Given an array a of n nonnegative integers, the maximum window area (MWA) of a is defined as the maximum of (j − i + 1) min i≤k≤j (a[k]), taken over all combinations i and j for which 0 ≤ i ≤ j ≤ n − 1. For example if a = 3, 3, 1, 7, 4, 2, 4, 6, 1, then MWA(a) = 10 via i = 3 and j = 7, since the minimum value in this range is a[5] = 2, and (7 − 3 + 1)(2) = 10. Clearly describe a divide-and-conquer algorithm for computing MWA(a). Provide a recurrence that describes its running time, and use it to provide your algorithm’s running time. Warning: credit will not be awarded for algorithms that seem vague or incomplete, or that do not achieve an o(n 2 ) running time

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Answer is (d) right aligned;

Public static int mystery(int a[], int x) {     int c = 0;     for(int i = 0; i < a.length; i++) {        if (a[i] == x) c++;     } return c; }the method counts how many times an integer shows up in an integer array.the c variable is declared in the method to keep track of how many times the certain value has shown up.the for-loop loops through the entire array.the line    if (a[i] == x) c++;   checks if each element of the array is equal to the inputted integer to check. for each number in the array that is equal to the inputted integer x, the c variable will be incremented by 1. otherwise it will do nothing.then c is returned at the end of the loop.for example, if we have, in the main method with the mystery method coded in the class:     int arraytest[] = {2, 23, 23, 23, 3};     int testme = 23;     int count = mystery(arraytest, testme);     system.out.println(count); then the print out will be    3 because there are three 23s in that array.