Question

on 2 Consider the following recursive method test: public static int test(String s, int last) if (last < 0) { return 0; if (s.charAt(last) == 0) { return 2 - test(s, last - 1); return 1 + 2 test(s, last - 1); What is the output of: test("11001", 4). TTT Arial 3 (12pt) - TEE 25

Answer 1

Please find the answer for the above given question ..

ANSWER :  
The Output of test("11001",4) is 25

OUTPUT :

Files : Main.java @ saved 1 class Main { //Test function definition public static int test(String s, int last) Add file Open JDK Runtime Environment (build 11.0.6 st-Ubuntu-lubuntul18.04.1) > javac -classpath ./run dir/junit-4.12, rget/dependency/* -d . Main.java > java -classpath ./run dir/junit-4.12. get/dependency/* Main if(last<0){ return 0; //Return o when the last value goes negavtive OUTPUT : Result : 25 // This return will get executed when the bit is "O" if(s.charAt(last) == '0'){ return 2*test(s,last-1); // This return will get executed when the bit is "1" return 1+2*test(s, last-1); //main method public static void main(String[] args) { int result;//To store the returned value from the function test result = test("11001",4);//Calling test function //To print the result on the console System.out.println("\nOUTPUT : \nResult : "+ result);

Explanation :

The entire code talk about the one of the function's feature called Recursion. So the above function will get returned to the main function only when the last value goes negative. So in total the test function will get executed for 5 times since the length of the given string is of 5. So all the function calls will get stored in a STACK.

when the last value is -1 and 0 will be returned back to the function call.

I will give the values of function returns based on the last value:

When last = -1 ,  function returns 0.

last = 0 , 1+2*0 = 1

last =1 , 1+2*1 = 3

last =2 , 2*3 = 6

last =3 , 2*6 = 12

last =4 , 1+2*12 = 1+24 =25

Hence the result is 25.

Thanks..