Sort an Array
<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 necessarily 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>
Array; Divide and Conquer; Bucket Sort; Counting Sort; Radix Sort; Sorting; Heap (Priority Queue); Merge Sort
class Solution:
    def sortArray(self, nums: List[int]) -> List[int]:
        def quick_sort(l, r):
            if l >= r:
                return
            x = nums[randint(l, r)]
            i, j, k = l - 1, r + 1, l
            while k < j:
                if nums[k] < x:
                    nums[i + 1], nums[k] = nums[k], nums[i + 1]
                    i, k = i + 1, k + 1
                elif nums[k] > x:
                    j -= 1
                    nums[j], nums[k] = nums[k], nums[j]
                else:
                    k = k + 1
            quick_sort(l, i)
            quick_sort(j, r)

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