<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]:
        
        # In-place merge two halves within nums
        def merge(nums: List[int], start: int, mid: int, end: int):
            # Create temporary arrays for the two halves
            left = nums[start:mid+1]
            right = nums[mid+1:end+1]
            
            i = j = 0  # Pointers for left and right arrays
            k = start  # Pointer for the main array
            
            # Merge the temporary arrays back into nums[start:end+1]
            while i < len(left) and j < len(right):
                if left[i] < right[j]:
                    nums[k] = left[i]
                    i += 1
                elif left[i] > right[j]:
                    nums[k] = right[j]
                    j += 1
                else:
                    # If elements are equal, take both
                    nums[k] = left[i]
                    i += 1
                    k += 1
                    nums[k] = right[j]
                    j += 1
                k += 1

            # Copy any remaining elements of left[]
            while i < len(left):
                nums[k] = left[i]
                i += 1
                k += 1

            # Copy any remaining elements of right[]
            while j < len(right):
                nums[k] = right[j]
                j += 1
                k += 1

        # Recursive merge sort
        def mergeSort(nums: List[int], start: int, end: int):
            if start < end:
                mid = (start + end) // 2
                
                # Sort first and second halves
                mergeSort(nums, start, mid)
                mergeSort(nums, mid + 1, end)
                
                # Merge the sorted halves
                merge(nums, start, mid, end)
        
        # Call the recursive mergeSort on the entire array
        mergeSort(nums, 0, len(nums) - 1)
        
        return nums

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