<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:
    # TODO: review, understand & implement(?) binary search solutions

    def maxIncreasingCells(self, mat: List[List[int]]) -> int:
        # group all matrix positions by their values
        val_pos_map = defaultdict(list)
        for i in range(len(mat)):
            for j in range(len(mat[0])): val_pos_map[mat[i][j]].append((i, j))

        plens_y, plens_x = defaultdict(int), defaultdict(int)   # track running max path-lengths by row & column
        for val, positions in sorted(val_pos_map.items()):      # iterate through all values in ascending order
            cur_plens_y, cur_plens_x = defaultdict(int), defaultdict(int)   # track max path-length of current value
            for y, x in positions:
                curr_max = max(plens_y[y], plens_x[x]) + 1      # max path-len from curr-pos is +1 of the global running max
                cur_plens_y[y] = max(cur_plens_y[y], curr_max)  # update max path-length for curr-y coordinate of curr-val
                cur_plens_x[x] = max(cur_plens_x[x], curr_max)  # update max path-length for curr-x coordinate of curr-val

            # NOTE: it's crucial that we don't update the global running maximums until after we check
            #       all positions of the current value, as this guarantees the global running maximums
            #       always hold path lengths ending at a value strictly smaller than the current value
            plens_y.update(cur_plens_y)         # update global running y-coordinate maxes using curr-val's max-y's
            plens_x.update(cur_plens_x)         # update global running x-coordinate maxes using curr-val's max-x's
        
        return max(max(plens_y.values()), max(plens_x.values()))
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