This problem is about the array-based binary heap implementation of a priority queue. let h denote the backing array, i. e h is a binary min-heap array. you should assume that the universe of possible keys is extremely large relative to the size of the heap (i. e., it is not practical to use a position array as described on p. 65 of the ) give an algorithm for a new operation decreasekey(index, delta) that subtracts delta from the key located at the given index in h, and restores the heap ordering. report the time taken by the algorithm.(b) suppose we also want an operation findkey(v) that returns the index of a key k in h (if k appears more than once, then it does not matter which index is returned). describe a modified implementation of the heap in which your findkey(v) operation is o(1), finding the minimum is still o(1), and the operations to insert a key or remove the minimum are still o(log n).

Yo can not change your age

Do you need medical attention?

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.