Answer:
One algorithm for finding both the minimum and maximum of n numbers using fewer than 3n/2 comparisons is the "Minimum-Maximum" algorithm. The basic idea is to compare pairs of numbers and keep track of both the minimum and maximum value seen so far. Here's a step-by-step description of the algorithm:
Initialize two variables, "min" and "max" to the first number in the list.
Iterate through the remaining n-1 numbers in the list.
For each number, compare it to the current "min" and "max" values. If the number is less than "min", update "min" to the current number. If the number is greater than "max", update "max" to the current number.
After iterating through all n numbers, "min" will contain the minimum value and "max" will contain the maximum value.
This algorithm uses 3/2*n-2 comparisons to find the minimum and maximum element of the input array.
In this algorithm, we are comparing the current number with both min and max at the same time, so the number of comparisons is less than 3/2*n.