<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]:
        # create max-heap 
        def heapify(nums, n, i):
            ptr = i 
            left = 2 * i + 1
            right = 2 * i + 2
            
            if left < n and nums[left] > nums[ptr]:
                ptr = left

            if right < n and nums[right] > nums[ptr]:
                ptr = right

            if ptr != i:
                nums[i], nums[ptr] = nums[ptr], nums[i]
                heapify(nums, n, ptr)

        n = len(nums)
        for i in range(n//2-1, -1, -1):
            heapify(nums, n, i)
        
        for i in range(n-1, -1, -1):
            nums[0], nums[i] = nums[i], nums[0]
            heapify(nums, i, 0)
        return nums

            
            
        
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