<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 countPaths(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        int count = 0;
        int mod = 1e9 + 7;
        vector<int> dr = {1, -1, 0, 0};
        vector<int> dc = {0, 0, 1, -1};
        set<pair<int, int>> path;
        vector<vector<int>> dp(m, vector<int>(n, -1));
        // considering every point as a starting point
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (dp[r][c] == -1) {
                    dfs(r, c, count, grid, path, mod, dr, dc, dp);
                }
                count = (count + dp[r][c]) % mod;
            }
        }
        return count % mod;
    }

private:
    int dfs(int r, int c, int& count, vector<vector<int>>& grid,
             set<pair<int, int>>& path, int mod, vector<int>& dr,
             vector<int>& dc, vector<vector<int>>& dp) {
        if (dp[r][c] != -1)
            return dp[r][c];
        int currCnt = 0;
        for (int i = 0; i < 4; i++) {
            int nr = r + dr[i];
            int nc = c + dc[i];
            if (nr >= 0 && nr < grid.size() && nc >= 0 && nc < grid[0].size() &&
                !path.count({nr, nc}) && grid[nr][nc] > grid[r][c]) {
                path.insert({nr, nc});
                currCnt = (currCnt + dfs(nr, nc, count, grid, path, mod, dr, dc, dp))%mod;
                path.erase({nr, nc});
            }
        }
        return dp[r][c] = (currCnt + 1)%mod;
    }
};
153130