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Heap Algo 4 You are using the HeapPriorityQueue class implemented with a binary min-heap, which is stored as an array (Python list) in level-order. The heap currently stores the following key-value pairs: [(2, 'A'), (4, 'B'), (6, 'C'), (8, 'D'), (5, 'E')] After calling remove_min() once, what will be the contents of the heap (in array/list format)? (Assume that the heap properties are restored correctly after the operation.) Pseudocode for remove_min() remove_min(): swap the root with the last element remove the last element (which is the min) call _downheap(0) Pseudocode for _downheap(j) _downheap(j): if left child exists: small_child = left if right child exists and right < left: small_child = right if small_child < j: swap(j, small_child) _downheap(small_child) In this implementation, for an element at index j: Left child is at index 2j + 1 Right child is at index 2j + 2 Parent is at index (j - 1) // 2

Options
A.[(4, 'B'), (8, 'D'), (6, 'C')]
B.[(5, 'E'), (4, 'B'), (6, 'C'), (8, 'D')]
C.[(4, 'B'), (8, 'D'), (6, 'C'), (5, 'E')]
D.[(4, 'B'), (5, 'E'), (6, 'C'), (8, 'D')]
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Step-by-Step Analysis
We start by restating the problem setup and then inspect each answer choice in light of how remove_min and _downheap operate on a binary min-heap stored in an array. First, the initial heap is [(2, 'A'), (4, 'B'), (6, 'C'), (8, 'D'), (5, 'E')]. When remove_min is called, the typical steps are to swap the root with the last element, remove the last element (the original min), and then restore the heap order by downheaping from the root. Step-by......Login to view full explanation

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