Discussion
_________ comparisons are necessary in the worst case to find both the maximum and minimum of n numbers.
Answer: C
The number of comparisons needed in the worst case to find both the maximum and the minimum of n numbers is about \(\lceil 3n/2 \rceil - 2\), and option C is the closest listed form. Pair the elements, compare each pair (n/2 comparisons), then find the maximum among the winners and the minimum among the losers (about n/2 - 1 each).
\(2n - 2\) is what the naive method needs, \(n + \lfloor \lg n \rfloor - 2\) is the bound for finding the two largest elements, and \(2 \lg n - 2\) is too small to be possible.
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