lp < len(left) and rp < len(right):
            if left[lp] <= right[rp]:
                output.append(left[lp])
                index += 1
                lp += 1
            else:
                output.append(right[rp])
                index += 1
                rp += 1
        if lp < len(left):
            output = output + left[lp:]
        else:
            output = output + right[rp:]
        return output
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