Flip Game II in C++

Suppose there are two players who are playing the flip game. Here we have a string that contains only these two characters: + and -, player1 and player2 take turns to flip two consecutive "++" into "--". The game ends when one player can no longer make a move and therefore the other one will be the winner. We have to define a function to check whether the starting player can guarantee a win.

So, if the input is like s = "++++", then the output will be true, as the starting player can guarantee a win by flipping the middle "++" to become "+--+".

To solve this, we will follow these steps −

• Define one map memo

• Define a function solve(), this will take s,

• if s in memo, then −

  • return memo[s]

• possible := false

• n := size of s

• for initialize i := 0, when i < n - 1, update (increase i by 1), do −

  • if s[i] is same as '+' and s[i + 1] is same as '+', then −

    • s[i] := '-', s[i + 1] := '-'

    • possible := possible OR inverse of solve(s)

    • s[i] := '+', s[i + 1] := '+'

    • if possible is non-zero, then −

      • return memo[s] := possible

• return memo[s] := possible

• From the main method do following −

• return solve(s)

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   unordered_map <string, bool> memo;
   bool solve(string s){
      if (memo.count(s))
         return memo[s];
      bool possible = false;
      int n = s.size();
      for (int i = 0; i < n - 1; i++) {
         if (s[i] == '+' && s[i + 1] == '+') {
            s[i] = '-';
            s[i + 1] = '-';
            possible |= !solve(s);
            s[i] = '+';
            s[i + 1] = '+';
            if (possible)
               return memo[s] = possible;
         }
      }
      return memo[s] = possible;
   }
   bool canWin(string s) {
      return solve(s);
   }
};
main(){
   Solution ob;
   cout << (ob.canWin("++++"));
}

Input

"++++"

Output

1