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The total comparisons in finding both smallest and largest elements are

(a) 2*n +2

(b) n + ((n+1)/2) -2

(c) n+logn

(d) n^2

Question is from Weak Heap topic in chapter Heap of Data Structures & Algorithms I

The question was posed to me in an interview.

1 Answer

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Best answer
The correct option is (b) n + ((n+1)/2) -2

The best explanation: The total comparisons in finding smallest and largest elements is n + ((n+1)/2) – 2.

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