Solution Explanation for Flip Game II

This problem is a classic example of a game theory problem that can be solved using several approaches, including backtracking, dynamic programming, and memoization. The core idea is to determine if the first player has a winning strategy.

Understanding the Game:

Two players take turns flipping two consecutive '+' characters in a string to '--'. The player who can't make a move loses. We need to determine if the first player can win, guaranteeing victory regardless of the second player's moves.

Solution 1: Memoized Backtracking (Recommended)

This approach uses backtracking to explore all possible game states. Memoization significantly optimizes the solution by storing the results of subproblems to avoid redundant calculations.

Algorithm:

1. Bit Manipulation: Represent the game state using a bitmask. Each bit represents a character in the string; 1 for '+' and 0 for '-'. This allows for efficient manipulation of the string.
2. Recursive Function (dfs): A recursive function explores all possible moves from a given game state.
   • Base Case: If no moves are possible (no consecutive '++'), the current player loses.
   • Recursive Step: For each possible move (flipping consecutive '++'), recursively call dfs with the updated bitmask. If the next player loses (returns false), the current player wins.
3. Memoization: A memo dictionary (or HashMap in Java) stores the result of each game state. If a state is encountered again, the stored result is returned directly, avoiding recalculation.

Time Complexity Analysis:

• The worst-case time complexity is O(2n), where n is the length of the string. However, memoization drastically reduces this in practice. Many game states will be explored only once. The actual runtime is heavily dependent on the input string. In the worst case (a string consisting entirely of '+'), it explores all possible game states.
• Space Complexity: O(2n) in the worst-case scenario due to the memoization table storing all possible game states. However, it's typically much smaller in practice.

Solution 2: Grundy's Game (Advanced)

This approach leverages the concept of the Grundy value (also known as the nimber) from combinatorial game theory. It is a more efficient solution for larger input strings.

Algorithm:

1. Grundy Value (sg): The Grundy value for a given substring length i represents the minimum excluded value (MEX) of the Grundy values of its possible sub-partitions. It is calculated recursively.
2. XOR Sum: The XOR sum of the Grundy values of all maximal '+' substrings in the currentState determines the overall game state. If this XOR sum is non-zero, the first player can win; otherwise, they lose.

Time Complexity Analysis:

• Time Complexity: O(n2). This is significantly faster than the memoized backtracking approach for large inputs. Calculating the Grundy values takes O(n2) time.
• Space Complexity: O(n) to store the Grundy values.

Code Examples (Python):

Solution 1 (Memoized Backtracking):

class Solution:
    def canWin(self, currentState: str) -> bool:
        @cache
        def dfs(mask):
            for i in range(n - 1):
                if (mask & (1 << i)) == 0 or (mask & (1 << (i + 1)) == 0):
                    continue
                if dfs(mask ^ (1 << i) ^ (1 << (i + 1))):
                    continue
                return True
            return False
 
        mask, n = 0, len(currentState)
        for i, c in enumerate(currentState):
            if c == '+':
                mask |= 1 << i
        return dfs(mask)
 

Solution 2 (Grundy's Game):

class Solution:
    def canWin(self, currentState: str) -> bool:
        def win(i):
            if sg[i] != -1:
                return sg[i]
            vis = [False] * n
            for j in range(i - 1):
                vis[win(j) ^ win(i - j - 2)] = True
            for j in range(n):
                if not vis[j]:
                    sg[i] = j
                    return j
            return 0
 
        n = len(currentState)
        sg = [-1] * (n + 1)
        sg[0] = sg[1] = 0
        ans = i = 0
        while i < n:
            j = i
            while j < n and currentState[j] == '+':
                j += 1
            ans ^= win(j - i)
            i = j + 1
        return ans > 0

The code examples in other languages (Java, C++, Go) follow similar logic and structures, adapting the data structures and syntax accordingly. Solution 1 is generally easier to understand, while Solution 2 offers better performance for large inputs but requires a more advanced understanding of game theory. Choose the solution that best suits your needs and understanding.