294: Flip Game II

You are playing the following Flip Game with your friend: Given a string that contains only these two characters: + and -, you and your friend take turns to flip twoconsecutive "++" into "--". The game ends when a person can no longer make a move and therefore the other person will be the winner.

Write a function to determine if the starting player can guarantee a win.

For example, given s = "++++", return true. The starting player can guarantee a win by flipping the middle "++" to become "+--+".

Follow up:

Derive your algorithm's runtime complexity.

public class Solution {
    public boolean canWin(String s) {
        if (s == null || s.length() == 0) {
            return false;
        }

        char[] arr = s.toCharArray();

        return canWinHelper(arr);
    }

    private boolean canWinHelper(char[] arr) {
        int i = 0;

        for (i = 0; i < arr.length - 1; i++) {
            if (arr[i] == '+' && arr[i + 1] == '+') {
                arr[i] = '-';
                arr[i + 1] = '-';

               boolean win = !canWinHelper(arr);

                arr[i] = '+';
                arr[i + 1] = '+';

                if (win) {
                    return true;
                }
            }
        }

        return false;
    }
}

For most interviews, this is the expected solution. Now let's check the time complexity: Suppose originally the board of size N contains only '+' signs, then roughly we have:

T(N) = T(N-2) + T(N-3) + [T(2) + T(N-4)] + [T(3) + T(N-5)] + ... [T(N-5) + T(3)] + [T(N-4) + T(2)] + T(N-3) + T(N-2) = 2 * sum(T[i]) (i = 3..N-2)

You will find that T(N) = 2^(N-1) satisfies the above equation. Therefore, this algorithm is at least exponential.