Maximum time of Extract-Min - Fibonacci heap

Question

What is the actual maximum time of executing extract-min on n-element Fibonacci heap?

Is it O(D(n) + t(H)), where D(n) = lg*n is the maximum degree of a node in n-element heap and t(H) = O(n) is a number of roots in heap H?

Does this mean that answer for question above is actually O(n) = Theta(n)? If no, please correct my thinking and answer.

score 1 · Accepted Answer · answered Feb 10 '14 at 17:33

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You're correct -- the maximum time complexity of a single call to deleteMin is O(n). The lower O(logn) bound on the operation is its amortized time complexity, and is the same between the best case and the worst case.

answered Feb 10 '14 at 17:33

Sneftel

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So the actual maximum time would be ``O(n)``. Can't there be any worse case? :) – Kamil Gosciminski Feb 10 '14 at 17:46
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Not sure what you're asking there. The worst situation for `deleteMin` is a heap with `n` roots, and that is processed in `O(n)` time. – Sneftel Feb 10 '14 at 17:50

Maximum time of Extract-Min - Fibonacci heap

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