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LeetCode 1563: Stone Game V Solution

Master LeetCode problem 1563 (Stone Game V), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming with memoization. We can define a recursive function that simulates the game for each possible split of stones. At each step, we calculate the score that Alice can obtain and store it in a memoization table to avoid redundant calculations. We need to consider all possible splits of stones to maximize Alice's score.

  1. Create a memoization table to store the maximum score Alice can obtain for a given range of stones.
  2. Define a recursive function that takes the range of stones as input and calculates the maximum score Alice can obtain.
  3. At each step, consider all possible splits of the range of stones, calculate the score for Alice, and store the maximum score in the memoization table.
  4. Return the maximum score that Alice can obtain starting from the entire range of stones.

Time Complexity: O(n^3) where n is the number of stones in the input array.

Space Complexity: O(n^2) for the memoization table.

:

Solution Code

class Solution {
    public int stoneGameV(int[] stoneValue) {
        int n = stoneValue.length;
        int[] prefixSum = new int[n + 1];
        for (int i = 0; i < n; i++) {
            prefixSum[i + 1] = prefixSum[i] + stoneValue[i];
        }
        
        int[][] memo = new int[n][n];
        return dfs(0, n - 1, prefixSum, memo);
    }
    
    private int dfs(int left, int right, int[] prefixSum, int[][] memo) {
        if (left == right) return 0;
        if (memo[left][right] > 0) return memo[left][right];
        
        int result = 0;
        for (int mid = left; mid < right; mid++) {
            int leftSum = prefixSum[mid + 1] - prefixSum[left];
            int rightSum = prefixSum[right + 1] - prefixSum[mid + 1];
            if (leftSum <= rightSum) {
                result = Math.max(result, leftSum + dfs(left, mid, prefixSum, memo));
            }
            if (leftSum >= rightSum) {
                result = Math.max(result, rightSum + dfs(mid + 1, right, prefixSum, memo));
            }
        }
        
        memo[left][right] = result;
        return result;
    }
}

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