<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>

class Solution:
    def maximumValueSum(self, board: List[List[int]]) -> int:
        n_r, n_c = len(board), len(board[0])

        best_threes = []
        for row in board:
            pq = []
            for i, val in enumerate(row):
                if len(pq) < 3:
                    heapq.heappush(pq, (val, i))
                else:
                    heapq.heappush(pq, (val, i))
                    heapq.heappop(pq)
            best_three = []
            while pq:
                best_three.append(heapq.heappop(pq))
            best_threes.append(best_three)
        
        @cache
        def dfs(i, rem, used_cols):
            if i == n_r:
                return -inf
            output = dfs(i + 1, rem, used_cols)
            if rem == 1:
                for c_i in range(3):
                    val, index = best_threes[i][c_i]
                    if index not in used_cols:
                        output = max(output, val)
            else:
                for c_i in range(3):
                    val, index = best_threes[i][c_i]
                    if index not in used_cols:
                        if used_cols:
                            if used_cols[0] < index:
                                new_used_cols = used_cols + (index, )
                            else:
                                new_used_cols = (index, ) + used_cols
                        else:
                            new_used_cols = (index, ) 
                        output = max(output, dfs(i + 1, rem - 1, new_used_cols) + val)
            
            return output


        return dfs(0, 3, tuple())
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