<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]:
        #quick sort
        n = len(nums)
        def partition(arr, start, end):
            if start >= end:
                return True

            mid = (start+end) // 2
            arr[mid], arr[end] = arr[end], arr[mid]

            flag = arr[end]
            left = start - 1
            for i in range(start, end):
                if arr[i] <= flag:
                    left += 1
                    arr[left], arr[i] = arr[i], arr[left]

            left+=1
            arr[left], arr[end] = arr[end], arr[left]
            return left

        def qs(arr, start, end):
            if start < end:
                pivot  = partition(arr, start, end)
                qs(arr, start, pivot-1)
                qs(arr, pivot+1, end)



        arr_nums = list(set(nums))
        m = len(arr_nums)
        qs(arr_nums, 0, m-1)

        dd = collections.Counter(nums)
        res = []
        for k in arr_nums:
            res += [k]*dd[k]

        return res

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