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Min-Max
Write an algorithm that finds both the smallest and
largest numbers in a list of n numbers and with complexity
T(n) is at most about (1.5)n comparisons.
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Answer Posted / oracle
Simple, first compare all elements pairwise, total n/2
comparisons. Then, you got two sets, first set has all
greater elements, and second all smaller ones. Find largest
in the first set (n/2) and smallest in the second set
(n/2). You got both largest and smallest ones, in total n/2
+ n/2 + n/2 = 3n/2 comparisons.
| Is This Answer Correct ? | 54 Yes | 11 No |
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