<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]:
        
        def mergeSort(nums): 
            if len(nums) > 1: 
                mid = len(nums)//2
                L = nums[:mid] 
                R = nums[mid:] 

                mergeSort(L)
                mergeSort(R)

                i = j = k = 0

                while i < len(L) and j < len(R): 
                    if L[i] < R[j]: 
                        nums[k] = L[i] 
                        i+=1
                    else: 
                        nums[k] = R[j] 
                        j+=1
                    k+=1
    
                while i < len(L): 
                    nums[k] = L[i] 
                    i+=1
                    k+=1

                while j < len(R): 
                    nums[k] = R[j] 
                    j+=1
                    k+=1
        
        mergeSort(nums)

        return nums


        
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