Answered: For this task, save your work in MinMax.java Consider the following problem: given an array of n numbers, we want to find both the minimum and the maximum of...

Transcribed Image Text:For this task, save your work in MinMax.java Consider the following problem: given an array of n numbers, we want to find both the minimum and the maximum of these numbers. For such a problem, we often measure the cost in terms of the number of comparisons made-that is, if we compare any two numbers from the input, that's one comparison. As an example, the following algorithm requires n-1 comparisons: // assume a.length > 0 int maxArray(int[] a) { int maxSoFar = a[0]; for (int i=1;i<a.length; i++) { } } if (a[i] maxSoFar) maxSoFar = a[i]%;B return maxSoFar; This is because in an array a of length n, only a[1], a[2],..., a[n-1] are compared with our maxSoFar in the if statement. Notice that a[0] is not involved in the if statement. You can use this algorithm to the find the maximum value and an almost-identical algorithm to find the minimum value. However, you'll need 2n-2 comparisons (n-1 for max and another n - 1 for min). Only comparisons between input integers (either directly or indirectly) matter here, which is why we don't count comparisons made by i<a.length in the above example. Your goal in this problem is to do better! Subtask I: First, implement a function public static double minMaxAverage (int[] numbers) { // your code goes here int myMin = ...; int myMax = return (myMin + myMax)/2.0; } that takes in an array of integer numbers, finds the minimum and the maximum among these numbers, and returns the average of the minimum and the maximum (as the code above shows). For full credit, if input contains n numbers, your function must use fewer than 3n/2 comparisons. Subtask II: As a comment block in your code file, make a logical argument-as close to an airtight mathematical proof as possible-for why your code is indeed using strictly fewer than 3n/2 comparisons. (Hint: Remember the maximum-number problem from class? What happens after one round in the pairing-up algorithm?)

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