<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:
        mylist = []
        rowlist = [0] * len(mat)
        collist = [0] * len(mat[0])
        for i in range(len(mat)):
            temp = [0] * len(mat[0])
            mylist.append(temp.copy())
        heap = []
        heapq.heapify(heap)
        for i in range(len(mat)):
            for j in range(len(mat[0])):
                heapq.heappush(heap, [mat[i][j], i, j])
        while len(heap) > 0:
            val, row, col = heapq.heappop(heap)
            templist = [[row, col]]
            while len(heap) > 0 and heap[0][0] == val:
                _, row, col = heapq.heappop(heap)
                templist.append([row, col])
            for row, col in templist:
                mylist[row][col] = max(mylist[row][col], rowlist[row]+1)
                mylist[row][col] = max(mylist[row][col], collist[col]+1)
            for row, col in templist:
                rowlist[row] = max(rowlist[row], mylist[row][col])
                collist[col] = max(collist[col], mylist[row][col])
        return max(max(rowlist), max(collist))
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