Question

Find the time complexity of the following code: i am confused because addA calls addB and so does addB.

   void addA (int n) {
       if (n == 0) {
           System.out.println("hi");
       } else {
           addA( n - 1);
           addB( n - 1);

       }
   }

   void addB(int n) {

       if (n == 0) {
           System.out.println("hello");
       } else {
           addA( n - 1);
           addB(n - 1);
       }

   }

Answer 1

Here is the solution to above program.

the recurrence relation for addA function is

addA(n) = addA(n-1) + addB(n-1)

the recurrence relation for addB is

addB(n) = addA(n-1) + addB(n-1)

since both are same we can denote them with single recurrence relation

T(n) =2T(n-1)

T(n) = 2 ( 2(T(n-2))

T(n) = 2³(T(n-3))

..

.

.

T(n) = 2ⁿT(0))

T(0) = 1 since it is the base case

now since each time we are reducing the value of n by one hence there will be total n times addA(n) will be there total time complexity of program is O(2ⁿ)