<p>You are given an <code>m x n</code> integer matrix <code>grid</code>, where you can move from a cell to any adjacent cell in all <code>4</code> directions.</p>

<p>Return <em>the number of <strong>strictly</strong> <strong>increasing</strong> paths in the grid such that you can start from <strong>any</strong> cell and end at <strong>any</strong> cell. </em>Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>Two paths are considered different if they do not have exactly the same sequence of visited cells.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/10/griddrawio-4.png" style="width: 181px; height: 121px;" />
<pre>
<strong>Input:</strong> grid = [[1,1],[3,4]]
<strong>Output:</strong> 8
<strong>Explanation:</strong> The strictly increasing paths are:
- Paths with length 1: [1], [1], [3], [4].
- Paths with length 2: [1 -&gt; 3], [1 -&gt; 4], [3 -&gt; 4].
- Paths with length 3: [1 -&gt; 3 -&gt; 4].
The total number of paths is 4 + 3 + 1 = 8.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1],[2]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The strictly increasing paths are:
- Paths with length 1: [1], [2].
- Paths with length 2: [1 -&gt; 2].
The total number of paths is 2 + 1 = 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 1000</code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def countPaths(self, grid: List[List[int]]) -> int:
        #dp?
        rows=len(grid)
        cols=len(grid[0])

        #memoization grid- initialized with 1 since each cell is treated length 1 path as default
        dp=[[-1]*cols for _ in range(rows)]
        #let dp[i][j] equal the number of strictly increasing paths starting from 
        #row i, col j

        #let dfs return the number of strictly increasing paths
        def dfs(r,c,prev):
            #base case- out of bounds or current cell val is less than previous cell val
            if (r<0 or c<0 or r>=rows or c>=cols or grid[r][c]<=prev):
                return 0

            #check dp grid
            if dp[r][c]==-1:
                #search 4 directions and sum the number of strictly increasing paths
                dp[r][c]=(1+dfs(r+1,c,grid[r][c])+
                            dfs(r-1,c,grid[r][c])+
                            dfs(r,c-1,grid[r][c])+
                            dfs(r,c+1,grid[r][c])
                        )
            return dp[r][c]
        
        res=0
        for i in range(rows):
            for j in range(cols):
                #dfs
                res+=dfs(i,j,-1)

        #print(dp)
        #print(res)
        return res%(10**9+7)
1232