Question

We have a sequence of N numbers, and we have them all in hand. We decide to sort those numbers using a minimum binary heap. What is the worst-case time complexity of producing a sorted sequence of these numbers? We will be as specific as we can here.

a. O(N log N)
b. O(N²)
c. O(log N)
d. O(N)

Answer 1

The time complexity for sorting a sequence of numbers using a minimum binary heap is O(N log N), considering both the heap construction and the sorting phases.

Step-by-step explanation:

The worst-case time complexity for sorting a sequence of N numbers using a minimum binary heap is O(N log N). The process of sorting with a heap consists of two main phases: building the heap and the actual sort. When you build a heap from an unsorted array, it has a time complexity of O(N). However, the sorting phase requires that you repeatedly delete the smallest element (the root of the heap) and then reheapify (which has a time complexity of O(log N) due to the height of the heap). Since you need to do this for each of the N elements, the total time complexity for the heap sort process is O(N log N).