What is the time complexity of Stone Game IX?

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

What data structures are used to solve Stone Game IX?

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

What are the key insights for solving Stone Game IX?

Insight 1: The core of the solution lies in understanding how the remainders of stone values when divided by 3 affect the game's outcome. Only the counts of stones with remainders 0, 1, and 2 matter. Insight 2: The number of stones with a remainder of 0 significantly impacts the game dynamic. An even count allows direct competition between stones with remainders 1 and 2, while an odd count introduces a slight advantage/disadvantage depending on the imbalance between counts of remainders 1 and 2.

Which companies ask LeetCode 2029. Stone Game IX in interviews?

This problem is frequently asked at Samsung.

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Stone Game IX - LeetCode 2029 Solution

Stone Game IX - Complete Solution Guide

Stone Game IX is LeetCode problem 2029, 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

Alice and Bob continue their games with stones. There is a row of n stones, and each stone has an associated value. You are given an integer array stones , where stones[i] is the value of the i th stone. Alice and Bob take turns, with Alice starting first. On each turn, the player may remove any stone from stones . The player who removes a stone loses if the sum of the values of all removed stones is divisible by 3 . Bob will win automatically if there are no remaining stones (even if it is Alic

Detailed Explanation

The problem presents a game between Alice and Bob involving a row of stones, each with a value. They take turns removing stones, and the player who makes the cumulative sum of removed stone values divisible by 3 loses. Alice goes first, and Bob wins if there are no stones left. The task is to determine, assuming optimal play by both, whether Alice wins (returns `true`) or Bob wins (returns `false`).

Solution Approach

The provided solutions efficiently determine the winner by counting the number of stones with remainders 0, 1, and 2 when divided by 3. Then, based on whether the count of stones with remainder 0 is even or odd, different winning conditions are evaluated. If the count of remainder 0 is even, Alice wins if there are both remainder 1 and 2 stones available. If the count of remainder 0 is odd, Alice wins if the absolute difference between the counts of stones with remainder 1 and 2 is greater than 2.

Step-by-Step Algorithm

Step 1: Count the number of stones that have remainders 0, 1, and 2 when divided by 3. This can be efficiently done by iterating through the `stones` array and using a HashMap/Counter to track the counts.
Step 2: Check if the count of stones with remainder 0 is even or odd.
Step 3: If the count of remainder 0 stones is even, Alice wins if there's at least one stone with remainder 1 and at least one stone with remainder 2. Return `true` if this condition is met, otherwise return `false`.
Step 4: If the count of remainder 0 stones is odd, Alice wins if the absolute difference between the count of stones with remainder 1 and the count of stones with remainder 2 is greater than 2. Return `true` if this condition is met, otherwise return `false`.

Key Insights

Insight 1: The core of the solution lies in understanding how the remainders of stone values when divided by 3 affect the game's outcome. Only the counts of stones with remainders 0, 1, and 2 matter.
Insight 2: The number of stones with a remainder of 0 significantly impacts the game dynamic. An even count allows direct competition between stones with remainders 1 and 2, while an odd count introduces a slight advantage/disadvantage depending on the imbalance between counts of remainders 1 and 2.
Insight 3: Optimal play involves understanding how to force a win or avoid a loss based on the current state of the counts and the total sum's remainder when divided by 3.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Greedy, Counting, Game Theory.

Companies

Asked at: Samsung.

Stone Game IX | LeetCode 2029

Medium

Array Math Greedy Counting Game Theory

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public boolean stoneGameIX(int[] stones) {
        Map<Integer, Integer> counts = new HashMap<>();
        for (int s : stones) {
            int remainder = s % 3;
            counts.put(remainder, counts.getOrDefault(remainder, 0) + 1);
        }

        int c0 = counts.getOrDefault(0, 0);
        int c1 = counts.getOrDefault(1, 0);
        int c2 = counts.getOrDefault(2, 0);

        if (c0 % 2 == 0) {
            return c1 > 0 && c2 > 0;
        } else {
            return Math.abs(c1 - c2) > 2;
        }
    }
}

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Problem Info

Description

Problem 2029: Stone Game IX - Detailed explanation and solution approaches.

Topics

Array Math Greedy Counting Game Theory

Companies

Samsung

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2029

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

4 steps

1
Step 1: Count the number of stones that have remainders 0, 1, and 2 when divided by 3. This can be efficiently done by iterating through the `stones` array and using a HashMap/Counter to track the counts.
2
Step 2: Check if the count of stones with remainder 0 is even or odd.
3
Step 3: If the count of remainder 0 stones is even, Alice wins if there's at least one stone with remainder 1 and at least one stone with remainder 2. Return `true` if this condition is met, otherwise return `false`.
4
Step 4: If the count of remainder 0 stones is odd, Alice wins if the absolute difference between the count of stones with remainder 1 and the count of stones with remainder 2 is greater than 2. Return `true` if this condition is met, otherwise return `false`.

💡

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 determined by the single loop used to iterate through the input array `stones`. The remaining operations involve simple comparisons and arithmetic, taking constant time. The space complexity is constant because the HashMap/Counter stores at most three key-value pairs (0, 1, and 2 as keys).

📖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

Minimax with memoization

This approach involves building a game tree and using minimax to determine the optimal move at each step. Memoization can be used to avoid recomputing the outcome of the same game states.

Time: Exponential in the worst case without proper memoization.Space: Exponential in the worst case without proper memoization due to the recursion depth.

Trade-offs: This approach is conceptually straightforward but computationally expensive and impractical for larger inputs. It's mainly useful for understanding the game's mechanics but is not efficient for solving the problem due to time limit constraints.

✨

Key Insights

1Insight 1: The core of the solution lies in understanding how the remainders of stone values when divided by 3 affect the game's outcome. Only the counts of stones with remainders 0, 1, and 2 matter.
2Insight 2: The number of stones with a remainder of 0 significantly impacts the game dynamic. An even count allows direct competition between stones with remainders 1 and 2, while an odd count introduces a slight advantage/disadvantage depending on the imbalance between counts of remainders 1 and 2.
3Insight 3: Optimal play involves understanding how to force a win or avoid a loss based on the current state of the counts and the total sum's remainder when divided by 3.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: All stones have a value divisible by 3 (only remainder 0 stones). If there are an even number of 0s, Bob wins, if odd Alice wins. The code handles this because if c1 and c2 are 0, the conditions `c1 > 0 && c2 > 0` and `abs(c1 - c2) > 2` will both evaluate to `false`.
Edge case 2: All stones have the same remainder (either all 1 or all 2). If the number of 0 stones is even, Alice loses if there are no other remainders. If the number of 0 stones is odd, Alice wins if the absolute difference between c1 and c2 is greater than 2 (which in this case, is n > 2 where n is the number of stones with same remainder).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider the impact of stones with values divisible by 3 (remainder 0). These stones significantly affect the turn parity and thus, the game's outcome.

✗Common mistake 2: Not fully understanding the optimal strategies for Alice and Bob. It is crucial to recognize that the goal is to force your opponent into making the sum divisible by 3, not just to maximize your own score.

✅

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: Game theory optimization: Understanding optimal strategies in competitive scenarios. This is applicable in resource allocation, bidding strategies, and negotiation.

🏦Real-world scenario 2: Turn-based decision-making: Analyzing and predicting outcomes based on a sequence of choices. Applicable in strategic planning, supply chain management, and simulations.

💡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

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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 20, 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

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

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Stone Game IX"?

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

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