2328-number-of-increasing-paths-in-a-grid.java

//Exactly similar to : 329. Longest Increasing Path in a Matrix

class Solution {

    long mod = (long) Math.pow(10, 9) + 7;

    public int countPaths(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        long[][] dp = new long[m][n];
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (dp[i][j] == 0) {
                    dfs(grid, dp, i, j, m, n, -1);
                }
            }
        }
        long sum = 0;
        for (long[] d : dp) {
            for (long i : d) {
                sum = (sum + i) % mod;
            }
        }
        return (int) sum;
    }

    public long dfs(
        int[][] grid,
        long[][] dp,
        int i,
        int j,
        int m,
        int n,
        int prevValue
    ) {
        if (
            i < 0 || j < 0 || i >= m || j >= n || prevValue >= grid[i][j]
        ) return 0;
        if (dp[i][j] != 0) return dp[i][j];
        prevValue = grid[i][j];
        long left = dfs(grid, dp, i, j - 1, m, n, prevValue) % mod;
        long bottom = dfs(grid, dp, i + 1, j, m, n, prevValue) % mod;
        long right = dfs(grid, dp, i, j + 1, m, n, prevValue) % mod;
        long top = dfs(grid, dp, i - 1, j, m, n, prevValue) % mod;
        dp[i][j] = (1 + left + top + bottom + right) % mod;
        return dp[i][j];
    }
}