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^2Question is from Weak Heap topic in chapter Heap of Data Structures & Algorithms IThe question was posed to me in an interview.

The total comparisons in finding both smallest and largest elements ar

1.	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^2Question is from Weak Heap topic in chapter Heap of Data Structures & Algorithms IThe question was posed to me in an interview.
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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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^2Question is from Weak Heap topic in chapter Heap of Data Structures & Algorithms IThe question was posed to me in an interview.

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