LeetCode 3256: Maximum Value Sum by Placing Three Rooks I Solution

Master LeetCode problem 3256 (Maximum Value Sum by Placing Three Rooks I), 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.

3256. Maximum Value Sum by Placing Three Rooks I

Hard

Array Dynamic Programming Matrix Enumeration

Problem Explanation

Explanation

To solve this problem, we need to find the maximum sum of cell values by placing three rooks on the chessboard without attacking each other. We can approach this problem by iterating through all possible combinations of three rook positions on the board and calculating the sum of cell values for each combination. We will keep track of the maximum sum found and return it as the result.

Algorithm:

Initialize a variable maxSum to store the maximum sum found.
Iterate through all combinations of three rook positions on the board.
For each combination, calculate the sum of cell values at those positions.
Update maxSum if the calculated sum is greater.
Return maxSum as the final result.

Time Complexity:

The time complexity of this approach is O(m*n choose 3), where m is the number of rows and n is the number of columns in the board.

Space Complexity:

The space complexity is O(1) as we are not using any extra space other than a few variables.

Solution Code

class Solution {
    public int maxThreeRooks(int[][] board) {
        int m = board.length;
        int n = board[0].length;
        int maxSum = 0;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < m; k++) {
                    for (int l = 0; l < n; l++) {
                        for (int p = 0; p < m; p++) {
                            for (int q = 0; q < n; q++) {
                                if (i != k && j != l && i != p && j != q && k != p && l != q) {
                                    maxSum = Math.max(maxSum, board[i][j] + board[k][l] + board[p][q]);
                                }
                            }
                        }
                    }
                }
            }
        }
        
        return maxSum;
    }
}

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