Problem Description

You're playing a two-player game called the Flip Game. The game starts with a string currentState that consists only of '+' and '-' characters.

The rules are:

• Players take turns making moves
• On each turn, a player must flip two consecutive "++" characters into "--"
• A player who cannot make a valid move loses (meaning their opponent wins)

Your task is to determine if the starting player can guarantee a win, assuming both players play optimally. Return true if the starting player has a winning strategy, and false otherwise.

For example:

• If currentState = "++++", the first player can flip positions 0-1 or 1-2 or 2-3 to get "--++", "+--+", or "++--" respectively
• The game continues until no player can find two consecutive "++" to flip
• You need to determine if the first player can force a win regardless of how the second player responds

The solution uses dynamic programming with memoization to explore all possible game states. It represents the current state as a bitmask where 1 indicates a '+' and 0 indicates a '-'. The recursive function dfs checks if the current player can win from a given state by trying all possible moves and seeing if any move leads to a losing position for the opponent.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: While we're exploring different game states, this isn't a traditional graph problem with explicit nodes and edges. We're dealing with a string that we modify through moves.

Need to solve for kth smallest/largest?

• No: We're not looking for any kth element. We need to determine if the first player can guarantee a win.

Involves Linked Lists?

• No: The problem works with a string, not linked lists.

Does the problem have small constraints?

• Yes: The problem involves exploring all possible game states from a given position. Since each move reduces the number of possible future moves (converting "++" to "--"), and strings in such games typically have reasonable lengths, we're dealing with a manageable state space that can be explored exhaustively.

Brute force / Backtracking?

• Yes: We need to try all possible moves from each game state and backtrack to explore different branches of the game tree. Each recursive call represents trying a move, and we backtrack when we've explored all possibilities from that state.

Conclusion: The flowchart correctly leads us to use a Backtracking approach. This makes sense because:

1. We need to explore all possible moves from each game state
2. We recursively check if making a particular move leads to a winning position
3. We backtrack to try different moves if one doesn't guarantee a win
4. The solution uses memoization (via @cache) to avoid recalculating the same game states, which is a common optimization for backtracking problems

Intuition

The key insight for this problem is recognizing it as a game theory problem where we need to determine if the current player has a winning strategy. In such problems, a player wins if they can make a move that puts their opponent in a losing position.

Think of it this way: if it's your turn and you can make a move that leaves your opponent with no winning moves, then you win. Conversely, if every move you make gives your opponent at least one winning path, then you're in a losing position.

This naturally leads to a recursive approach:

1. From the current state, try all possible valid moves (flipping any "++" to "--")
2. For each move, check if the resulting state is a losing state for the opponent
3. If we find even one move that puts the opponent in a losing state, we can win
4. If all our moves lead to winning states for the opponent, we lose

The challenge is that checking all possibilities can lead to revisiting the same game states multiple times. For example, flipping positions 0-1 then 3-4 leads to the same state as flipping 3-4 then 0-1. This is where memoization becomes crucial - we cache the results for each unique game state.

The solution cleverly uses a bitmask to represent the game state, where each bit indicates whether that position has a '+' (1) or '-' (0). This allows us to:

• Efficiently check for consecutive "++" patterns using bit operations
• Create new states by flipping specific bits
• Use the bitmask as a unique identifier for memoization

The recursive function returns true if the current player can win from this state, and false otherwise. By starting with the initial state and working our way through all possible game branches, we can determine if the first player has a guaranteed winning strategy.

Solution Approach

The solution implements the game theory approach using backtracking with memoization. Let's walk through the implementation step by step:

1. State Representation with Bitmask

First, we convert the string into a bitmask where each bit represents a position in the string:

• '+' → bit value 1
• '-' → bit value 0

mask, n = 0, len(currentState)
for i, c in enumerate(currentState):
    if c == '+':
        mask |= 1 << i

For example, "++++" becomes 1111 in binary (mask = 15), and "--++" becomes 0011 in binary (mask = 3).

2. Recursive Function with Memoization

The core logic is in the dfs(mask) function, which determines if the current player can win from the given state:

@cache
def dfs(mask):
    for i in range(n - 1):
        # Check if positions i and i+1 both have '+'
        if (mask & (1 << i)) == 0 or (mask & (1 << (i + 1)) == 0):
            continue
      
        # Make the move: flip "++" to "--"
        if dfs(mask ^ (1 << i) ^ (1 << (i + 1))):
            continue
      
        # Found a move that puts opponent in losing position
        return True
  
    # No winning move found
    return False

3. The Algorithm Flow

• Finding Valid Moves: We iterate through positions 0 to n-2, checking if both position i and i+1 have '+' (bit value 1). This is done using bitwise AND operations.

• Making a Move: When we find a valid "++", we flip it to "--" by XORing the mask with (1 << i) and (1 << (i + 1)). This flips bits at positions i and i+1 from 1 to 0.

• Game Theory Logic:

  • If dfs(new_mask) returns True, it means the opponent can win from that state, so we continue searching for other moves
  • If dfs(new_mask) returns False, it means the opponent cannot win from that state, so we've found a winning move and return True
  • If we've tried all possible moves and none put the opponent in a losing position, we return False

4. Base Case

The base case is implicit: when no valid moves exist (no consecutive "++" found), the loop completes without returning True, so the function returns False, indicating the current player loses.

5. Memoization Optimization

The @cache decorator ensures that once we've computed the result for a particular mask, we don't recalculate it. This is crucial because the same game state can be reached through different sequences of moves, significantly reducing the time complexity from exponential to manageable bounds.

The final answer is simply dfs(mask) called with the initial state's bitmask representation.

Example Walkthrough

Let's walk through a small example with currentState = "++++" to illustrate how the solution works.

Step 1: Convert to Bitmask

• String: "++++"
• Bitmask: 1111 (binary) = 15 (decimal)
• Each '+' becomes a 1-bit

Step 2: First Player's Turn - dfs(1111)

The first player can make three possible moves:

1. Flip positions 0-1: "+++" → "--++" (mask: 1111 → 0011)
2. Flip positions 1-2: "+++" → "+--+" (mask: 1111 → 1001)
3. Flip positions 2-3: "+++" → "++--" (mask: 1111 → 1100)

Let's trace Move 1 (flip positions 0-1):

• New state: "--++" (mask = 0011)
• Call dfs(0011) to see if opponent can win

Step 3: Second Player's Turn - dfs(0011)

From "--++", the second player has only one valid move:

• Flip positions 2-3: "--++" → "----" (mask: 0011 → 0000)
• Call dfs(0000)

Step 4: First Player's Turn - dfs(0000)

From "----", there are no consecutive "++" to flip.

• No valid moves available
• Return False (first player loses from this state)

Step 5: Backtrack to Second Player

Since dfs(0000) returned False:

• The second player found a move that puts the first player in a losing position
• So dfs(0011) returns True (second player can win)

Step 6: Backtrack to First Player

Since dfs(0011) returned True:

• Move 1 leads to a state where the opponent can win
• We need to check the other moves (Move 2 and Move 3)

After checking all three moves:

• Move 1 (→ "--++") gives opponent a winning position
• Move 2 (→ "+--+") gives opponent a winning position
• Move 3 (→ "++--") gives opponent a winning position

Since all moves lead to states where the opponent can win, the first player cannot guarantee a win from "++++". The function returns False.

Key Observations:

• The algorithm explores all possible game branches
• Memoization prevents recalculating the same states (e.g., "----" might be reached through different move sequences)
• A player wins if they can find at least one move that puts the opponent in a losing position
• The bitmask representation allows efficient state manipulation using XOR operations

Solution Implementation

Time and Space Complexity

Time Complexity: O(2^n * n) where n is the length of the input string.

The algorithm uses dynamic programming with memoization through a bitmask. The mask can have at most 2^n different states (each bit position represents whether a '+' at that position has been flipped or not). For each unique state, we iterate through n-1 possible consecutive pairs to check if we can make a move. Due to memoization with @cache, each state is computed only once. Therefore, the overall time complexity is O(2^n * n).

Space Complexity: O(2^n) where n is the length of the input string.

The space complexity consists of:

• The recursion call stack depth, which in the worst case can go up to O(n) levels deep (when we flip pairs sequentially)
• The memoization cache storing up to 2^n different states, each state being an integer mask
• The dominant factor is the cache storage of O(2^n)

Therefore, the total space complexity is O(2^n).

Common Pitfalls

1. Incorrect Bit Manipulation When Checking Consecutive Positions

Pitfall: A common mistake is incorrectly checking if two consecutive positions both contain '+'. Developers might write:

# WRONG: This checks if EITHER position has a '+'
if (mask & (1 << i)) or (mask & (1 << (i + 1))):
    # Make move...

Why it's wrong: This condition is true when at least one position has '+', but we need BOTH positions to have '+' to make a valid move.

Correct approach:

# Check that BOTH positions have '+'
if (mask & (1 << i)) != 0 and (mask & (1 << (i + 1))) != 0:
    # Make move...

2. Forgetting to XOR Both Bit Positions When Making a Move

Pitfall: When flipping two consecutive '+' to '-', forgetting to flip both bits:

# WRONG: Only flips one position
new_mask = mask ^ (1 << i)

Why it's wrong: The move requires flipping TWO consecutive positions from '+' to '-', not just one.

Correct approach:

# Flip both positions i and i+1
new_mask = mask ^ (1 << i) ^ (1 << (i + 1))

3. Misunderstanding the Game Theory Logic

Pitfall: Incorrectly interpreting what the recursive call's return value means:

# WRONG: Returns True if opponent wins
if dfs(new_mask):
    return True

Why it's wrong: If dfs(new_mask) returns True, it means the opponent (who becomes the current player in that state) can win. This is BAD for us, so we should continue searching for other moves.

Correct approach:

# If opponent CANNOT win from new state, we found a winning move
if not dfs(new_mask):
    return True

4. Integer Overflow with Large Strings

Pitfall: For strings longer than 32 or 64 characters (depending on system), using bit manipulation can cause overflow issues:

# Potential overflow for large n
mask |= 1 << i  # If i >= 64, this might overflow

Solution: Add a length check or use alternative approaches for very long strings:

def canWin(self, currentState: str) -> bool:
    n = len(currentState)
  
    # For very long strings, use string-based memoization instead
    if n > 30:  # Adjust threshold based on system
        @cache
        def dfs(state: str) -> bool:
            for i in range(len(state) - 1):
                if state[i:i+2] == '++':
                    new_state = state[:i] + '--' + state[i+2:]
                    if not dfs(new_state):
                        return True
            return False
        return dfs(currentState)
  
    # Original bitmask approach for shorter strings
    # ... (rest of the code)

5. Not Using Memoization or Using It Incorrectly

Pitfall: Forgetting the @cache decorator or trying to manually implement memoization with mutable default arguments:

# WRONG: Mutable default argument persists across function calls
def dfs(mask, memo={}):  # Don't do this!
    if mask in memo:
        return memo[mask]
    # ...

Correct approach: Use @cache decorator or properly initialize memoization:

# Option 1: Using @cache (recommended)
@cache
def dfs(mask):
    # ...

# Option 2: Manual memoization
def canWin(self, currentState: str) -> bool:
    memo = {}  # Initialize inside the main function
  
    def dfs(mask):
        if mask in memo:
            return memo[mask]
        # ... compute result ...
        memo[mask] = result
        return result