Question

3. What does mystery2 compute on an input n? Please explain how you found this. thanks!

public static int problem1(int n) {

int begin = 0;

int end = n+1;

while (begin+1 < end){

int mid = (begin+end)/2;

if(112*mid <= n){

begin = mid;

}

else{

end = mid;

}

}

return begin;

}

Answer 1

ANSWER: I created a program to test your method we have put n value is higher to get the result greater than 0 because this method always returning begin value which 0 and it is updating when if condition becomes true. below is the explanation in the code please like it.

CODE:

public class Test

{
   public static int problem1(int n) {

       int begin = 0;// here begin is initialized with 0

       int end = n+1; // end vaue will be 1 more than n

       while (begin+1 < end){ // this while will start from 1 to end means n+1

       int mid = (begin+end)/2; // mid value for each iteration

       if(112*mid <= n){ // condition which will be true if mid*112 > n then only begin will be equal to mid

       begin = mid;

       }

       else{

       end = mid; // otherwise end will be mid

       }

       }

       return begin; // returning the begin value always

       }
public static void main(String [] args)
  
{
   System.out.println("problem1 value when n=500: "+problem1(500));

   }

  

}

output: