<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:
            left_arr = nums[:len(nums)//2]
            right_arr = nums[len(nums)//2:]

            # recursion
            self.sortArray(left_arr)
            self.sortArray(right_arr)

            # merge
            i, j, k = 0, 0, 0 # left, right, merged idx
            while i < len(left_arr) and j<len(right_arr):
                if left_arr[i] < right_arr[j]:
                    nums[k] = left_arr[i]
                    i+=1
                else:
                    nums[k] = right_arr[j]
                    j+=1
                k+=1
            # leftovers
            while i < len(left_arr):
                nums[k] = left_arr[i]
                i+=1
                k+=1
            while j < len(right_arr):
                nums[k] = right_arr[j]
                j+=1
                k+=1
        return nums
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