<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;= 500</code></li>
	<li><code>3 &lt;= n == board[i].length &lt;= 500</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:
        lst = []
        m = len(board)
        n = len(board[0])
        for i in range(m):
            for j in range(n):
                lst.append((board[i][j], (i, j)))

        lst.sort()
        lst.reverse()

        # print(lst)

        cur_best = -inf
        def dfs(i, selected, curval):
            # print(i, selected, curval)
            nonlocal cur_best
            if len(selected) == 3:
                cur_best = max(cur_best, curval)
                return
            if len(lst) - i + len(selected) < 3:
                return
            val = lst[i][0]
            if len(selected) == 2 and curval + val <= cur_best:
                return
            if len(selected) == 1 and curval + val + lst[i + 1][0] <= cur_best:
                return
            if len(selected) == 0 and val + lst[i + 1][0] + lst[i + 1][0] <= cur_best:
                return
            x, y = lst[i][1]
            for idx in selected:
                x2, y2 = lst[idx][1]
                if x == x2 or y == y2:
                    dfs(i + 1, selected, curval)
                    return
            dfs(i + 1, selected + [i], curval + val)
            dfs(i + 1, selected, curval)

        # print(cur_best)
        dfs(0, [], 0)
        return cur_best
67993