<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 not nums:
            return []
        self.merge_sort(nums, 0, len(nums) - 1)
        return nums 
    
    def merge_sort(self, nums, start, end):
        if start >= end:
            return 
        
        mid = (start + end) // 2
        
        self.merge_sort(nums, start, mid)
        self.merge_sort(nums, mid + 1, end)
        
        tmp = []
        leftindex = start 
        rightindex = mid + 1 
        
        while leftindex <= mid or rightindex <= end:
            if leftindex > mid or (rightindex <= end and nums[rightindex] < nums[leftindex]):
                tmp.append(nums[rightindex])
                rightindex += 1
            else:
                tmp.append(nums[leftindex])
                leftindex += 1
                
        nums[start: end + 1] = tmp 
        
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