<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, m = len(board), len(board[0])
        
        arr = []
        for i in range(n):
            for j in range(m):
                arr.append((board[i][j], i, j))
        arr.sort()
        arr.reverse()
        mx = min(90, max(n, m) * 3)
        thres = arr[mx-1][0]
        while arr and arr[-1][0] < thres:
            arr.pop()
       
        
        
        def rev(idx):
            # return board[idx//m][idx%m]
            return arr[idx][0]
        
        @functools.lru_cache(None)
        def dfs(d, tup):
            if len(tup) == 3:
                return sum(rev(val) for val in tup)
            # if d == n*m:
            if d == len(arr):
                return -10 ** 12
            nxt1 = dfs(d+1, tup)
            
            # r, c = d // m, d % m
            r = arr[d][1]
            c = arr[d][2]
            for val in tup:
                if arr[val][1] == r or arr[val][2] == c:
                    return nxt1
                # if val // m == r or val % m == c:
                #     return nxt1            
             
            tup2 = (*tup, d)
            nxt2 = dfs(d+1, tup2)
            return max(nxt1, nxt2)
        
        empty = tuple()
        return dfs(0, empty)
67575