<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 dfs(self, row, col, grid):

        # print('We at row: ' + str(row) + ' and col: ' + str(col))

        if self.cache[row][col] != 0:
            return self.cache[row][col]

        # ans = 0
        ans = 1

        for oneDir in self.dirs:

            nextRow = row + oneDir[0]
            nextCol = col + oneDir[1]

            # bounds and incrment check
            if (nextRow >= 0 and nextCol >= 0 and 
               nextRow < self.numRows and nextCol < self.numCols and
               grid[row][col] < grid[nextRow][nextCol]):

            #    oneAns = grid[row][col] + dfs(nextRow, nextCol, grid) + 1
            #    oneAns = ans + self.dfs(nextRow, nextCol, grid) 
            #+ 1
            #    ans = max(oneAns, ans)

            #    ans = dfs(nextRow, nextCol, grid) + 1
            #    ans = dfs(nextRow, nextCol, grid)

                oneAns = self.dfs(nextRow, nextCol, grid) 
                ans = ans + oneAns
                # ans = ans%1000000007

        self.cache[row][col] = ans

        return ans

    def countPaths(self, grid: List[List[int]]) -> int:
        self.numRows = len(grid)
        self.numCols = len(grid[0])

        self.dirs = [(1, 0), (-1, 0), (0, -1), (0, 1)]
        self.cache = [[0]* self.numCols for i in range(0, self.numRows) ]

        # print(grid)
        # print(self.cache)

        ans = 0
        for row in range(0, self.numRows):
            for col in range(0, self.numCols):
                oneAns = self.dfs(row, col, grid) # + 1
                # ans = max(oneAns, ans)
                ans += oneAns 
                ans = ans%1000000007
                # pass

        # print()
        # print(self.cache)

        return ans
61015