Maximum Strictly Increasing Cells in a Matrix
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2700-2799/2713.Maximum%20Strictly%20Increasing%20Cells%20in%20a%20Matrix/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2700-2799/2713.Maximum%20Strictly%20Increasing%20Cells%20in%20a%20Matrix/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2700-2799/2713.Maximum%20Strictly%20Increasing%20Cells%20in%20a%20Matrix/images/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>
Memoization; Array; Hash Table; Binary Search; Dynamic Programming; Matrix; Ordered Set; Sorting
func maxIncreasingCells(mat [][]int) (ans int) {
	m, n := len(mat), len(mat[0])
	g := map[int][][2]int{}
	for i, row := range mat {
		for j, v := range row {
			g[v] = append(g[v], [2]int{i, j})
		}
	}
	nums := make([]int, 0, len(g))
	for k := range g {
		nums = append(nums, k)
	}
	sort.Ints(nums)
	rowMax := make([]int, m)
	colMax := make([]int, n)
	for _, k := range nums {
		pos := g[k]
		mx := make([]int, len(pos))
		for i, p := range pos {
			mx[i] = max(rowMax[p[0]], colMax[p[1]]) + 1
			ans = max(ans, mx[i])
		}
		for i, p := range pos {
			rowMax[p[0]] = max(rowMax[p[0]], mx[i])
			colMax[p[1]] = max(colMax[p[1]], mx[i])
		}
	}
	return
}