What is the time complexity of Remove Colored Pieces if Both Neighbors are the Same Color?

The optimal solution for Remove Colored Pieces if Both Neighbors are the Same Color runs in O(n) time and uses O(1) space. The complexity is O(n) because we only need a single pass through the string to count the valid moves for both players.

What is the brute force solution for Remove Colored Pieces if Both Neighbors are the Same Color?

The brute force approach for Remove Colored Pieces if Both Neighbors are the Same Color is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we simulate the game by repeatedly checking for valid moves for both players until no moves are left. This is straightforward but inefficient as it involves checking the entire string multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove Colored Pieces if Both Neighbors are the Same Color?

Remove Colored Pieces if Both Neighbors are the Same Color uses the following concepts: Math, String, Greedy, Game Theory. The recommended patterns to study are: Greedy, String Manipulation.

What are the interview tips for Remove Colored Pieces if Both Neighbors are the Same Color?

Always consider edge cases and how they might affect the outcome. Explain your thought process clearly; interviewers appreciate understanding your reasoning. Practice counting patterns in strings, as many problems involve similar logic.

What are common mistakes when solving Remove Colored Pieces if Both Neighbors are the Same Color?

Failing to account for edge cases where there are not enough pieces to make a move. Not recognizing that the game can be solved by counting valid moves instead of simulating each turn.

#2038

Remove Colored Pieces if Both Neighbors are the Same Color

Medium

MathStringGreedyGame TheoryGreedyString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

Instead of simulating the game, we can count the maximum number of valid moves each player can make based on the initial configuration. This allows us to determine the winner without simulating each turn.

⚙️

Algorithm

3 steps

1Step 1: Count the number of valid moves Alice can make by iterating through the string and checking for 'AAA' patterns.
2Step 2: Count the number of valid moves Bob can make by checking for 'BBB' patterns.
3Step 3: Compare the counts: if Alice's moves are greater than Bob's, Alice wins; otherwise, Bob wins.

solution.py4 lines

1def winner(colors):
2    alice_moves = sum(1 for i in range(1, len(colors) - 1) if colors[i] == 'A' and colors[i - 1] == 'A' and colors[i + 1] == 'A')
3    bob_moves = sum(1 for i in range(1, len(colors) - 1) if colors[i] == 'B' and colors[i - 1] == 'B' and colors[i + 1] == 'B')
4    return alice_moves > bob_moves

ℹ

Complexity note: The complexity is O(n) because we only need a single pass through the string to count the valid moves for both players.

1The game is determined by the number of valid moves each player can make.
2Players cannot remove pieces at the edges, which limits their options.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.