Problem Description

In the Flip Game, you are given a string currentState which consists of only '+' and '-'. A move in this game consists of turning two adjacent '+' into '--'. The game ends when a player can't make any more moves, implying the last person to make a valid move wins.

Your objective is to determine if the starting player can always win, regardless of the moves the opposing player makes. If the starting player can guarantee a win, your function should return true. If there's no strategy that guarantees a win for the starting player, return false.

It's a game of strategic choice, where each decision can affect the outcome. The challenge is to find whether there exists a sequence of moves that, if played optimally, ensures victory for the first player.

Flowchart Walkthrough

Let's analyze the problem with the help of the Flowchart. Here's a detailed examination based on our flowchart:

Is it a graph?

• No: Flip Game II does not involve graph-like structures with nodes and edges typical to graph theory.

Need to solve for kth smallest/largest?

• No: The problem is not about finding a solution based on ranking or order such as kth smallest or largest.

Involves Linked Lists?

• No: The game primarily involves strings manipulations, not linked lists.

Does the problem have small constraints?

• Yes: Despite possibly having many configurations, each game sequence due only works from a finite, manageable list of string manipulations, which can be considered having small enough constraints to attempt all possibilities.

Brute force / Backtracking?

• Yes: The need to explore every possible flip and its consequences recursively points towards the use of backtracking. Each choice affects subsequent choices, making it necessary to backtrack and try different paths to find a winning strategy.

Conclusion: The flowchart and the nature of the problem suggest using a backtracking approach to solve the Flip Game II. Every alteration to the string (game state) potentially alters the subsequent plays, thereby necessitating a backtrack to explore all different game paths to assure if a winning strategy exists.

Intuition

The solution to this problem lies in using backtracking to consider all possible moves of the game. This is done by using Depth First Search (DFS) to explore every potential game state that can result from a series of valid moves.

The backtracking approach iterates over the string, and for each pair of consecutive '+' symbols, flips them to '--' and then recursively checks if the other player would lose from the new game state. The key observation is that if there is at least one move after which the other player is guaranteed to lose (they cannot make any move that guarantees their win), then the current player is guaranteed to win.

The code uses a bitmask to represent the game state, where a '1' bit represents a '+', and '0' represents a '-'. A mask is created by setting bits for each '+' in the input currentState. The dfs function then checks each position in the string. If the current and next positions are '++', it flips them and recursively checks if this new state would lead to a losing state when it's the opponent's turn. If the recursive call returns false, meaning the opponent can't win from that state, then the current player can win by making this move.

The use of memoization (@cache) avoids re-computation of the states that have already been computed, significantly optimizing the solution. Without memoization, the solution would be too slow, as it would explore the same states multiple times.

Through this approach, we can determine whether the starting player can guarantee a win by meticulously searching through all possible game states and making the optimal move at each step.

Learn more about Memoization, Math, Dynamic Programming and Backtracking patterns.

Solution Approach

The initial approach to solving the problem involves a recursive Depth First Search (DFS) strategy to explore each possible state of the game after a move has been made, combined with bitmasking to efficiently represent and manipulate the game states, and memoization to avoid recalculating results for previously explored game states.

Here's an in-depth walk-through of the code provided:

1. Bitmask Representation: A mask is created where each bit represents a position in the currentState string. A '1' in the bitmask corresponds to a '+' in the string, and a '0' corresponds to a '-'.

2. Recursive DFS: The core of the solution is the dfs function, which tries to simulate each possible move recursively.

   • For each call to dfs, we iterate through all positions in the string (except the last one since we are checking pairs of characters).
   • If the current bit and the next bit are set to '1' (indicating "++" in the original string), we flip these two bits to '0' (making them "--") and call dfs on the new mask.
   • The flip is executed using the XOR operation mask ^ (1 << i) ^ (1 << (i + 1)) which flips the current bit and the bit next to it.
   • If the recursive dfs call returns false, it means that the opponent can't win from that modified game state, so the current player will indeed win if they make this move.

3. Memoization: The @cache decorator is Python’s built-in way to employ memoization. With the help of this memoization, any call to the dfs function with the same mask is computed only once, and the result is stored for subsequent calls with the same argument. This dramatically reduces the number of computations required and is essential for making this recursive solution feasible.

4. Winning Condition: Once all possible moves are explored by recursive calls, if the function finds a move that guarantees a win (as the opponent loses), it returns true. If no such move exists (i.e., every move results in a winning position for the opponent), the dfs returns false.

5. Result: The canWin function initializes the setup (bitmask, string length) and makes the initial call to dfs(mask). It returns the result of this call. If dfs returns true, it means the starting player can guarantee a victory with the right strategy.

In summary, this solution uses DFS to simulate the game, bitmasking for efficient state representation, and memoization to optimize the search process. It relies on the principle that if a player can force at least one scenario where the opponent cannot win after their next move, then the current player can secure a win.

Example Walkthrough

Let’s walk through a small example to illustrate the solution approach described in the content provided. Assume we are given the currentState string as "++--+++".

1. Bitmask Representation: We begin by representing this currentState as a bitmask:

   • ++--+++ corresponds to the bitmask 1100111. Each '1' represents a '+' and '0' represents a '-'.

2. Initial Call: Call the dfs function on this initial mask 1100111.

3. Recursive DFS: Inside the dfs function, we explore all possible moves:

   • We check pairs of bits from left to right.
   • On encountering 11 (which means "++"), we flip them to 00 (creating "--") and recursively call dfs.

4. Exploration:

   • First Recursion: Let's flip the first two pluses. Our mask becomes 0000111.
   • dfs(0000111) is called, and then it will check for moves within this new state.
   • Second Recursion: In the new state, we flip the last two pluses. Our mask becomes 0000001.
   • dfs(0000001) is called but no moves can be made within this state, so this returns false, which implies that the opponent cannot win from this state.

5. Backtracking:

   • Since dfs(0000001) returned false, that means dfs(0000111) should return true, indicating that there is a move that ensures the parent state win.
   • The process continues by backtracking and exploring other paths to check if there is any sequence of moves that would lead to the starting player's loss.
   • However, because we can guarantee at least one winning move from our initial state, in this case, flipping the first two pluses, we already know the starting player can secure a win.

6. Result: After exploring all possible moves and finding that there is at least one sequence of moves leading the opposing player to a state where they cannot move, we determine that the starting player can indeed guarantee a win. Thus, the dfs(mask) will ultimately return true for the initial mask 1100111.

In this case, the function canWin with the input "++--+++" would return true, signifying that the starting player has a winning strategy.

Solution Implementation

Time and Space Complexity

The given Python code uses a depth-first search algorithm with memoization to resolve a variant of the Nim game, where we check if we can remove two adjacent '+' symbols in a string representing a game state.

Time Complexity

The time complexity of the DFS algorithm is generally O(2^N), where N is the length of the input string currentState. This represents the number of possible states that the recursive function dfs might explore, as each position in the mask could be a '+' or a '-' after a move, effectively leading to a binary tree of depth N.

However, memoization is used, courtesy of the @cache decorator, which stores the result of each unique state of the mask. This means that each distinct game state would only be computed once. There are at most 2^N different states for the mask (since each of the N positions in the mask can either be 0 or 1).

As a result, the total unique calls to the dfs function would not exceed 2^N. Combined with the fact that for each call to dfs we iterate over the N positions for possible moves, the total time complexity is O(N * 2^N).

Space Complexity

The space complexity includes the storage for the recursive call stack and the memoization cache. In the worst case, the call stack could grow up to N as the depth of the recursion could be N in the case of a sequence of '+' signs. Therefore, the recursive call stack could contribute O(N).

The memoization cache could hold up to 2^N entries, each requiring a constant amount of space. Therefore, the space used by memoization is O(2^N).

Combining the call stack space and memoization cache, the total space complexity is O(N + 2^N) which is dominated by O(2^N) for large N.

Hence, the overall space complexity of the code is O(2^N).

Note: If we consider the length of the input string currentState constant or insignificant compared to the size of the state space, some may argue that the complexity can also be described as O(1) for time or space, as the number of operations or space doesn't increase with input size but rather with the number of possible states derived from the input.

Learn more about how to find time and space complexity quickly using problem constraints.