LeetCode 3257: Maximum Value Sum by Placing Three Rooks II Solution

Master LeetCode problem 3257 (Maximum Value Sum by Placing Three Rooks II), 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.

3257. Maximum Value Sum by Placing Three Rooks II

Hard

Array Dynamic Programming Matrix Enumeration

Problem Explanation

Explanation:

To maximize the sum of values in the cells where the rooks are placed, we need to strategically place the rooks such that they do not attack each other. In this problem, we are asked to place three rooks on a chessboard with given values in each cell.

We can solve this problem by iterating through each possible combination of three non-attacking positions for the rooks on the chessboard. For each combination, we calculate the sum of values in those cells and keep track of the maximum sum found.

To ensure that the rooks do not attack each other, we can check if any two rooks are in the same row or column. If they are, we skip that combination.

After considering all possible combinations, we return the maximum sum found.

Time Complexity: O(m^3 * n^3), where m is the number of rows and n is the number of columns in the chessboard. Space Complexity: O(1)

Solution Code

class Solution {
    public int maxThreeRooks(int[][] board) {
        int m = board.length;
        int n = board[0].length;
        int maxSum = Integer.MIN_VALUE;

        for (int r1 = 0; r1 < m; r1++) {
            for (int c1 = 0; c1 < n; c1++) {
                for (int r2 = 0; r2 < m; r2++) {
                    for (int c2 = 0; c2 < n; c2++) {
                        for (int r3 = 0; r3 < m; r3++) {
                            for (int c3 = 0; c3 < n; c3++) {
                                if (r1 != r2 && r2 != r3 && r1 != r3 && c1 != c2 && c2 != c3 && c1 != c3) {
                                    maxSum = Math.max(maxSum, board[r1][c1] + board[r2][c2] + board[r3][c3]);
                                }
                            }
                        }
                    }
                }
            }
        }

        return maxSum;
    }
}

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