Description

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 two consecutive "++" 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.

Example:

Input: s = "++++"
Output: true
Explanation: The starting player can guarantee a win by flipping the middle "++" to become "+--+".

Follow up:
Derive your algorithm's runtime complexity.

Solution

Backtracking, O(n!), S(n)

時(shí)間復(fù)雜度有人說(shuō)是O(n!!)犀盟？待研究其做。

The idea is try to replace every "++" in the current string s to "--" and see if the opponent can win or not, if the opponent cannot win, great, we win!

class Solution {
    public boolean canWin(String s) {
        if (s == null || s.length() < 2) {
            return false;
        }
        
        return canWin(s.toCharArray());
    }
    
    private boolean canWin(char[] arr) {
        // try to find a solution making nextCanNotWin
        boolean nextCanWin = true;
        
        for (int i = 0; i < arr.length - 1 && nextCanWin; ++i) {
            if (arr[i] == '+' && arr[i + 1] == '+') {
                arr[i] = '-';
                arr[i + 1] = '-';
                nextCanWin = canWin(arr);
                arr[i] = '+'; // don't forget to revert, even if nextCanNotWin!
                arr[i + 1] = '+';
            }
        }
        return !nextCanWin;
    }
}

Backtracking with memo, O(?), S(?)

class Solution {
    public boolean canWin(String s) {
        if (s == null || s.length() < 2) {
            return false;
        }
        
        return canWin(s.toCharArray(), new HashMap<>());
    }
    
    private boolean canWin(char[] arr, Map<String, Boolean> memo) {
        if (memo.containsKey(new String(arr))) {
            return memo.get(new String(arr));
        }
        
        // try to find a solution making nextCanNotWin
        boolean nextCanWin = true;
        
        for (int i = 0; i < arr.length - 1 && nextCanWin; ++i) {
            if (arr[i] == '+' && arr[i + 1] == '+') {
                arr[i] = '-';
                arr[i + 1] = '-';
                nextCanWin = canWin(arr, memo);
                arr[i] = '+'; // don't forget to revert, even if nextCanNotWin!
                arr[i + 1] = '+';
            }
        }
        
        boolean canWin = !nextCanWin;
        memo.put(new String(arr), canWin);
        return canWin;
    }
}