How do you solve Stone Game V on LeetCode?

Stone Game V (LeetCode #1563) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Dynamic programming helps avoid recalculating overlapping subproblems.

What is the brute force solution for Stone Game V?

The brute force approach for Stone Game V is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we explore all possible ways to split the array into two non-empty parts and calculate Alice's score for each split. This method is straightforward but inefficient as it recalculates scores for overlapping subarrays multiple times. The optimal Optimal Solution approach improves this to O(n²).

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

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

What are the interview tips for Stone Game V?

Clearly explain your thought process and how you plan to optimize the brute force solution. Discuss the importance of dynamic programming in reducing time complexity. Practice breaking down the problem into smaller parts and using prefix sums.

What are common mistakes when solving Stone Game V?

Not using memoization or dynamic programming, leading to excessive recomputation. Failing to consider edge cases, such as arrays with equal values.

#1563

Stone Game V

Hard

ArrayMathDynamic ProgrammingGame TheoryDynamic ProgrammingPrefix Sums

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

The optimal solution uses dynamic programming to store the maximum score that can be achieved for each subarray. By avoiding redundant calculations, we significantly reduce the time complexity.

⚙️

Algorithm

3 steps

1Step 1: Create a DP array where dp[i] represents the maximum score obtainable from the first i stones.
2Step 2: Iterate through all possible split points and calculate the left and right sums using prefix sums for efficiency.
3Step 3: Update the DP array based on the scores from the splits, considering which row Bob would discard.

solution.py17 lines

1def stoneGameV(stoneValue):
2    n = len(stoneValue)
3    prefix_sum = [0] * (n + 1)
4    for i in range(n):
5        prefix_sum[i + 1] = prefix_sum[i] + stoneValue[i]
6    dp = [0] * n
7    for i in range(1, n):
8        for j in range(i):
9            left_sum = prefix_sum[j + 1]
10            right_sum = prefix_sum[i + 1] - left_sum
11            if left_sum > right_sum:
12                dp[i] = max(dp[i], dp[j] + left_sum)
13            elif left_sum < right_sum:
14                dp[i] = max(dp[i], dp[j] + right_sum)
15            else:
16                dp[i] = max(dp[i], dp[j] + left_sum, dp[j] + right_sum)
17    return dp[-1]

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops for calculating scores from splits, but we use O(n) space for the DP array and prefix sums, which optimizes our calculations.

1Dynamic programming helps avoid recalculating overlapping subproblems.
2Using prefix sums allows for efficient sum calculations.

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