<p>You are given a <code>m x n</code> 2D array <code>board</code> representing a chessboard, where <code>board[i][j]</code> represents the <strong>value</strong> of the cell <code>(i, j)</code>.</p>

<p>Rooks in the <strong>same</strong> row or column <strong>attack</strong> each other. You need to place <em>three</em> rooks on the chessboard such that the rooks <strong>do not</strong> <strong>attack</strong> each other.</p>

<p>Return the <strong>maximum</strong> sum of the cell <strong>values</strong> on which the rooks are placed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = </span>[[-3,1,1,1],[-3,1,-3,1],[-3,2,1,1]]</p>

<p><strong>Output:</strong> 4</p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/08/rooks2.png" style="width: 294px; height: 450px;" /></p>

<p>We can place the rooks in the cells <code>(0, 2)</code>, <code>(1, 3)</code>, and <code>(2, 1)</code> for a sum of <code>1 + 1 + 2 = 4</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = [[1,2,3],[4,5,6],[7,8,9]]</span></p>

<p><strong>Output:</strong> <span class="example-io">15</span></p>

<p><strong>Explanation:</strong></p>

<p>We can place the rooks in the cells <code>(0, 0)</code>, <code>(1, 1)</code>, and <code>(2, 2)</code> for a sum of <code>1 + 5 + 9 = 15</code>.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = [[1,1,1],[1,1,1],[1,1,1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>We can place the rooks in the cells <code>(0, 2)</code>, <code>(1, 1)</code>, and <code>(2, 0)</code> for a sum of <code>1 + 1 + 1 = 3</code>.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= m == board.length &lt;= 100</code></li>
	<li><code>3 &lt;= n == board[i].length &lt;= 100</code></li>
	<li><code>-10<sup>9</sup> &lt;= board[i][j] &lt;= 10<sup>9</sup></code></li>
</ul>

import heapq
from functools import cache
class Solution:
    def maximumValueSum(self, board: List[List[int]]) -> int:
        largest = []
        rows, cols = len(board), len(board[0])
        for i, row in enumerate(board):
            largest.append(heapq.nlargest(3,range(cols), lambda x: row[x]))
        @cache
        def max_three(start_row):
            if len(largest) - start_row < 3:
                return float("-inf")
            cand1 = max_three(start_row+1)
            cand2 = max( board[start_row][c] + max_two(start_row+1, c)   for c in largest[start_row]) 
            #print(f"max_three:{start_row}: {max(cand1,cand2)}")
            return max(cand1, cand2)
        @cache
        def max_two(start_row, used_c):
            if len(largest) - start_row < 2:
                return float("-inf")
            cand1 = max_two(start_row+1, used_c)
            cand2 = max((board[start_row][c]+max_one(start_row+1, c, used_c) )   for c in largest[start_row] if c != used_c)
            #print(f"max_two:{start_row},used_c{used_c}: {max(cand1,cand2)}")
            return max(cand1,cand2)
        @cache
        def max_one(start_row, used_c1,used_c2):
            if start_row >= len(largest):
                return float("-inf")
            cand1 = max_one(start_row+1, used_c1, used_c2)
            cand2 = max(board[start_row][c]    for c in largest[start_row] if c not in [used_c1,used_c2])
            #print(f"one:{start_row},{used_c1},{used_c2}: {max(cand1,cand2)}")
            return max(cand1, cand2)
        return max_three(0)
86003