<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:
        m = len(board)
        n = len(board[0])
        
        # Check every combo of rows (2 * 10^7)
        # DP: store the top three indices in each row
        top = [[(float('-inf'), -1)] * 3 for _ in range(m)]
        for i in range(m):
            for j in range(n):
                if board[i][j] > top[i][0][0]:
                    top[i][2] = top[i][1]
                    top[i][1] = top[i][0]
                    top[i][0] = (board[i][j], j)
                elif board[i][j] > top[i][1][0]:
                    top[i][2] = top[i][1]
                    top[i][1] = (board[i][j], j)
                elif board[i][j] > top[i][2][0]:
                    top[i][2] = (board[i][j], j)
                    
        #print(top)
        # Then basically get those nine elements and take the top 3 until they are not all in the same row
        lst = list(itertools.combinations(range(m), 3))
        
        res = float('-inf')
        for x, y, z in lst:
            lst1 = top[x]
            lst2 = top[y]
            lst3 = top[z]
            
            for i in lst1:
                for j in lst2:
                    for k in lst3:
                        if i[1] != j[1] and i[1] != k[1] and j[1] != k[1]:
                            res = max(res, i[0] + j[0] + k[0])
        
        return res
        
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