Hints for the solution

#1

Could you please tell me how to approch the solution

#2

hey @indrabijaynarayan

If observed carefully , we can identify that this is a problem for fibonacci series. This is because at the nth place , there are two possibilities.
Possibility 1 : We can choose to place the current character as ‘a’. If so , then it doesn’t matter whether we placed ‘a’ or ‘b’ at the previous position. The total number of ways in this possibility would equal to f(n-1)
Possibility 2 : We can place the current character as ‘b’. However we can only do it if the previous character was not ‘b’ . Hence the total number of ways for this case must be f(n-2)
We add these two possibilities up and obtain the recursive relation
f(n) = f(n-1) + f(n-2)
This is clearly the recursive relation for Fibonacci Series .

If you are not able to write a code. I will help
Try yourself first

#3

please help me with pseudo code

#4

main() {
answer(n + 2);

	}



public static int answer(int n) {
	if (n == 0 || n == 1) {
		return n;
	}
	return answer(n - 1) + answer(n - 2);
}