<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>

class Solution:
    def maxIncreasingCells(self, mat: List[List[int]]) -> int:
        pass


class Solution:
    def maxIncreasingCells(self, mat: List[List[int]]) -> int:
        R, C = len(mat), len(mat[0])

        hmap = defaultdict(list)
        for i in range(R) :
            for j in range(C) :
                hmap[mat[i][j]].append((i, j))
        
        dp = [[0 for j in range(C)] for i in range(R)]
        row_res, col_res = [0 for i in range(R)], [0 for j in range(C)]

        for num in sorted(hmap.keys()) :
            for i, j in hmap[num] :
                dp[i][j] = max(row_res[i], col_res[j]) + 1
            
            for i, j in hmap[num] :
                row_res[i] = max(row_res[i], dp[i][j])
                col_res[j] = max(col_res[j], dp[i][j])
        
        return max(max(row_res), max(col_res))



class Solution:
    def maxIncreasingCells(self, mat: List[List[int]]) -> int:
        m, n = len(mat), len(mat[0])
        
        vals = collections.defaultdict(list)
        for i in range(m):
            for j in range(n):
                vals[mat[i][j]].append((i, j))
        
        dp = [[0] * n for _ in range(m)]
        track_row = [0] * m
        track_col = [0] * n

        for num in sorted(vals.keys()):
            for i, j in vals[num]:
                dp[i][j] = max(track_row[i], track_col[j]) + 1
            
            for i, j in vals[num]:
                track_row[i] = max(dp[i][j], track_row[i])
                track_col[j] = max(dp[i][j], track_col[j])
        return max(max(track_row), max(track_col))
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