<p>Given a <strong>1-indexed</strong>&nbsp;<code>m x n</code> integer matrix <code>mat</code>, you can select any cell in the matrix as your <strong>starting cell</strong>.</p>

<p>From the starting cell, you can move to any other cell <strong>in the</strong> <strong>same row or column</strong>, but only if the value of the destination cell is <strong>strictly greater</strong> than the value of the current cell. You can repeat this process as many times as possible, moving from cell to cell until you can no longer make any moves.</p>

<p>Your task is to find the <strong>maximum number of cells</strong> that you can visit in the matrix by starting from some cell.</p>

<p>Return <em>an integer denoting the maximum number of cells that can be visited.</em></p>

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

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/23/diag1drawio.png" style="width: 200px; height: 176px;" /></strong></p>

<pre>
<strong>Input:</strong> mat = [[3,1],[3,4]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The image shows how we can visit 2 cells starting from row 1, column 2. It can be shown that we cannot visit more than 2 cells no matter where we start from, so the answer is 2. 
</pre>

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

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/23/diag3drawio.png" style="width: 200px; height: 176px;" /></strong></p>

<pre>
<strong>Input:</strong> mat = [[1,1],[1,1]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> Since the cells must be strictly increasing, we can only visit one cell in this example. 
</pre>

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

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/23/diag4drawio.png" style="width: 350px; height: 250px;" /></strong></p>

<pre>
<strong>Input:</strong> mat = [[3,1,6],[-9,5,7]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The image above shows how we can visit 4 cells starting from row 2, column 1. It can be shown that we cannot visit more than 4 cells no matter where we start from, so the answer is 4. 
</pre>

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

<ul>
	<li><code>m == mat.length&nbsp;</code></li>
	<li><code>n == mat[i].length&nbsp;</code></li>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>5</sup>&nbsp;&lt;= mat[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

from collections import defaultdict

class Solution:
    def maxIncreasingCells(self, mat: List[List[int]]) -> int:
        nums_with_indices = []
        for row_idx in range(len(mat)):
            for col_idx in range(len(mat[0])):
                nums_with_indices.append((mat[row_idx][col_idx], row_idx, col_idx))
        nums_with_indices = sorted(nums_with_indices)

# N^2 vals * N
        dp = defaultdict(int)
        best_in_col = defaultdict(int)
        best_in_row = defaultdict(int)
        val_idx = 0
        while val_idx < len(nums_with_indices):
            best_candidates = []
            prev = nums_with_indices[val_idx][0]
            while val_idx < len(nums_with_indices) and nums_with_indices[val_idx][0] == prev:
                val, row_idx, col_idx = nums_with_indices[val_idx]
                best_can_get = max(best_in_col[col_idx], best_in_row[row_idx])
                dp[(row_idx, col_idx)] = best_can_get + 1
                val_idx += 1
                best_candidates.append((row_idx, col_idx))
            # print(dp)
            for cand_row, cand_col in best_candidates:
                best_in_row[cand_row] = max(dp[(cand_row, cand_col)], best_in_row[cand_row])
                best_in_col[cand_col] = max(dp[(cand_row, cand_col)], best_in_col[cand_col])
            # print(dp, best_in_row, best_in_col)
        return max(dp.values())
67700