<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 heap_sort():
            n = len(nums)
            for i in range(n//2-1, -1, -1):
                heapify(i, n)
            for i in range(n-1, -1, -1):
                nums[0], nums[i] = nums[i], nums[0]
                heapify(0, i)
            
        def heapify(i, n):
            root = i
            left = i * 2 + 1
            right = i * 2 + 2
            if left < n and nums[root] < nums[left]:
                root = left
            if right < n and nums[root] < nums[right]:
                root = right
            if root != i:
                nums[i], nums[root] = nums[root], nums[i]
                heapify(root, n)

        heap_sort()
        return nums
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