<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 {
public:
    int mod_val = pow(10,9) + 7;

    int dfs_on_node(vector<vector<int>>& grid, vector<vector<int>>& dp, int n, int m, int x, int y){
        // Already visited
        if (dp[x][y] != -1) return dp[x][y];

        // Check all 4 directions
        int paths[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        long result = 1;

        for (auto path : paths){
            int next_x = x + path[0];
            int next_y = y + path[1]; 

            // Check in bounds and greater
            if (next_x >= 0 && next_x < n && next_y >= 0 && next_y < m 
                && grid[next_x][next_y] > grid[x][y])       
            {
                result = (result + dfs_on_node(grid, dp, n, m, next_x, next_y)) % mod_val;
            }
        }

        return dp[x][y] = result;
    }

    int countPaths(vector<vector<int>>& grid) {
        /*  Idea use 2d array that is the same size as the board and
            calculate the increasing paths for each one in a dp approach.
            Use this to determine the number of increasing paths.
        */

        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> dp(n, vector<int>(m, -1));

        // Iterate over dp
        long num_paths = 0;
        for (int i = 0; i < n; ++i){
            for (int j = 0; j < m; ++j){

                if (dp[i][j] == -1){
                    dfs_on_node(grid, dp, n, m, i, j);
                }

                num_paths = (num_paths + dp[i][j]) % mod_val;
            }
        }

        return num_paths;
    }
};
82969