Question

1. Suppose A is (10, 2, 5, 9, 1, 8, 2, 4). Consider the function:

int BBOX(int n, int k)

            if (n <= 0) return 0;

            else if (A[n] < k) return (1+ 2*BBOX(n-1,k+1));

            else return BBOX(n-1,k-2);

            Find BBOX(8, 5)

Answer 1

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Explaination of the function BBOX(int n, int k):

1. The value of n is checked first. If the value of n is less than or equal to 0, then the programs returns zero. If the value of n is not less than or equal to 0, then go to step 2.
2. Here, the value of A[n] is compared with the value of k. If A[n] is less than the value of k, then the program recursively calls itself, i.e BBOX(n-1, k+1). Then multiplies its value with 2, adds 1 and retuens the final value. If the condition is not true, i.e if A[n] is not less than k then go to step 3.
3. Here, the function calls itself BB)X(n-1, k-2)

The working of the function call BBOX(8,5) is shown below