How do you solve Stone Game III on LeetCode?

Stone Game III (LeetCode #1406) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Alice and Bob both play optimally, meaning they will always make the best possible move to maximize their score.

What is the brute force solution for Stone Game III?

The brute force approach for Stone Game III is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we explore all possible ways Alice can take 1, 2, or 3 stones and recursively calculate the scores for both players. This method checks every possible outcome, which can be quite slow. The optimal Optimal Solution approach improves this to O(n).

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

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

What are the interview tips for Stone Game III?

Always clarify the rules of the game and how players score before diving into coding. Consider edge cases, such as when the stoneValue array is empty or contains negative values. Explain your thought process clearly while coding, as it helps interviewers understand your reasoning.

What are common mistakes when solving Stone Game III?

Not considering the optimal strategy for both players, leading to incorrect assumptions about the game's outcome. Failing to use memoization in the recursive approach, resulting in excessive computations.

#1406

Stone Game III

Hard

ArrayMathDynamic ProgrammingGame TheoryDynamic ProgrammingMinimax Strategy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to store the maximum score difference Alice can achieve from each index. This avoids redundant calculations and allows us to build the solution efficiently.

⚙️

Algorithm

3 steps

1Step 1: Create a DP array where dp[i] represents the maximum score difference Alice can achieve starting from index i.
2Step 2: Iterate backwards through the stoneValue array, calculating the maximum score Alice can achieve for taking 1, 2, or 3 stones.
3Step 3: Use the DP values to determine the final result based on the score difference.

solution.py16 lines

1def stoneGameIII(stoneValue):
2    n = len(stoneValue)
3    dp = [float('-inf')] * (n + 1)
4    dp[n] = 0
5    for i in range(n - 1, -1, -1):
6        total = 0
7        for x in range(1, 4):
8            if i + x <= n:
9                total += stoneValue[i + x - 1]
10                dp[i] = max(dp[i], total - dp[i + x])
11    if dp[0] > 0:
12        return 'Alice'
13    elif dp[0] < 0:
14        return 'Bob'
15    else:
16        return 'Tie'

ℹ

Complexity note: The time complexity is O(n) because we iterate through the stoneValue array once, and for each index, we perform a constant amount of work (up to 3 moves). The space complexity is O(n) due to the DP array.

1Alice and Bob both play optimally, meaning they will always make the best possible move to maximize their score.
2The game can be viewed as a minimax problem where one player's gain is the other's loss.

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