What is the time complexity of Stone Game VII?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Stone Game VII?

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

What are the key insights for solving Stone Game VII?

Insight 1: Dynamic Programming is crucial. The optimal strategy for a sub-array depends on the optimal strategies for its sub-sub-arrays. Insight 2: The prefix sum helps in calculating the sum of a subarray in O(1) time, avoiding repeated calculations within the DP.

Which companies ask LeetCode 1690. Stone Game VII in interviews?

This problem is frequently asked at Dunzo.

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Stone Game VII - LeetCode 1690 Solution

Q: What is the space complexity of Stone Game VII?

The space complexity is O(n^2).

Stone Game VII - Complete Solution Guide

Stone Game VII is LeetCode problem 1690, 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 arranged in a row. On each player's turn, they can remove either the leftmost stone or the rightmost stone from the row and receive points equal to the sum of the remaining stones' values in the row. The winner is the one with the higher score when there are no stones left to remove. Bob found that he will always lose this game (poor Bob, he always loses), so he decided to minimize the score's difference . Al

Detailed Explanation

The Stone Game VII presents a game between Alice and Bob where they remove stones from either end of a row. Each player's score is the sum of the remaining stones after their removal. Alice aims to maximize the difference between her score and Bob's, while Bob wants to minimize it. The goal is to determine the optimal score difference Alice can achieve, assuming both players play perfectly. The input is an array `stones` representing the values of the stones. The output is the maximum possible score difference between Alice and Bob.

Solution Approach

The provided code implements a dynamic programming solution. It calculates a prefix sum array to efficiently compute the sum of stone values in a range. A 2D DP table `dp[i][j]` stores the maximum score difference Alice can achieve when playing optimally on the subarray `stones[i...j]`. The table is filled in order of increasing subarray length, from length 2 up to n. For each subarray, Alice has two choices: remove the leftmost stone or the rightmost stone. The `dp[i][j]` value is then the maximum of (sum of remaining stones - dp[i+1][j]) and (sum of remaining stones - dp[i][j-1]), representing the optimal choice Alice can make. The final result is stored in `dp[0][n-1]`.

Step-by-Step Algorithm

Step 1: Calculate the prefix sum array `prefix`. `prefix[i]` stores the sum of the first `i` stones.
Step 2: Initialize a 2D DP table `dp` of size n x n with all values set to 0. `dp[i][j]` will store the maximum score difference for the subarray stones[i...j].
Step 3: Iterate over the length of the subarrays from 2 to n. This is the outer loop: `for length in range(2, n + 1)`
Step 4: For each length, iterate over all possible starting positions `i` of the subarray. This is the inner loop: `for i in range(n - length + 1)`
Step 5: Calculate the ending position `j` of the subarray: `j = i + length - 1`
Step 6: Calculate the sum of the stones in the subarray [i+1...j] if removing the leftmost stone: `sum_without_left = prefix[j + 1] - prefix[i + 1]`
Step 7: Calculate the sum of the stones in the subarray [i...j-1] if removing the rightmost stone: `sum_without_right = prefix[j] - prefix[i]`
Step 8: Calculate the score difference if removing the leftmost stone: `option1 = sum_without_left - dp[i + 1][j]`
Step 9: Calculate the score difference if removing the rightmost stone: `option2 = sum_without_right - dp[i][j - 1]`
Step 10: Store the maximum of `option1` and `option2` in `dp[i][j]`. This represents Alice's optimal choice for this subarray: `dp[i][j] = max(option1, option2)`
Step 11: After the loops complete, return `dp[0][n - 1]`, which represents the maximum score difference for the entire array.

Key Insights

Insight 1: Dynamic Programming is crucial. The optimal strategy for a sub-array depends on the optimal strategies for its sub-sub-arrays.
Insight 2: The prefix sum helps in calculating the sum of a subarray in O(1) time, avoiding repeated calculations within the DP.
Insight 3: The core recurrence relation relates the optimal score difference for a sub-array to the optimal score differences of sub-arrays formed by removing the left or right element.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

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

Companies

Asked at: Dunzo.

Stone Game VII | LeetCode 1690

Medium

Array Math Dynamic Programming Game Theory

Time: O(n^2)Space: O(n^2)

Solution Code

Available in:

class Solution {
    public int stoneGameVII(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];
        }

        int[][] dp = new int[n][n];

        for (int length = 2; length <= n; length++) {
            for (int i = 0; i <= n - length; i++) {
                int j = i + length - 1;

                // Option 1: remove the leftmost stone
                int sum_without_left = prefix[j + 1] - prefix[i + 1];
                int option1 = sum_without_left - dp[i + 1][j];

                // Option 2: remove the rightmost stone
                int sum_without_right = prefix[j] - prefix[i];
                int option2 = sum_without_right - dp[i][j - 1];

                dp[i][j] = Math.max(option1, option2);
            }
        }

        return dp[0][n - 1];
    }
}

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Problem Info

Description

Problem 1690: Stone Game VII - Detailed explanation and solution approaches.

Topics

Array Math Dynamic Programming Game Theory

Companies

Dunzo

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1690

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

11 steps

1
Step 1: Calculate the prefix sum array `prefix`. `prefix[i]` stores the sum of the first `i` stones.
2
Step 2: Initialize a 2D DP table `dp` of size n x n with all values set to 0. `dp[i][j]` will store the maximum score difference for the subarray stones[i...j].
3
Step 3: Iterate over the length of the subarrays from 2 to n. This is the outer loop: `for length in range(2, n + 1)`
4
Step 4: For each length, iterate over all possible starting positions `i` of the subarray. This is the inner loop: `for i in range(n - length + 1)`
5
Step 5: Calculate the ending position `j` of the subarray: `j = i + length - 1`
6
Step 6: Calculate the sum of the stones in the subarray [i+1...j] if removing the leftmost stone: `sum_without_left = prefix[j + 1] - prefix[i + 1]`
7
Step 7: Calculate the sum of the stones in the subarray [i...j-1] if removing the rightmost stone: `sum_without_right = prefix[j] - prefix[i]`
8
Step 8: Calculate the score difference if removing the leftmost stone: `option1 = sum_without_left - dp[i + 1][j]`
9
Step 9: Calculate the score difference if removing the rightmost stone: `option2 = sum_without_right - dp[i][j - 1]`
10
Step 10: Store the maximum of `option1` and `option2` in `dp[i][j]`. This represents Alice's optimal choice for this subarray: `dp[i][j] = max(option1, option2)`
11
Step 11: After the loops complete, return `dp[0][n - 1]`, which represents the maximum score difference for the entire array.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is O(n^2) because the dynamic programming solution iterates over all possible sub-arrays of the input array. The space complexity is O(n^2) because the DP table used to store the intermediate results has dimensions n x n, where n is the size of the input array. The prefix sum array contributes an additional O(n) space, but this is dominated by the DP table.

📖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

Recursion with Memoization

A recursive solution with memoization could also solve this problem. Instead of iteratively filling the DP table, we can recursively calculate the optimal score difference for each subarray and store the results in a memoization table to avoid redundant calculations.

Time: O(n^2)Space: O(n^2)

Trade-offs: The recursive approach might be more intuitive for some, but it can be less efficient due to function call overhead compared to the iterative DP approach. The space complexity remains the same as it also requires an n x n memoization table.

✨

Key Insights

1Insight 1: Dynamic Programming is crucial. The optimal strategy for a sub-array depends on the optimal strategies for its sub-sub-arrays.
2Insight 2: The prefix sum helps in calculating the sum of a subarray in O(1) time, avoiding repeated calculations within the DP.
3Insight 3: The core recurrence relation relates the optimal score difference for a sub-array to the optimal score differences of sub-arrays formed by removing the left or right element.

📚

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: n = 2. The code correctly handles the base case when the array has only two elements. The loop starts from length 2, and the calculations for option1 and option2 are accurate.
Edge case 2: All stones have the same value. The algorithm still correctly computes the optimal difference, as the prefix sum and DP calculations remain valid even with identical stone values.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using prefix sum. Calculating the sum of the remaining stones in each turn without a prefix sum array leads to O(n) time complexity for each calculation, resulting in an overall time complexity of O(n^3), which is too slow.

✗Common mistake 2: Incorrect DP transition. Forgetting to subtract the result of the next player's optimal play (i.e., `dp[i+1][j]` or `dp[i][j-1]`) from the current player's score can result in a wrong solution. The problem is a minimax problem, so we should consider that the other player also plays optimally.

✅

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. Imagine allocating resources between two parties where each party can choose to take from either end of a line of resources. This problem's solution strategy could inform optimal allocation to maximize one party's advantage.

🏦Real-world scenario 2: Game strategy. This type of dynamic programming solution can be adapted to other competitive games where players make sequential moves with the goal of maximizing their own score or minimizing their opponent's score. Auctions and negotiations can have similar underlying structures.

💡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.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 10, 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

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

@DevExpert92Jul 14

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

@AlgoMasterJul 7

❓

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 VII"?

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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