The idea is to turn your original Array into a Heap using "heapify", which takes $O(n)$ time.

You can then repeatedly pop the smallest value from the Heap, and transfer it over to a new Array. The new Array will end up sorted when you're done.

def heapSort(arr):
    # turn array into a min heap
    heapq.heapify(arr)

    # transfer smallest value from Heap to Sorted Array
    sortedArr = []
    for _ in range(len(arr)):
        sortedArr.append(heapq.heappop(arr))
    
    return sortedArr

Time Complexity $O(n \log n)$

We first use heapify which takes $O(n)$. We then pop the heap $n$ times, and popping takes $O(\log n)$ each time, since the heap has $O(n)$ elements in it. This takes $O(n)+O(n\log n) = O(n \log n)$ time.

Space Complexity $O(1)$

We're just transferring elements from the Heap to the sorted array. As the sorted array grows, the Heap shrinks. This means the space complexity is $O(1)$ - there's no extra space needed.