Introduction

In this blog post, we will explore how to solve the LeetCode problem "3256" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a m x n 2D array board representing a chessboard, where board[i][j] represents the value of the cell (i, j). Rooks in the same row or column attack each other. You need to place three rooks on the chessboard such that the rooks do not attack each other. Return the maximum sum of the cell values on which the rooks are placed. Example 1: Input: board = [[-3,1,1,1],[-3,1,-3,1],[-3,2,1,1]] Output: 4 Explanation: We can place the rooks in the cells (0, 2), (1, 3), and (2, 1) for a sum of 1 + 1 + 2 = 4. Example 2: Input: board = [[1,2,3],[4,5,6],[7,8,9]] Output: 15 Explanation: We can place the rooks in the cells (0, 0), (1, 1), and (2, 2) for a sum of 1 + 5 + 9 = 15. Example 3: Input: board = [[1,1,1],[1,1,1],[1,1,1]] Output: 3 Explanation: We can place the rooks in the cells (0, 2), (1, 1), and (2, 0) for a sum of 1 + 1 + 1 = 3. Constraints: 3 <= m == board.length <= 100 3 <= n == board[i].length <= 100 -109 <= board[i][j] <= 109

Explanation

Here's the breakdown of the solution:

• Greedy Approach with Permutations: The core idea is to generate all possible placements of the three rooks such that they don't attack each other (i.e., they are in different rows and columns). This can be efficiently done by considering all possible permutations of columns for the given rows.

• Calculate Sum and Maximize: For each valid rook placement (permutation), calculate the sum of the cell values. Keep track of the maximum sum found so far.

• Optimization: We avoid unnecessary computations by directly generating valid placements using permutations, ensuring we only evaluate rook positions that satisfy the non-attacking condition.

• Runtime Complexity: O(m!), where m is the number of rows. In this case, m is bounded by 100, and since we need to place 3 rooks, the algorithm considers at most 100 99 98 combinations in the worst case, which simplifies to O(1). Storage Complexity: O(1)

Code

    def max_rook_sum(board):
    """
    Calculates the maximum sum of cell values for three non-attacking rooks on a chessboard.

    Args:
        board: A 2D array representing the chessboard.

    Returns:
        The maximum sum of cell values.
    """

    m = len(board)
    n = len(board[0])
    max_sum = float('-inf')

    # Iterate through all possible column permutations for placing 3 rooks in the first 3 rows (if available).
    import itertools
    for cols in itertools.permutations(range(n), 3):
        if m >= 3:
            current_sum = board[0][cols[0]] + board[1][cols[1]] + board[2][cols[2]]
            max_sum = max(max_sum, current_sum)
        else:
            return 0  # Not enough rows to place 3 rooks

    return max_sum