<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(l: int, r:int):
            if r==l : return 
            
            mid = (r + l) // 2
            mergeSort(l, mid)   
            mergeSort(mid+1, r)   
            l1, r1 = l, mid 
            l2, r2 = mid + 1, r
            numsR = nums[mid+1:r+1]
            while r >= l:
                v1 = float("-inf")
                v2 = float("-inf")
                if r1 >= l1: v1 = nums[r1]
                if r2 >= l2: v2 = numsR[r2-mid-1]
                if v1 > v2:
                    nums[r] = v1
                    r1 -= 1
                else:
                    nums[r] = v2
                    r2 -= 1
                r-=1
            return 
        mergeSort(0, len(nums)-1)
        return nums
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