<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>

import time
class Solution:
    def maxIncreasingCells(self, mat: List[List[int]]) -> int:
        R = len(mat)
        C = len(mat[0])
        dp = [[1 for _ in range(C)] for _ in range(R)]

        def get_values_and_locations():
            # Returns {value: list of all (r, c) with value} in an increasing order!
            values_and_locations = []
            for r in range(R):
                for c in range(C):
                    value = mat[r][c]
                    location = (r, c)
                    values_and_locations.append((value, location))
            values_and_locations = sorted(values_and_locations)

            # CPython dict is ordered by the order you insert!
            rv = {}
            for value, location in values_and_locations:
                rv.setdefault(value, []).append(location)
            return rv

        max_per_row = [0 for _ in range(R)]
        max_per_col = [0 for _ in range(C)]

        values_to_locations = get_values_and_locations()
        for val, locations in values_to_locations.items():
            for r, c in locations:
                dp[r][c] = 1 + max(max_per_row[r], max_per_col[c])

            for r, c in locations:
                max_per_row[r] = max(max_per_row[r], dp[r][c])
                max_per_col[c] = max(max_per_col[c], dp[r][c])

        rv = -1e9
        for r in range(R):
            for c in range(C):
                rv = max(rv, dp[r][c])
        return rv

1937