#2038

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Remove Colored Pieces if Both Neighbors are the Same Color

Alice and Bob play a game removing colored pieces; Alice wins if she makes the last valid move.

Problem Statement

In a game, Alice and Bob take turns removing pieces from a string of colored pieces, where each piece is either 'A' or 'B'. The goal is to remove pieces that are surrounded by two matching neighbors. Alice starts first, and they alternate moves until no more valid moves are possible. The winner is the player who makes the last valid move. Your task is to return true if Alice wins, or false if Bob wins.

You are given a string colors representing the pieces, and each character in the string corresponds to a colored piece. The game proceeds by Alice and Bob removing pieces based on the rule mentioned above. The challenge is to determine who will win if both players play optimally. The string will only contain 'A' and 'B' characters, and its length is between 1 and 105.

Examples

Example 1

Input: colors = "AAABABB"

Output: true

AAABABB -> AABABB Alice moves first. She removes the second 'A' from the left since that is the only 'A' whose neighbors are both 'A'.

Now it's Bob's turn. Bob cannot make a move on his turn since there are no 'B's whose neighbors are both 'B'. Thus, Alice wins, so return true.

Example 2

Input: colors = "AA"

Output: false

Alice has her turn first. There are only two 'A's and both are on the edge of the line, so she cannot move on her turn. Thus, Bob wins, so return false.

Example 3

Input: colors = "ABBBBBBBAAA"

Output: false

ABBBBBBBAAA -> ABBBBBBBAA Alice moves first. Her only option is to remove the second to last 'A' from the right.

ABBBBBBBAA -> ABBBBBBAA Next is Bob's turn. He has many options for which 'B' piece to remove. He can pick any.

On Alice's second turn, she has no more pieces that she can remove. Thus, Bob wins, so return false.

Constraints

1 <= colors.length <= 105
colors consists of only the letters 'A' and 'B'

Solution Approach

Greedy Strategy

The problem can be solved using a greedy strategy. The key observation is that a player removes a piece only when it has the same color on both its neighbors. By following this approach, players can quickly evaluate which pieces are removable during their turn. The game boils down to counting how many valid moves each player can make and determining who has the last valid move.

Turn Simulation

Simulate the game by alternating between Alice and Bob's turns. At each step, check if there are valid pieces for a player to remove. Once a player cannot make a valid move, the game ends. This simulation can be optimized by scanning the string for removable pieces and updating the game state dynamically. This approach works well when the constraints are managed by efficient piece checking.

Optimal Move Calculation

To improve performance, count the possible valid moves initially before starting the simulation. By reducing the number of checks for each turn and eliminating unnecessary recalculations, the problem can be solved in linear time relative to the size of the string. This ensures the solution can handle the maximum constraints efficiently.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach used to check valid moves. In the worst case, it could be O(n) if each piece is checked individually. Space complexity is O(1) as we only need a constant amount of extra space for simulation purposes.

What Interviewers Usually Probe

Look for efficient ways to simulate the turn-based nature of the game.
The candidate should demonstrate an understanding of greedy algorithms and their application to this problem.
Consider how optimizations can reduce the time complexity for large input sizes.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the game mechanics can lead to incorrect simulations or missed moves.
Failing to optimize the search for valid moves can result in inefficient solutions for large inputs.
Not correctly handling the alternating nature of the game can lead to errors in turn simulation.

Follow-up variants

What if the number of valid moves changes dynamically due to prior removals?
How would the solution change if players could remove pieces that have neighbors of different colors?
What if players could remove any piece that is a neighbor to a matching color, not just surrounded pieces?

FAQ

What is the main strategy used to solve 'Remove Colored Pieces if Both Neighbors are the Same Color'?

The main strategy is to use a greedy approach, where players aim to remove pieces surrounded by two matching neighbors, alternating turns until no valid moves remain.

How do Alice and Bob play optimally in this game?

Alice and Bob play optimally by always removing a piece when possible and making moves that maximize their chances of winning while minimizing the opponent's opportunities.

What makes this problem a greedy algorithm problem?

This problem is greedy because players take the best possible move at each step by removing pieces that have matching neighbors, without considering future moves.

How do I simulate the game turns for Alice and Bob?

You simulate the game by alternating between Alice and Bob, checking for valid pieces to remove on each player's turn. The game ends when no valid pieces remain.

What are the key constraints of the 'Remove Colored Pieces if Both Neighbors are the Same Color' problem?

The string length is between 1 and 105, and the string only contains 'A' and 'B'. These constraints require efficient simulation and valid move checking for large inputs.

terminal

Solution

Solution 1: Counting

We count the number of times that the string `colors` contains three consecutive `'A'`s or three consecutive `'B'`s, denoted as $a$ and $b$, respectively.

class Solution:
    def winnerOfGame(self, colors: str) -> bool:
        a = b = 0
        for c, v in groupby(colors):
            m = len(list(v)) - 2
            if m > 0 and c == 'A':
                a += m
            elif m > 0 and c == 'B':
                b += m
        return a > b

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