Leetcode: 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 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.

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.

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

For the time complexity, here is what I thought, let's say the length of the input string s is n, there are at most n - 1 ways to replace "++" to "--" (imagine s is all "+++..."), once we replace one "++", there are at most (n - 2) - 1 ways to do the replacement, it's a little bit like solving the N-Queens problem, the time complexity is (n - 1) x (n - 3) x (n - 5) x ..., so it'sO(n!!), double factorial.

 1 public class Solution {
 2     public boolean canWin(String s) {
 3         if (s==null || s.length()<=1) return false;
 4         for (int i=0; i<s.length()-1; i++) {
 5             if (s.charAt(i)=='+' && s.charAt(i+1)=='+' && !canWin(s.substring(0,i) + "--" + s.substring(i+2))) 
 6                 return true;
 7         }
 8         return false;
 9     }
10 }

Better Solution: (205ms -> 19ms)

but the time complexity of the backtracking method is high. During the process of searching, we could encounter duplicate computation as the following simple case.

One search path:

Input s = "++++++++"

Player 0: "--++++++"

Player 1: "----++++"

Player 0: "----+--+"

Player0 can win for the input string as "----++++".

Another search path:

Player 0: "++--++++"

Player 1: "----++++"

Player 0: "----+--+"

(Duplicate computation happens. We have already known anyone can win for the

input string as "----++++".)

Use a HashMap to avoid duplicate computation

Key : InputString.

Value: can win or not.

 1 public boolean canWin(String s) {
 2     if (s == null || s.length() < 2) {
 3         return false;
 4     }
 5     HashMap<String, Boolean> winMap = new HashMap<String, Boolean>();
 6     return helper(s, winMap);
 7 }
 8 
 9 public boolean helper(String s, HashMap<String, Boolean> winMap) {
10     if (winMap.containsKey(s)) {
11         return winMap.get(s);
12     }
13     for (int i = 0; i < s.length() - 1; i++) {
14         if (s.startsWith("++", i)) {
15             String t = s.substring(0, i) + "--" + s.substring(i+2);
16             if (!helper(t, winMap)) {
17                 winMap.put(s, true);
18                 return true;
19             }
20         }
21     }
22     winMap.put(s, false);
23     return false;
24 }

Reference: https://leetcode.com/discuss/64291/share-my-java-backtracking-solution

https://leetcode.com/discuss/64486/backtracking-solution-time-optimization-through-205ms-19ms