<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:
        directions = [(-1, 0), (1, 0), (0, 1), (0, -1)] # Left, right, top, bottom
        mod = 10 ** 9 + 7
        num_rows, num_cols = len(grid), len(grid[0])
        cache = [[-1] * num_cols for i in range(num_rows)]
        total_inc_paths = 0
        # Iterate over all cells grid[i][j] and sum over dfs(r, c).
        for r in range(num_rows):
            for c in range(num_cols):
                total_inc_paths += self.dfs(grid, r, c, cache, directions) 
        
        return total_inc_paths % mod

    def dfs(self, grid, r, c, cache, directions):
        mod = 10 ** 9 + 7

        # If cache[i][j] is non-zero, it means we have got the value of dfs(i, j),
        # so just return dp[i][j].
        if cache[r][c] != -1:
            return cache[r][c]

        # Otherwise, set answer = 1, the path made of grid[i][j] itself.
        inc_paths = 1

        # Check its four neighbor cells, if a neighbor cell grid[prevI][prevJ] 
        # has a smaller value, we move to this cell and solve the subproblem: 
        # dfs(prevI, prevJ).
        for (r_dir, c_dir) in directions:
            if r+r_dir >= 0 and c+c_dir >= 0 and \
                r+r_dir < len(grid) and c+c_dir < len(grid[0]) and \
                grid[r+r_dir][c+c_dir] < grid[r][c]:
                inc_paths += self.dfs(grid, r+r_dir, c+c_dir, cache, directions) % mod
        # Update cache[i][j], so that we don't recalculate its value later.
        cache[r][c] = inc_paths
        return inc_paths
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