<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:

// BFS gave TLE
    int mod = 1000000007;
    int dp[1001][1001];
    int n,m;
    int dfs(int i, int j, vector<vector<int>>& grid, int pre){
       if(i<0 || i>=n || j<0 || j>=m || pre >= grid[i][j]) return 0;
       if(dp[i][j] !=-1) return dp[i][j];

       long long temp = 1;
       temp += (dfs(i+1,j,grid,grid[i][j]) + dfs(i,j+1,grid,grid[i][j]) +
               dfs(i-1,j,grid,grid[i][j]) + dfs(i,j-1,grid,grid[i][j]))%mod;
        return dp[i][j] = temp%mod;       
    }

    int countPaths(vector<vector<int>>& grid) {
        n = grid.size();
        m = grid[0].size();
        long long ans = 0;
        memset(dp,-1,sizeof(dp));
        for(int i=0; i<n; i++) for(int j=0; j<m; j++) ans +=dfs(i,j,grid,-1);
        return ans%mod; 
    }
};









// class Solution {
// public:

// int dx[4] = {0,-1,0,1};
// int dy[4] = {1,0,-1,0};
// int mod = 1e9+7;

// bool isValid(int i, int j, int n, int m){
//     if(i<0 || i>=n || j<0 || j>=m) return false;
//     return true;
// }
//     int countPaths(vector<vector<int>>& grid) {
//         int n = grid.size();
//         int m = grid[0].size();

//         queue<vector<int>> q;
//         for(int i=0; i<n; i++) for(int j=0; j<m; j++) q.push({i,j});

//         int ans = 0;
//         while(!q.empty()){
//             int sz = q.size();
//             ans +=sz;
//             ans = ans%mod;

//             while(sz--){
//                 int i = q.front()[0];
//                 int j = q.front()[1];
//                 q.pop();

//                 for(int k=0; k<4; k++){
//                     int x = i+dx[k];
//                     int y = j+dy[k];
//                     if(isValid(x,y,n,m)){
//                         if(grid[x][y] > grid[i][j]) q.push({x,y});
//                     } 
//                 }
//             }
//         }
//         return ans;
//     }

// };
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