Python Coding challenge - Day 722| What is the output of the following Python Code?

 

Code Explanation:

1) import heapq

Loads Python’s heapq module — utilities for working with a min-heap implemented on top of a regular list.

In a min-heap, the smallest element is always at index 0.

2) nums = [5, 1, 8, 3]

Creates a normal Python list with those four values.

At this point it is not yet a heap — just an ordinary list.

3) heapq.heapify(nums)

Converts the list in-place into a valid min-heap (O(n) operation).

After heapify the smallest element moves to index 0 and the list satisfies the heap property.

One possible internal arrangement after heapify is:

[1, 3, 8, 5]

(1 is the smallest and placed at index 0).

4) heapq.heappush(nums, 0)

Inserts 0 into the heap while maintaining the heap property.

Internally it appends the new element and “sifts it up” to the correct position.

Step-by-step (starting from [1, 3, 8, 5]):

append → [1, 3, 8, 5, 0]

sift up: 0 swaps with 3 → [1, 0, 8, 5, 3]

sift up: 0 swaps with 1 → [0, 1, 8, 5, 3]

Final heap after push:

[0, 1, 8, 5, 3]

5) print(heapq.heappop(nums), nums[0])

heapq.heappop(nums) removes and returns the smallest element (root) from the heap.

From [0, 1, 8, 5, 3] it removes 0.

To rebalance, it moves the last element (3) to the root and sifts it down:

put 3 at root → [3, 1, 8, 5]

sift down: swap 3 with smaller child 1 → [1, 3, 8, 5]

result heap after pop: [1, 3, 8, 5]

nums[0] after the pop is now the new smallest element 1.

Final printed output

0 1

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