<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 helper(beg, end):
            if end - beg < 2:
                return
            pivot = nums[(beg + end) // 2]
            l, curr, g = beg, beg, end - 1
            while curr <= g:
                if nums[curr] < pivot:
                    nums[curr], nums[l] = nums[l], nums[curr]
                    curr += 1
                    l += 1
                elif nums[curr] > pivot:
                    nums[curr], nums[g] = nums[g], nums[curr]
                    g -= 1
                else:
                    curr += 1
            helper(beg, l)
            helper(g + 1, end)
        
        helper(0, len(nums))
        return nums
35975