<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 m;
    int n; 
    int dp[1001][1001];
    long long MOD = 1e9 + 7;

    vector<vector<int>> directions = {
            {-1, 0}, // Top
            {1, 0},  // Bottom
            {0, -1}, // Left
            {0, 1},  // Right
    };
    bool isValidIndex(int i , int j){
        if(i >= 0 && i < m && j >= 0 && j < n){
            return true;
        }
        return false;
    }
    // DFS on gird
    int solveDFS(vector<vector<int>>& grid, int i, int j){
        int ans = 1; // Why 1 because single cell itself is an increasing path

        // Memoization
        if(dp[i][j] != -1){
            return dp[i][j];
        }

        for(auto &x : directions){
            int new_i = i + x[0];
            int new_j = j + x[1];

            if(isValidIndex(new_i, new_j) && grid[new_i][new_j] < grid[i][j]){
                // Further calling DFS
                ans = (ans + solveDFS(grid, new_i, new_j)) % MOD; // dry run to understand
            }
        }
        // cout << "grid[" << i << "][" << j << "] --> " <<  ans << endl;
        return dp[i][j] = ans;
    }
    int countPaths(vector<vector<int>>& grid) {
        // Incerasing paths of Length 1: 1,1,3,4
        // Increasing paths of Length 2: 1->3, 3->4, 1->4
        // Increasing paths of Length 3: 1->3->4

        // e.g i,j is 1,1 which is cell of 4, will call DFS in decreasing order --> single cell 4 itself is a path, then 4->1 is another decreasing path, and the last 4->3 is also decreasing path, in short for the i,j there are 3 strictly increasing 
        
        memset(dp, -1, sizeof(dp));

        m = grid.size();
        n = grid[0].size();

        int result = 0;
        for(int i = 0; i < m; i++){

            for(int j = 0; j < n; j++){
                result = (result + solveDFS(grid, i,j)) % MOD;
            }
        }

        return result;
    }
};
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