3257. Maximum Value Sum by Placing Three Rooks II
Description
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 <= 5003 <= n == board[i].length <= 500-109 <= board[i][j] <= 109
Solutions
Solution 1
Thinking
Same as I, but the side length is \(500\), so even a few candidates per row make row-triple enumeration tight. We further keep top cells per column, or enumerate only globally large cells.
The rooks still need distinct rows and columns; candidates come from the top few of each row or column. There is no implementation in the tree yet; the reasoning is to shrink candidates, then check conflicts.
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