<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
        # find the middle 
        mid = len(nums) //2

        #recursively split and sort both halves
        left_half = self.sortArray(nums[:mid])
        right_half = self.sortArray(nums[mid:])

        #return the sorted halves
        return self.merge(left_half, right_half)

    def merge(self, left, right):
        sorted_list = []
        i, j = 0,0
        while i < len(left) and j < len(right):
            if left[i] <= right[j]:
                sorted_list.append(left[i])
                i+=1
            else:
                sorted_list.append(right[j])
                j+=1
        sorted_list.extend(left[i:])
        sorted_list.extend(right[j:])
        return sorted_list
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