Problem Description

You have a line of n colored pieces, where each piece is colored either 'A' or 'B'. The colors are represented by a string colors of length n, where colors[i] indicates the color of the i-th piece.

Alice and Bob play a game taking turns removing pieces from this line, with Alice going first.

Game Rules:

• Alice's moves: She can only remove a piece colored 'A' if both of its neighbors are also colored 'A'. She cannot remove 'B' pieces.
• Bob's moves: He can only remove a piece colored 'B' if both of its neighbors are also colored 'B'. He cannot remove 'A' pieces.
• Edge restriction: Neither player can remove pieces from the edges (first or last position) of the line.
• Winning condition: The first player who cannot make a valid move on their turn loses, and the other player wins.

Both players play optimally (making the best possible moves). Your task is to determine who wins the game - return true if Alice wins, or false if Bob wins.

Example interpretation: If we have "AAABABB", Alice can remove the middle 'A' from the first group (since it has 'A' neighbors on both sides), but Bob cannot remove any 'B' initially since no 'B' has two 'B' neighbors. The game continues with players making optimal moves until one cannot move.

Intuition

The key insight is recognizing that once a player removes a piece, they don't create new opportunities for their opponent - they only potentially create more opportunities for themselves.

When Alice removes an 'A' from a sequence like "AAAA", it becomes "A_AA" (where _ represents the removed piece). This doesn't help Bob at all since Bob needs consecutive 'B's. However, it might help Alice in future turns if the gap gets filled or connected somehow.

More importantly, we need to count how many moves each player can potentially make. For a consecutive sequence of the same color, we can determine the number of removable pieces:

• In a sequence of 3 same-colored pieces: 1 piece can be removed (the middle one)
• In a sequence of 4 same-colored pieces: 2 pieces can be removed
• In a sequence of k same-colored pieces: k - 2 pieces can be removed (since edge pieces cannot be removed)

The crucial realization is that the game is essentially a race - whoever runs out of moves first loses. Since the players can only remove their own color and removing pieces doesn't create opportunities for the opponent, we can simply count the total number of possible moves for each player at the start of the game.

For each group of consecutive 'A's of length k where k ≥ 3, Alice can make k - 2 moves from that group. Similarly, for each group of consecutive 'B's of length k where k ≥ 3, Bob can make k - 2 moves.

Since Alice goes first and both play optimally, Alice wins if and only if she has more total available moves than Bob. If Alice has a total moves and Bob has b total moves, Alice wins when a > b.

Learn more about Greedy and Math patterns.

Solution Approach

The solution uses a grouping technique to identify consecutive sequences of the same color and count the available moves for each player.

Implementation Steps:

1. Group consecutive colors: We use groupby() function to group consecutive identical characters in the string. This gives us groups like ['A', 'A', 'A'] or ['B', 'B'].

2. Count available moves for each group: For each group:

   • Convert the group to a list and get its length
   • Calculate the number of removable pieces as length - 2
   • Only count groups where length - 2 > 0 (groups of 3 or more pieces)

3. Accumulate moves by player:

   • If the group color is 'A' and has 3+ pieces, add length - 2 to Alice's move count a
   • If the group color is 'B' and has 3+ pieces, add length - 2 to Bob's move count b

4. Determine the winner: Compare the total moves:

   • Return true if a > b (Alice has more moves and wins)
   • Return false otherwise (Bob has equal or more moves and wins)

Example walkthrough: For colors = "AAABABB":

• Groups: ['AAA'], ['B'], ['A'], ['BB']
• 'AAA': length 3, Alice gets 3-2 = 1 move
• 'B': length 1, no moves (less than 3)
• 'A': length 1, no moves (less than 3)
• 'BB': length 2, no moves (less than 3)
• Result: a = 1, b = 0, so a > b returns true (Alice wins)

The algorithm runs in O(n) time where n is the length of the string, as we iterate through the string once to form groups and count moves. The space complexity is O(1) as we only store the move counts.

Example Walkthrough

Let's walk through the solution with colors = "AABBBBA":

Step 1: Identify consecutive groups

• Group 1: "AA" (2 A's)
• Group 2: "BBBB" (4 B's)
• Group 3: "A" (1 A)

Step 2: Calculate available moves for each group

• Group 1 ("AA"): Length = 2, moves = 2 - 2 = 0 (too short, need at least 3)
• Group 2 ("BBBB"): Length = 4, moves = 4 - 2 = 2 (Bob can remove 2 pieces)
• Group 3 ("A"): Length = 1, moves = 1 - 2 = -1 (too short, no moves)

Step 3: Count total moves per player

• Alice's total moves (from A groups): 0
• Bob's total moves (from B groups): 2

Step 4: Determine winner

• Compare: Alice has 0 moves, Bob has 2 moves
• Since 0 is not greater than 2, return false (Bob wins)

Why Bob wins: Bob can make 2 moves from the "BBBB" group. He could remove the piece at index 3 (turning "BBBB" into "BB_B"), and later remove the piece at index 2 (if positions allow). Alice cannot make any moves since she has no group of 3 or more consecutive A's. Since Alice goes first but has no valid moves, she loses immediately.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the string colors. The groupby function iterates through the entire string once to create groups of consecutive identical characters. For each group, we convert the iterator v to a list using list(v), which takes time proportional to the size of that group. Since the sum of all group sizes equals n, the total time across all groups is O(n).

The space complexity is O(n) rather than O(1). While variables a and b use constant space, the critical issue is the list(v) operation inside the loop. When we convert the iterator v to a list, we create a list that can be as large as n in the worst case (when all characters in the string are the same). Therefore, the space complexity is O(n), not O(1) as stated in the reference answer.

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

Common Pitfalls

1. Consuming the Iterator Multiple Times

A critical pitfall in the provided code is calling len(list(group)) on the groupby iterator. The group object from groupby() is an iterator that can only be consumed once. After converting it to a list to get the length, the iterator is exhausted and cannot be reused.

Why this happens: groupby() returns an iterator of (key, group) pairs where group is also an iterator. Once you consume it (like converting to a list), it's empty.

Solution: Store the list conversion result in a variable:

for character, group in groupby(colors):
    group_list = list(group)  # Convert once and store
    group_length = len(group_list)
    possible_moves = group_length - 2
    # ... rest of the code

2. Misunderstanding the Game Rules

A common misconception is thinking that once a player removes a piece, the neighboring pieces immediately become new neighbors. This could lead to incorrect move counting by trying to dynamically recalculate available moves after each removal.

Why this is wrong: The problem asks for the total number of moves each player can make, not a simulation of the actual game. The key insight is that removing a piece from a group of size n still leaves n-3 pieces that can be removed (if any remain).

Solution: Count all possible moves at once for each group without simulating the actual game:

# Correct: Count all moves for a group at once
possible_moves = max(0, group_length - 2)

# Incorrect: Trying to simulate piece-by-piece removal
# This overcomplicates and likely miscounts

3. Off-by-One Errors in Move Calculation

It's easy to make mistakes when calculating the number of removable pieces. Some might think it's length - 1 or just length for groups of 3 or more.

Why length - 2 is correct: In a group of n identical characters, the first and last cannot be removed (edge restriction). So only the middle n - 2 pieces are removable.

Solution: Always remember the formula: removable_pieces = max(0, length - 2)

4. Not Handling Edge Cases Properly

Forgetting to check if possible_moves > 0 before adding to the counters could lead to negative move counts for small groups.

Example problem case:

• Group "AA" has length 2, so 2 - 2 = 0 moves (correct)
• Group "A" has length 1, so 1 - 2 = -1 moves (incorrect if added)

Solution: Always validate before adding:

if possible_moves > 0:
    if character == 'A':
        alice_moves += possible_moves
    # ...