How do you solve Stone Game VII on LeetCode?

Stone Game VII (LeetCode #1690) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n²) space complexity. Key insight: Both players will always play optimally, making the game a perfect information game.

What is the brute force solution for Stone Game VII?

The brute force approach for Stone Game VII is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves recursively calculating all possible outcomes of the game by removing stones from either end. This method explores every possible game state, ensuring we consider all choices made by both players. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Stone Game VII?

Stone Game VII uses the following concepts: Array, Math, Dynamic Programming, Game Theory. The recommended patterns to study are: Dynamic Programming, Game Theory.

What are the interview tips for Stone Game VII?

Clearly explain your thought process and the reasoning behind each decision. Consider edge cases and how they might affect your solution. Practice explaining your code as you write it, as this simulates the interview environment.

What are common mistakes when solving Stone Game VII?

Not considering both players' optimal strategies, leading to incorrect assumptions about the game. Failing to properly implement the DP table, which can lead to incorrect results.

#1690

Stone Game VII

Medium

ArrayMathDynamic ProgrammingGame TheoryDynamic ProgrammingGame Theory

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n²)

💡

Intuition

Time O(n²)Space O(n²)

The optimal solution uses dynamic programming to store previously computed results, avoiding redundant calculations. This allows us to efficiently determine the maximum score difference by considering the optimal choices of both players.

⚙️

Algorithm

3 steps

1Step 1: Create a DP table where dp[i][j] represents the maximum score difference when considering stones from index i to j.
2Step 2: Initialize the DP table for the base cases where only one stone is left.
3Step 3: Fill the DP table iteratively based on the choices of removing stones from either end.

solution.py9 lines

1def stoneGameVII(stones):
2    n = len(stones)
3    dp = [[0] * n for _ in range(n)]
4    total = sum(stones)
5    for length in range(2, n + 1):
6        for i in range(n - length + 1):
7            j = i + length - 1
8            dp[i][j] = max(total - stones[i] - dp[i + 1][j], total - stones[j] - dp[i][j - 1])
9    return dp[0][n - 1]

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops filling the DP table, while the space complexity is also O(n²) for storing the results in the DP table.

1Both players will always play optimally, making the game a perfect information game.
2The total score of remaining stones is crucial for determining the best moves.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.