<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:
    vector<vector<int>> dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    const int MOD = 1e9+7;

    int solve(int i, int j, vector<vector<int>>& grid, vector<vector<int>>& dp, int n, int m) {
        if(dp[i][j] != -1) return dp[i][j];  // If already computed, return stored result

        int cnt = 1;  // Each cell is at least one path by itself

        for(auto dir : dirs) {
            int x = dir[0] + i;
            int y = dir[1] + j;

            // Check bounds and ensure the new cell has a strictly larger value
            if(x >= 0 && y >= 0 && x < n && y < m && grid[x][y] > grid[i][j]) {
                cnt = (cnt + solve(x, y, grid, dp, n, m)) % MOD;
            }
        }

        return dp[i][j] = cnt;  // Store result in dp and return it
    }   

    int countPaths(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();

        vector<vector<int>> dp(n, vector<int>(m, -1));  // Memoization array

        int totalPaths = 0;

        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                totalPaths = (totalPaths + solve(i, j, grid, dp, n, m)) % MOD;  // Accumulate total paths
            }
        }

        return totalPaths;
    }
};

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