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

    def heapify(self, arr: List[int], N: int, i: int) -> None:
        largest = i  
        l = 2 * i + 1
        r = 2 * i + 2
        if l < N and arr[largest] < arr[l]:
            largest = l
        if r < N and arr[largest] < arr[r]:
            largest = r
        if largest != i:
            arr[i], arr[largest] = arr[largest], arr[i]
            self.heapify(arr, N, largest)

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