Problem Description

In the given problem, we are playing a game called Flip Game with a friend. The game is played with a string currentState that can consist of only two characters: '+' and '-'. In each turn, a player must flip exactly two consecutive '+' characters into '-' characters. This is the only valid move in the game. The game proceeds with each player taking turns until no more valid moves can be made, at which point the last player to have made a valid move wins.

Our objective is to find all possible states of the currentState after one valid move has been made by the current player. We must provide a list of these possible states, or an empty list if no valid moves are available. The resulting states can be returned in any order.

Intuition

The solution to this problem is straightforward and involves iterating through the string to look for pairs of consecutive '+' characters because that's the only condition where a flip is possible. Here's the logical approach:

1. Convert the string currentState into a list of characters, so we can easily modify individual characters during our iteration.

2. Initialize an empty list ans to store the possible next states.

3. Loop through the list starting from index 0 and going to the second to last index (since we are always looking for pairs, we don't need to check the last character by itself).

4. At each iteration, check if the current character and the one following it are both '+'.

   • If they are, we flip them to '-' by setting them to '-'.
   • Then, we take the modified list, convert it back into a string, and add this state to our list of possible next moves ans.
   • After this operation, reverting the characters back to '+' is important because we are exploring all possible moves from the original state, one at a time.

5. Continue this process until we have checked all possible pairs of '+'.

6. Return the ans list containing all the possible next states. If no pairs of '+' were found, the ans list would be empty.

Solution Approach

The solution takes advantage of the linear nature of the currentState to perform a single-pass scan. Here is how the algorithm unfolds:

1. Data Structure:

   • The function converts the currentState string into a list of characters, referred to as s in the code, to enable easy modification of individual characters.
   • An empty list, ans, is created for storing the possible states after making valid moves.

2. Algorithm:

   • The function iterates through the list s with a loop, starting from the first character and stops at the second-last character using enumerate(s[:-1]). This range is used because the last character cannot start a pair of two '+'.
   • At each iteration, the function checks if the current character and the one immediately following it form a pair of consecutive '+' using if c == "+" and s[i + 1] == "+".
   • When such a pair is found:
     • Both characters are flipped from '+' to '-' by setting s[i] and s[i + 1] to '-'.
     • The updated list s is then joined to form a string representing the new state and is added to the ans list.
     • Immediately after recording the new state, the characters at index i and i + 1 are reset back to '+' to restore the original state before moving to the next iteration. This is crucial as it allows us to explore all possible moves independently.

3. Return Value:

   • After the end of the loop, the list ans will contain all the unique states that result from performing valid moves on the currentState.
   • The function returns the ans list. If no valid moves exist, ans will be returned as an empty list.

The key pattern used here is iteration over the list, combined with simple character manipulation to explore different move outcomes. This solution is efficient as it runs in O(n) time, where n is the length of the currentState, and operates with constant space complexity beyond the input and output lists.

Example Walkthrough

Let's use a small example to illustrate the solution approach.

Suppose the currentState string is "+++-++". Our task is to find all possible states after one valid move. A valid move entails flipping two consecutive '+' characters into '-'.

Following the solution approach:

1. Convert the string into a list of characters: s = ['+', '+', '+', '-', '+', '+']
2. Initialize an empty list to store possible next states: ans = []
3. Iterate over the list (ignoring the last character as a start since there can't be a pair):
   • Check s[0] and s[1]: Both are '+', so flip them to get ['-', '-', '+', '-', '+', '+'], convert back to string "--+-++", and add to ans.
   • Revert s[0] and s[1] back to '+', so s is now ['+', '+', '+', '-', '+', '+'] again.
   • Move to the next pair.
   • Check s[1] and s[2]: Both are '+', so flip them to get ['+', '-', '-', '-', '+', '+'], convert back to string "+-- -++", and add to ans.
   • Revert s[1] and s[2] back to '+'.
   • Continue in this fashion.
   • The next pair s[2] and s[3] are not both '+', so no flip.
   • Check s[3] and s[4]: Not both '+', so no flip.
   • Finally, check s[4] and s[5]: Both are '+', so flip to get ['+', '+', '+', '-', '-', '-'], convert back to string "+++- --", and add to ans.
4. We're left with ans containing the strings: "--+-++", "+-- -++", and "+++- --".
5. Return ans, which is ["--+-++", "+-- -++", "+++- --"].

The players continue the game with each one using the states from ans until no more moves can be played. The last move before the game state has no more consecutive '+' is the winning move.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n), where n is the length of the input string currentState. This is due to the fact that we traverse the string once, with each iteration checking a pair of adjacent characters and modifying the string if a match is found. The replacement operation for a single pair of characters is O(1), therefore the overall time taken is proportional to the length of the string.

The space complexity of the code is also O(n). This is because:

1. We convert the input string into a list, which takes O(n) space.
2. A new string is created in each iteration when a valid move is found (the ''.join(s) operation), and in the worst case, this could happen for every pair of adjacent characters. However, these strings are not all stored in memory at the same time; they are added to the list ans and then the original list s is restored to its previous state. Since the longest string appended to ans has the same length as the input string, ans will also take up O(n) space.
3. The auxiliary space for the loop and variables is O(1) (constant).

It's important to note that, while we do create multiple strings in the process, they are not all in memory at the same time. Each generated result is added to ans then the original list s is modified back. Nevertheless, in the worst case, where you can flip every pair, you'd have n/2 results stored in ans which constitute O(n^2) space if you consider all potential flips together. But considering space complexity in terms of additional space required by the program, we disregard the space taken by the input and output. Hence, we typically consider the space complexity of the additional space required, which remains O(n).

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