Question

Consider a binary heap (min heap) with N elements in it. What is the worst case time complexity of the "delMin" operation?

a) O(log N)
b) O(N)
c) O(N²)
d) O(1)

Answer 1

The worst-case time complexity of the 'delMin' operation in a binary heap (min heap) is O(log N), which is due to the process of heapifying the new root to maintain the heap property.

Step-by-step explanation:

In a binary heap, particularly a min heap, the delMin operation involves two primary steps to maintain the heap property after the removal of the minimum element (which is at the root).

1. The last element in the heap is moved to the root position, taking the place of the minimum element that was just removed.
2. The new root is then heapified, which involves recursively adjusting the heap by moving this element down the tree to its proper position. This step ensures that the heap maintains its min heap property where the key of each node is greater than or equal to the key of its parent, with the minimum key at the root.

The worst-case time complexity for the heapify process, which is the most time-consuming part of the delMin operation, is O(log N) because this process may potentially have to touch all levels of the heap, and the height of a binary heap is №log N.