What is the time complexity of Stone Game VIII?

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

What data structures are used to solve Stone Game VIII?

This problem uses Array, Math, Dynamic Programming, Prefix Sum, Game Theory. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Stone Game VIII?

Insight 1: Recognizing the problem can be solved using dynamic programming to determine the optimal strategy for both players. Insight 2: Using prefix sums to efficiently calculate the sum of stones removed in each move reduces the complexity of the algorithm.

Which companies ask LeetCode 1872. Stone Game VIII in interviews?

This problem is frequently asked at Infosys.

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Stone Game VIII - LeetCode 1872 Solution

Stone Game VIII - Complete Solution Guide

Stone Game VIII is LeetCode problem 1872, a Hard 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 arranged in a row. On each player's turn, while the number of stones is more than one , they will do the following: Choose an integer x > 1 , and remove the leftmost x stones from the row. Add the sum of the removed stones' values to the player's score. Place a new stone , whose value is equal to that sum, on the left side of the row. The game stops when only one stone is left in the row. The score difference

Detailed Explanation

The Stone Game VIII presents a game played by Alice and Bob on a row of stones, each with a value. Players take turns removing a number of stones (greater than 1) from the left, adding the sum of their values to their score, and replacing them with a new stone whose value is the sum of the removed stones. The game ends when only one stone remains. Alice aims to maximize the score difference (Alice's score - Bob's score), while Bob aims to minimize it. The goal is to determine the score difference if both players play optimally. The input is an array `stones` representing the stone values, and the output is the maximum score difference achievable.

Solution Approach

The provided solution uses a dynamic programming approach. It calculates a prefix sum array to optimize the computation of the sum of removed stones for each move. The core idea is to determine the maximum score difference the current player can achieve given the remaining stones. This is done by iterating backwards from the second-to-last state (two stones remaining) to the initial state. At each step, the current player considers all possible moves (removing x stones), calculating the score difference resulting from that move, and choosing the move that maximizes their score difference. Because only the result from the previous step is used, space is optimized by only storing the previous `dp` value rather than an entire `dp` array.

Step-by-Step Algorithm

Step 1: Calculate the prefix sum of the `stones` array. This allows for constant-time calculation of the sum of the first x stones.
Step 2: Initialize `dp` with the result of taking all stones. This represents the score if only 2 stones exist, in other words, when `i = n-1`.
Step 3: Iterate backward from `i = n - 2` to `i = 1`. In each iteration, `dp` is the maximum score the player can achieve if they act optimally, so `dp[i] = max(dp[i+1], prefix[i+1] - dp[i+1])`
Step 4: At each step, calculate the potential score difference by subtracting the opponent's best score (`dp`) from the current move's score (`prefix[i + 1]`).
Step 5: Update the `dp` with the maximum value between current `dp` and potential score difference, this captures that each player act optimally.
Step 6: Return the final `dp` value, which represents the maximum score difference Alice can achieve.

Key Insights

Insight 1: Recognizing the problem can be solved using dynamic programming to determine the optimal strategy for both players.
Insight 2: Using prefix sums to efficiently calculate the sum of stones removed in each move reduces the complexity of the algorithm.
Insight 3: The optimal strategy can be found by working backwards from the end of the game (when there are only two stones left), using previously computed results to make optimal decisions.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Dynamic Programming, Prefix Sum, Game Theory.

Companies

Asked at: Infosys.

Stone Game VIII | LeetCode 1872

Hard

Array Math Dynamic Programming Prefix Sum Game Theory

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

Solution Code

Available in:

class Solution {
    public int stoneGameVIII(int[] stones) {
        int n = stones.length;
        
        int[] prefix = new int[n + 1];
        for (int i = 0; i < n; i++) {
            prefix[i + 1] = prefix[i] + stones[i];
        }
        
        // Let dp[i] be the maximum score difference the current player can achieve
        // from a state where the stones are [prefix[i], stones[i], ..., stones[n-1]].
        //
        // Alice's first move, taking k stones (2 <= k <= n), results in a score of
        // prefix[k] and puts Bob in state k. Bob will achieve a difference of dp[k].
        // Alice's net difference for this move is prefix[k] - dp[k].
        // Alice maximizes this over all k. The final answer is max_{k=2..n} (prefix[k] - dp[k]),
        // which is equivalent to the definition of dp[1].
        //
        // The recurrence for dp[i] is:
        // dp[i] = max_{k=i+1..n} (prefix[k] - dp[k])
        // This can be simplified to:
        // dp[i] = max(dp[i+1], prefix[i+1] - dp[i+1])
        //
        // The base case is dp[n-1]. In state n-1, there are two stones. The only move
        // is to take both, resulting in a score of prefix[n]. So, dp[n-1] = prefix[n].
        
        // We can compute dp values from n-1 down to 1. Since dp[i] only depends on
        // dp[i+1], we can use a single variable instead of an array.
        
        // Initialize dp with the value for dp[n-1]
        int dp = prefix[n];
        
        // Iterate from i = n-2 down to 1 to compute dp[n-2], ..., dp[1].
        // In each iteration, the variable `dp` holds the value of dp[i+1].
        // We calculate dp[i] and update `dp`.
        for (int i = n - 2; i > 0; i--) {
            dp = Math.max(dp, prefix[i + 1] - dp);
        }
        
        return dp;
    }
}

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Problem Info

Description

Problem 1872: Stone Game VIII - Detailed explanation and solution approaches.

Topics

Array Math Dynamic Programming Prefix Sum Game Theory

Companies

Infosys

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1872

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

6 steps

1
Step 1: Calculate the prefix sum of the `stones` array. This allows for constant-time calculation of the sum of the first x stones.
2
Step 2: Initialize `dp` with the result of taking all stones. This represents the score if only 2 stones exist, in other words, when `i = n-1`.
3
Step 3: Iterate backward from `i = n - 2` to `i = 1`. In each iteration, `dp` is the maximum score the player can achieve if they act optimally, so `dp[i] = max(dp[i+1], prefix[i+1] - dp[i+1])`
4
Step 4: At each step, calculate the potential score difference by subtracting the opponent's best score (`dp`) from the current move's score (`prefix[i + 1]`).
5
Step 5: Update the `dp` with the maximum value between current `dp` and potential score difference, this captures that each player act optimally.
6
Step 6: Return the final `dp` value, which represents the maximum score difference Alice can achieve.

💡

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 O(n) because the code iterates through the stones array once to compute the prefix sum and then iterates backwards from n-2 to 1 in the second loop. The space complexity is O(1) because it only uses a constant amount of extra space, regardless of the size of the input array. Only the prefix array is O(n), but the single variable dp overrides this.

📖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

Top-Down Dynamic Programming (Memoization)

Instead of iterating backward in bottom up fashion, one could solve the problem using top down dynamic programming with memoization. This would involves creating a recursive function that calculates the maximum score difference with given remaining stones, storing intermediate results to avoid recalculations.

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

Trade-offs: Top-down approach can be more intuitive for some, but it introduces the overhead of recursive calls and may consume more memory for the memoization table and call stack. In this instance, bottom-up (iterative) dynamic programming implementation is more efficient due to the fact that it can be implemented with O(1) space.

✨

Key Insights

1Insight 1: Recognizing the problem can be solved using dynamic programming to determine the optimal strategy for both players.
2Insight 2: Using prefix sums to efficiently calculate the sum of stones removed in each move reduces the complexity of the algorithm.
3Insight 3: The optimal strategy can be found by working backwards from the end of the game (when there are only two stones left), using previously computed results to make optimal decisions.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Input array with only two stones. The algorithm correctly calculates the score difference by simply taking the sum of both stones. The loop does not execute and the initially set value of 'dp' which is prefix[n] is returned.
Edge case 2: Input array with all negative stones. The algorithm will still correctly compute the optimal strategy by maximizing the score difference, which may be a negative value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using prefix sums, leading to O(n^2) time complexity, thus exceeding time limit for larger inputs.

✗Common mistake 2: Failing to correctly formulate the DP recurrence relation, resulting in incorrect calculations of the optimal score difference. Need to take opponent's optimality into account when maximizing current player's 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, where algorithms like minimax or dynamic programming are used to determine the optimal strategy in games or competitive situations.

🏦Real-world scenario 2: Resource allocation problems, where the goal is to maximize profit or minimize cost by making optimal decisions at each stage, considering the impact of those decisions on future outcomes.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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 12, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 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.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 VIII"?

For this hard-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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