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

The optimal solution has a time complexity of O(n).

What data structures are used to solve Remove Colored Pieces if Both Neighbors are the Same Color?

This problem uses Math, String, Greedy, Game Theory. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Remove Colored Pieces if Both Neighbors are the Same Color?

Insight 1: The number of moves available to each player is independent of the order in which they make their moves. Therefore, we only need to count the total possible moves for Alice and Bob. Insight 2: A player can make a move if and only if the piece is surrounded by two pieces of the same color as itself. Identifying such sequences is the key to determining the winner.

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Remove Colored Pieces if Both Neighbors are the Same Color - LeetCode 2038 Solution

Q: Which companies ask LeetCode 2038. Remove Colored Pieces if Both Neighbors are the Same Color in interviews?

This problem is frequently asked at J.P. Morgan, MathWorks, Roblox, Unity, Yelp.

Remove Colored Pieces if Both Neighbors are the Same Color - Complete Solution Guide

Remove Colored Pieces if Both Neighbors are the Same Color is LeetCode problem 2038, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are n pieces arranged in a line, and each piece is colored either by 'A' or by 'B' . You are given a string colors of length n where colors[i] is the color of the i th piece. Alice and Bob are playing a game where they take alternating turns removing pieces from the line. In this game, Alice moves first . Alice is only allowed to remove a piece colored 'A' if both its neighbors are also colored 'A' . She is not allowed to remove pieces that are colored 'B' . Bob is only allowed to remove a

Detailed Explanation

The problem presents a game played between Alice and Bob using colored pieces ('A' or 'B') arranged in a line. Alice can only remove 'A' pieces if both its neighbors are also 'A's. Bob can only remove 'B' pieces if both its neighbors are also 'B's. They take turns removing pieces, and the first player unable to make a move loses. The goal is to determine who wins assuming both players play optimally, given the initial arrangement of colored pieces.

Solution Approach

The solution calculates the number of possible moves for Alice and Bob independently. It iterates through the `colors` string, checking each piece (excluding the first and last) to see if it is surrounded by two pieces of the same color. If so, it increments the move count for the corresponding player ('A' for Alice, 'B' for Bob). Finally, it compares the number of moves available to each player and returns `true` if Alice has more moves than Bob (meaning Alice wins), and `false` otherwise (meaning Bob wins or they draw and Bob wins).

Step-by-Step Algorithm

Step 1: Initialize two counters, `alice_moves` and `bob_moves`, to 0. These counters will store the number of valid moves for each player.
Step 2: Iterate through the `colors` string from the second element (index 1) to the second-to-last element (index n-2), where n is the length of the string. This is because the first and last elements cannot be removed.
Step 3: For each element in the string, check if the element and its two neighbors are of the same color. Specifically, check if `colors[i-1] == colors[i] == colors[i+1]`.
Step 4: If the condition in Step 3 is true, determine which player can remove the piece. If `colors[i]` is 'A', increment `alice_moves`. If `colors[i]` is 'B', increment `bob_moves`.
Step 5: After iterating through the string, compare `alice_moves` and `bob_moves`. If `alice_moves > bob_moves`, return `true` (Alice wins). Otherwise, return `false` (Bob wins or draw and Bob wins).

Key Insights

Insight 1: The number of moves available to each player is independent of the order in which they make their moves. Therefore, we only need to count the total possible moves for Alice and Bob.
Insight 2: A player can make a move if and only if the piece is surrounded by two pieces of the same color as itself. Identifying such sequences is the key to determining the winner.
Insight 3: The game is deterministic with perfect information, so the player with more available moves will win, unless they have the same number of moves, in which case the second player wins

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Math, String, Greedy, Game Theory.

Companies

Asked at: J.P. Morgan, MathWorks, Roblox, Unity, Yelp.

Remove Colored Pieces if Both Neighbors are the Same Color | LeetCode 2038

Medium

Math String Greedy Game Theory

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean winnerOfGame(String colors) {
        int alice_moves = 0;
        int bob_moves = 0;
        
        for (int i = 1; i < colors.length() - 1; i++) {
            if (colors.charAt(i - 1) == colors.charAt(i) && colors.charAt(i) == colors.charAt(i + 1)) {
                if (colors.charAt(i) == 'A') {
                    alice_moves++;
                } else {
                    bob_moves++;
                }
            }
        }
        
        return alice_moves > bob_moves;
    }
}

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Problem Info

Description

Problem 2038: Remove Colored Pieces if Both Neighbors are the Same Color - Detailed explanation and solution approaches.

Topics

Math String Greedy Game Theory

Companies

J.P. MorganMathWorksRobloxUnityYelp

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:2038

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize two counters, `alice_moves` and `bob_moves`, to 0. These counters will store the number of valid moves for each player.
2
Step 2: Iterate through the `colors` string from the second element (index 1) to the second-to-last element (index n-2), where n is the length of the string. This is because the first and last elements cannot be removed.
3
Step 3: For each element in the string, check if the element and its two neighbors are of the same color. Specifically, check if `colors[i-1] == colors[i] == colors[i+1]`.
4
Step 4: If the condition in Step 3 is true, determine which player can remove the piece. If `colors[i]` is 'A', increment `alice_moves`. If `colors[i]` is 'B', increment `bob_moves`.
5
Step 5: After iterating through the string, compare `alice_moves` and `bob_moves`. If `alice_moves > bob_moves`, return `true` (Alice wins). Otherwise, return `false` (Bob wins or draw and Bob wins).

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is directly proportional to the number of characters in the input string as we perform a single pass. The space complexity is constant as the algorithm uses only a fixed number of variables, regardless of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using regular expressions

An alternative approach could use regular expressions to find sequences of three or more consecutive 'A's and 'B's. The number of moves for each player would then be one less than the length of each sequence.

Time: O(n) in the average case, but potentially higher depending on the regex engine.Space: O(1) or O(n) depending on how regex is implemented.

Trade-offs: While regex can be more concise, it might be less efficient and harder to understand for beginners. The current solution is more straightforward and easier to debug.

✨

Key Insights

1Insight 1: The number of moves available to each player is independent of the order in which they make their moves. Therefore, we only need to count the total possible moves for Alice and Bob.
2Insight 2: A player can make a move if and only if the piece is surrounded by two pieces of the same color as itself. Identifying such sequences is the key to determining the winner.
3Insight 3: The game is deterministic with perfect information, so the player with more available moves will win, unless they have the same number of moves, in which case the second player wins

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `colors` string has length less than 3. In this case, neither player can make a move, so Alice loses, and the solution correctly returns `false` because the loop won't execute.
Edge case 2: `colors` string has only 'A's or only 'B's. The algorithm correctly identifies the number of moves available and determines the winner based on whether alice_moves > bob_moves
Edge case 3: `colors` string alternates between 'A' and 'B'. In this case, neither player can make a move because no piece has two neighbors of the same color, so both alice_moves and bob_moves will be 0 and the solution correctly returns false since 0 is not greater than 0

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not excluding the first and last pieces when counting moves. The problem states that players cannot remove pieces from the edge of the line.

✗Common mistake 2: Assuming the order of moves matters. The problem statement implies that the total number of valid moves for each player determines the winner, not the specific sequence of moves.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation in a constrained environment. The 'A' and 'B' pieces could represent different types of resources, and the removal rule could represent constraints on how these resources can be used. Determining the winner helps optimize allocation strategies.

🏦Real-world scenario 2: Game development. This type of game logic can be used as a component in more complex board games or puzzle games, where players strategically remove elements to gain an advantage.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 29, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Remove Colored Pieces if Both Neighbors are the Same Color"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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