<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]:
        if len(nums)<2:
            return nums
        def qsort(l,r):
            if l<r:
                mean=nums[(l+r)//2]
                i,j,k=l,l,r
                while j<=k:
                    if nums[j]<mean:
                        nums[i],nums[j]=nums[j],nums[i]
                        i+=1
                        j+=1
                    elif nums[j]>mean:
                        nums[j],nums[k]=nums[k],nums[j]
                        k-=1
                    else:
                        j+=1
                qsort(l,i-1)
                qsort(k+1,r)
        random.shuffle(nums)
        qsort(0,len(nums)-1)
        return nums        
34295