Saimadhav Heblikar added the comment:
Hi,
I dont think its a bug.
The textbook definition of a min(or max) heap is
"Heaps are binary trees for which every parent node has a value less than or
equal to any of its children."
Therefore,
when lista = [1,6,5,4], and heapify is run on it,it can be viewed as
1
/ \
4 5 6
or [1,4,5,6]
When listb=[1,6,5], running heapify on it,gives
1
/ \
6 5
or [1,6,5]
heapify maintains the heap-invariant - Every key is smaller than its children.
This invariant is not violated in the above example.
The heappop maintains the heap-invariant after popping. The line in your code
[heapq.heappop(listb) for i in range(len(listb))]
is the heapsort algorithm!.
Hopefully, this clears your question.
----------
nosy: +sahutd
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