<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 partition(self, nums: list[int], left: int, right: int) -> int:
        pivot = nums[(right - left) // 2 + left]
        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 quicksort(self, nums: list[int], left: int, right: int):
        index = self.partition(nums, left, right)
        if left < index - 1:
            self.quicksort(nums, left, index - 1)
        if right > index:
            self.quicksort(nums, index, right)

    def sortArray(self, nums: List[int]) -> List[int]:
        self.quicksort(nums, 0, len(nums) - 1)
        return nums
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