What is the time complexity of Stone Game VI?

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

What data structures are used to solve Stone Game VI?

This problem uses Array, Math, Greedy, Sorting, Heap (Priority Queue), Game Theory. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Stone Game VI?

Insight 1: The optimal strategy involves considering the combined value of each stone for both players (aliceValues[i] + bobValues[i]). Picking stones with higher combined values benefits the picker more relative to their opponent, as it deprives the opponent of a potentially valuable stone from both their perspectives. Insight 2: Sorting the stones based on the combined value allows us to prioritize picking the stones that offer the most significant advantage to the current player (Alice or Bob) in terms of depriving the opponent.

Which companies ask LeetCode 1686. Stone Game VI in interviews?

This problem is frequently asked at Arcesium.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Stone Game VI - LeetCode 1686 Solution

Stone Game VI - Complete Solution Guide

Stone Game VI is LeetCode problem 1686, 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 take turns playing a game, with Alice starting first. There are n stones in a pile. On each player's turn, they can remove a stone from the pile and receive points based on the stone's value. Alice and Bob may value the stones differently . You are given two integer arrays of length n , aliceValues and bobValues . Each aliceValues[i] and bobValues[i] represents how Alice and Bob, respectively, value the i th stone. The winner is the person with the most points after all the stones

Detailed Explanation

The problem Stone Game VI presents a scenario where Alice and Bob are playing a game involving a pile of stones. Each stone has two values associated with it: one representing its value to Alice (aliceValues[i]) and another representing its value to Bob (bobValues[i]). Players take turns picking stones, and their score increases by the value of the stone they pick. The goal is to determine who wins the game, assuming both players play optimally. Alice starts first. The outcome can be Alice winning (1), Bob winning (-1), or a draw (0).

Solution Approach

The solution employs a greedy approach. First, it calculates the sum of Alice's and Bob's values for each stone (aliceValues[i] + bobValues[i]). Then, it sorts the indices of the stones in descending order based on this combined value. Finally, it iterates through the sorted indices, assigning stones to Alice on even turns and to Bob on odd turns, and calculating their scores accordingly. The player with the higher score wins.

Step-by-Step Algorithm

Step 1: Calculate the sum of aliceValues[i] and bobValues[i] for each stone and store this sum along with the stone's original index.
Step 2: Sort the indices of the stones in descending order based on the calculated sum (aliceValues[i] + bobValues[i]). This is the crucial greedy step.
Step 3: Initialize alice_score and bob_score to 0.
Step 4: Iterate through the sorted indices. If the index `i` is even, it's Alice's turn, so add aliceValues[indices[i]] to alice_score. If the index `i` is odd, it's Bob's turn, so add bobValues[indices[i]] to bob_score.
Step 5: After iterating through all stones, compare alice_score and bob_score. Return 1 if alice_score > bob_score, -1 if bob_score > alice_score, and 0 if they are equal.

Key Insights

Insight 1: The optimal strategy involves considering the combined value of each stone for both players (aliceValues[i] + bobValues[i]). Picking stones with higher combined values benefits the picker more relative to their opponent, as it deprives the opponent of a potentially valuable stone from both their perspectives.
Insight 2: Sorting the stones based on the combined value allows us to prioritize picking the stones that offer the most significant advantage to the current player (Alice or Bob) in terms of depriving the opponent.
Insight 3: This greedy approach works because by prioritizing stones with the highest combined value, we ensure that each player maximizes their relative gain compared to the other player during their turn. Even if a particular stone has a low personal value, picking a stone with high combined value deprives the other player of a larger total potential benefit.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Math, Greedy, Sorting, Heap (Priority Queue), Game Theory.

Companies

Asked at: Arcesium.

Stone Game VI | LeetCode 1686

Medium

Array Math Greedy Sorting Heap (Priority Queue)Game Theory

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    public int stoneGameVI(int[] aliceValues, int[] bobValues) {
        int n = aliceValues.length;
        
        Integer[] indices = new Integer[n];
        for (int i = 0; i < n; i++) {
            indices[i] = i;
        }
        
        Arrays.sort(indices, (i, j) -> (bobValues[j] + aliceValues[j]) - (aliceValues[i] + bobValues[i]));
        
        int alice_score = 0;
        int bob_score = 0;
        
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0) { 
                alice_score += aliceValues[indices[i]];
            } else { 
                bob_score += bobValues[indices[i]];
            }
        }
        
        if (alice_score > bob_score) {
            return 1;
        } else if (bob_score > alice_score) {
            return -1;
        } else {
            return 0;
    }
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1686: Stone Game VI - Detailed explanation and solution approaches.

Topics

Array Math Greedy Sorting Heap (Priority Queue)Game Theory

Companies

Arcesium

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1686

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: Calculate the sum of aliceValues[i] and bobValues[i] for each stone and store this sum along with the stone's original index.
2
Step 2: Sort the indices of the stones in descending order based on the calculated sum (aliceValues[i] + bobValues[i]). This is the crucial greedy step.
3
Step 3: Initialize alice_score and bob_score to 0.
4
Step 4: Iterate through the sorted indices. If the index `i` is even, it's Alice's turn, so add aliceValues[indices[i]] to alice_score. If the index `i` is odd, it's Bob's turn, so add bobValues[indices[i]] to bob_score.
5
Step 5: After iterating through all stones, compare alice_score and bob_score. Return 1 if alice_score > bob_score, -1 if bob_score > alice_score, and 0 if they are equal.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Sorting algorithms generally have a time complexity of O(n log n). The space complexity comes from the auxiliary space used to store the sorted indices. The python solution uses `list(range(n))` which creates a list of size `n`, and similarly for the other language solutions.

📖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

Dynamic Programming

While not efficient, one could theoretically formulate a dynamic programming approach to explore all possible move sequences and determine the optimal strategy. This would be extremely complex and computationally expensive.

Time: ExponentialSpace: Exponential

Trade-offs: Not practical for this problem due to its high complexity. DP would involve exploring all possible combinations which is not efficient.

✨

Key Insights

1Insight 1: The optimal strategy involves considering the combined value of each stone for both players (aliceValues[i] + bobValues[i]). Picking stones with higher combined values benefits the picker more relative to their opponent, as it deprives the opponent of a potentially valuable stone from both their perspectives.
2Insight 2: Sorting the stones based on the combined value allows us to prioritize picking the stones that offer the most significant advantage to the current player (Alice or Bob) in terms of depriving the opponent.
3Insight 3: This greedy approach works because by prioritizing stones with the highest combined value, we ensure that each player maximizes their relative gain compared to the other player during their turn. Even if a particular stone has a low personal value, picking a stone with high combined value deprives the other player of a larger total potential benefit.

📚

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.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: When aliceValues and bobValues are identical arrays. The solution still correctly determines the winner based on alternating turns.
Edge case 2: Small input sizes (e.g., n = 1 or n = 2). The solution works correctly for small inputs due to the correct sorting and turn-based score accumulation.
Edge case 3: When all combined values are the same. The sorting will result in an arbitrary order and the turns dictate outcome.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly assuming that a player should always pick the stone that maximizes their individual value. This ignores the impact of depriving the opponent of a valuable stone.

✗Common mistake 2: Failing to sort the stones correctly based on the combined value, which leads to a suboptimal and incorrect strategy.

✗Common mistake 3: Not accounting for the alternating turns correctly during score accumulation.

✅

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 problems where two parties compete for resources, and each resource has different values for each party. This approach helps in determining the optimal resource allocation strategy.

🏦Real-world scenario 2: Negotiation scenarios where two parties negotiate over multiple items, each with different values for each party. The algorithm can help determine which items to prioritize to maximize gains while minimizing losses.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: August 6, 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

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

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time 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 "Stone Game VI"?

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.

🤖 Need help with this problem?