<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 partition(nums, lo, hi):
            left, right = lo, hi
            pivot = nums[(lo + hi) // 2]
            while left <= right:
                while nums[left] < pivot:
                    left += 1

                while nums[right] > pivot:
                    right -= 1

                if left <= right:
                    nums[left], nums[right] = nums[right], nums[left]
                    left += 1
                    right -= 1
            return left
            

        def qsort(nums, lo, hi):
            if lo < hi:
                p = partition(nums, lo, hi)

                qsort(nums, lo, p-1)
                qsort(nums, p, hi)

        qsort(nums, 0, len(nums)-1)
        return nums
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