<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 dx[4]={-1,1,0,0};
    int dy[4]={0,0,-1,1};
    const int mod=1e9+7;
    bool check(int i,int j,int n,int m){
        if(i>=0 && j>=0 && i<n && j<m) 
        return true;
        return false;
    }
    void  dfs(int node,vector<int>&dp,vector<int>adj[]){
        if(dp[node]!=-1)
        return;
        dp[node]=1;
        for(auto child:adj[node]){
           dfs(child,dp,adj);
           dp[node]=(dp[node]+dp[child])%mod;
        }
    }
    int countPaths(vector<vector<int>>& matrix) {
        int n=matrix.size();
        int m=matrix[0].size();
        int N=n*m;
        vector<int>adj[N];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                for(int k=0;k<4;k++){
                    int ni=i+dx[k];
                    int nj=j+dy[k];
                    if(check(ni,nj,n,m) && matrix[i][j]<matrix[ni][nj]){
                        int u=(i*m) + j;
                        int v=(ni*m)+ nj;
                        adj[u].push_back(v);
                    }
                }
            }
        }
        vector<int>dp(N,-1);
        int ans=0;
        for(int i=0;i<N;i++){
            if(dp[i]==-1){
                dfs(i,dp,adj);
            }
        }
        for(int i=0;i<N;i++){
            ans=(ans+dp[i])%mod;
        }
        return ans;
    }
}; 
109516