The minimum number of comparisons required to find the minimum and the maximum…

Key idea: use pairwise comparisons to reduce the total number of comparisons.

• If n is even: Compare the first two elements (1 comparison) to initialize the current minimum and maximum. Then process the remaining n−2 elements in pairs. Each pair requires 3 comparisons (compare the two elements, then compare the larger to the current max and the smaller to the current min). Total = 1 + 3*(n−2)/2 = 3n/2 − 2.

• If n is odd: Initialize min and max using the first element (0 comparisons). Process the remaining n−1 elements in pairs, each requiring 3 comparisons. Total = 3*(n−1)/2.

Apply to n = 100 (even):

Comparisons = 3*100/2 − 2 = 150 − 2 = 148.

Therefore the minimum number of comparisons required to find both the minimum and maximum of 100 numbers is 148.