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Flip Game II

Medium

Asked by:

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Topics:

RecursionDynamic Programming

You are playing the following Flip Game with your friend:

Given a string currentState that contains only '+' and '-' characters, you and your friend take turns to flip two consecutive "++" into "--". The game ends when no more moves are possible. You are the first person to move.

Return true if the starting player can guarantee a win; otherwise, return false.

Example 1:

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

Example 2:

Input: currentState = "+"
Output: false

Constraints:

1 <= currentState.length <= 60
currentState[i] is either '+' or '-'.

Try coding it in LeetCode

Flip Game II

Medium

Asked by:

8 views

Topics:

RecursionDynamic Programming

You are playing the following Flip Game with your friend:

Return true if the starting player can guarantee a win; otherwise, return false.

Example 1:

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

Example 2:

Input: currentState = "+"
Output: false

Constraints:

1 <= currentState.length <= 60
currentState[i] is either '+' or '-'.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum length of the input string 's'? Can I assume the string will only contain '+' and '-' characters?
If no valid move exists for the current player, should I return false immediately?
Is the input string guaranteed to be non-null and contain at least two characters?
If multiple sequences of moves lead to the first player winning, do I need to find a specific winning sequence, or can I return true as soon as I find any such sequence?
Can you provide a few example inputs and their corresponding expected outputs to further illustrate the problem?

Brute Force Solution

Approach

The brute force strategy explores every possible outcome of the game. It involves trying every single valid move a player can make and then recursively evaluating if that move leads to a win. This approach guarantees finding a winning strategy if one exists, but it can be very slow for more complex scenarios.

Here's how the algorithm would work step-by-step:

Consider all possible pairs of consecutive '+' signs in the input.
For each of these pairs, imagine flipping them to '--'.
After making this flip, check if the other player can win the game, assuming they play optimally.
If, after any of your possible flips, the other player *cannot* win, then you have found a winning move.
If you have tried all possible flips, and after each flip, the other player *can* win, then there is no winning move for you from the current state.

Code Implementation

def can_win(current_state):
    # Iterate through all possible moves
    for i in range(len(current_state) - 1):
        if current_state[i:i+2] == '++':
            # Flip the signs and generate next state
            next_state = current_state[:i] + '--' + current_state[i+2:]

            # Check if the opponent loses from the next state
            if not can_win(next_state):
                return True

    # If no winning move is found, return False
    return False

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores all possible moves recursively. In the worst case, nearly every position in the input string of length n could potentially be part of a "++" pair. For each such pair, a recursive call is made. This branching factor leads to an exponential number of recursive calls. Because each '+' can be flipped or not flipped, there are potentially 2^n branches that are explored. Therefore, the time complexity is O(2^n).

Space Complexity

O(N) – The dominant space complexity comes from the recursion. In the worst-case scenario, each call to the function 'canWin' flips a pair of '++' and then recursively calls itself with the modified string. The maximum depth of the recursion is bounded by the number of possible flips, which, in the worst case (e.g., '++++++++++++++++'), is related to the length of the input string. Therefore, the maximum depth of the recursion stack is proportional to N, where N is the length of the input string, resulting in O(N) space due to the call stack.

Optimal Solution

Approach

The core idea is to recognize this problem as a game where players take turns flipping consecutive '++' to '--'. We can determine if the first player can win by figuring out all the possible game states and whether those states are winning or losing for the first player.

Here's how the algorithm would work step-by-step:

First, find all possible moves the current player can make by identifying all locations where '++' exists.
For each possible move, imagine the current player makes that move. This creates a new game state.
Figure out if the new game state is a winning or losing state *for the next player*. We can reuse the same logic we're using for the current player to figure this out.
If *any* of the new game states are losing states for the next player, then the current state is a winning state for the current player. This is because the current player can make a move that forces the next player into a losing position.
If *all* of the new game states are winning states for the next player, then the current state is a losing state for the current player. No matter what the current player does, the next player will win.
To avoid recalculating the win/loss status of the same game states multiple times, store (memoize) the results. This speeds up the process significantly.
The game is over when there are no more '++' sequences. The player who cannot move loses.

Code Implementation

def canWin(currentState):
    game_states_cache = {}
    
    def canWinHelper(current_board_state):
        if current_board_state in game_states_cache:
            return game_states_cache[current_board_state]

        # Find all possible moves
        for i in range(len(current_board_state) - 1):
            if current_board_state[i:i+2] == "++":
                next_board_state = current_board_state[:i] + "--" + current_board_state[i+2:]

                # Check if the next player loses from this state
                if not canWinHelper(next_board_state):
                    game_states_cache[current_board_state] = True
                    return True

        # If no move leads to the next player losing, the current player loses.
        game_states_cache[current_board_state] = False
        return False

    return canWinHelper(currentState)

Big(O) Analysis

Time Complexity

O(n!) – The algorithm explores all possible game states using recursion with memoization. In the worst-case scenario, where memoization doesn't significantly reduce redundant calculations, the number of possible moves grows factorially with the input size 'n' because each '++' pair flipped creates a new potentially unique game state and the number of such pairs in a string of length n can grow linearly with n. Thus, in the worst case, the algorithm explores close to n! states. Therefore, the time complexity is approximately O(n!).

Space Complexity

O(N^2) – The space complexity is dominated by the memoization used to store the results of already computed game states. Each unique game state, represented as a string, can be a key in the memoization table (e.g., a hash map or dictionary). In the worst-case scenario, every possible board configuration could be a unique key. The number of possible game states is related to the different arrangements of '+' and '-' in the board string, where the length of the board string is N. Since each position can be either '+' or '-', the total possible combinations can approach O(2^N). However, the number of game states which matters are only those reachable by flipping "++" to "--". In the worst case, we might explore all possible substrings of length N, leading to O(N^2) possible unique board configurations. Therefore, the memoization table requires O(N^2) space.

Edge Cases

Case	How to Handle
Null or empty input string	Return false immediately, as no moves are possible.
Input string of length 1	Return false immediately, as no moves are possible.
Input string of length 2 with no '++'	Return false, as the starting string has no valid move.
Input string of length 2 with '++'	Return true, as the first player can immediately win.
Input string with many '++' sequences clustered together	Memoization is crucial to avoid redundant computations of the same substring state.
Input string with very long sequences of '+' characters	The recursive calls may become very deep, so ensure the memoization avoids exponential growth.
Input string with no '++' sequences	The first player cannot move and loses; return false.
Input string filled with only '+', which leads to extremely long search space	Memoization or dynamic programming strategies are vital for efficient processing to avoid timeouts.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum length of the input string 's'? Can I assume the string will only contain '+' and '-' characters?
If no valid move exists for the current player, should I return false immediately?
Is the input string guaranteed to be non-null and contain at least two characters?
If multiple sequences of moves lead to the first player winning, do I need to find a specific winning sequence, or can I return true as soon as I find any such sequence?
Can you provide a few example inputs and their corresponding expected outputs to further illustrate the problem?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider all possible pairs of consecutive '+' signs in the input.
For each of these pairs, imagine flipping them to '--'.
After making this flip, check if the other player can win the game, assuming they play optimally.
If, after any of your possible flips, the other player *cannot* win, then you have found a winning move.
If you have tried all possible flips, and after each flip, the other player *can* win, then there is no winning move for you from the current state.

Code Implementation

def can_win(current_state):
    # Iterate through all possible moves
    for i in range(len(current_state) - 1):
        if current_state[i:i+2] == '++':
            # Flip the signs and generate next state
            next_state = current_state[:i] + '--' + current_state[i+2:]

            # Check if the opponent loses from the next state
            if not can_win(next_state):
                return True

    # If no winning move is found, return False
    return False

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, find all possible moves the current player can make by identifying all locations where '++' exists.
For each possible move, imagine the current player makes that move. This creates a new game state.
Figure out if the new game state is a winning or losing state *for the next player*. We can reuse the same logic we're using for the current player to figure this out.
If *any* of the new game states are losing states for the next player, then the current state is a winning state for the current player. This is because the current player can make a move that forces the next player into a losing position.
If *all* of the new game states are winning states for the next player, then the current state is a losing state for the current player. No matter what the current player does, the next player will win.
To avoid recalculating the win/loss status of the same game states multiple times, store (memoize) the results. This speeds up the process significantly.
The game is over when there are no more '++' sequences. The player who cannot move loses.

Code Implementation

def canWin(currentState):
    game_states_cache = {}
    
    def canWinHelper(current_board_state):
        if current_board_state in game_states_cache:
            return game_states_cache[current_board_state]

        # Find all possible moves
        for i in range(len(current_board_state) - 1):
            if current_board_state[i:i+2] == "++":
                next_board_state = current_board_state[:i] + "--" + current_board_state[i+2:]

                # Check if the next player loses from this state
                if not canWinHelper(next_board_state):
                    game_states_cache[current_board_state] = True
                    return True

        # If no move leads to the next player losing, the current player loses.
        game_states_cache[current_board_state] = False
        return False

    return canWinHelper(currentState)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input string	Return false immediately, as no moves are possible.
Input string of length 1	Return false immediately, as no moves are possible.
Input string of length 2 with no '++'	Return false, as the starting string has no valid move.
Input string of length 2 with '++'	Return true, as the first player can immediately win.
Input string with many '++' sequences clustered together	Memoization is crucial to avoid redundant computations of the same substring state.
Input string with very long sequences of '+' characters	The recursive calls may become very deep, so ensure the memoization avoids exponential growth.
Input string with no '++' sequences	The first player cannot move and loses; return false.
Input string filled with only '+', which leads to extremely long search space	Memoization or dynamic programming strategies are vital for efficient processing to avoid timeouts.