Number of Increasing Paths in a Grid

// Accepted solution for LeetCode #2328: Number of Increasing Paths in a Grid
class Solution {
    private int[][] f;
    private int[][] grid;
    private int m;
    private int n;
    private final int mod = (int) 1e9 + 7;

    public int countPaths(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        f = new int[m][n];
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans = (ans + dfs(i, j)) % mod;
            }
        }
        return ans;
    }

    private int dfs(int i, int j) {
        if (f[i][j] != 0) {
            return f[i][j];
        }
        int ans = 1;
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[i][j] < grid[x][y]) {
                ans = (ans + dfs(x, y)) % mod;
            }
        }
        return f[i][j] = ans;
    }
}

// Accepted solution for LeetCode #2328: Number of Increasing Paths in a Grid
func countPaths(grid [][]int) (ans int) {
	const mod = 1e9 + 7
	m, n := len(grid), len(grid[0])
	f := make([][]int, m)
	for i := range f {
		f[i] = make([]int, n)
	}
	var dfs func(int, int) int
	dfs = func(i, j int) int {
		if f[i][j] != 0 {
			return f[i][j]
		}
		f[i][j] = 1
		dirs := [5]int{-1, 0, 1, 0, -1}
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < m && y >= 0 && y < n && grid[i][j] < grid[x][y] {
				f[i][j] = (f[i][j] + dfs(x, y)) % mod
			}
		}
		return f[i][j]
	}
	for i, row := range grid {
		for j := range row {
			ans = (ans + dfs(i, j)) % mod
		}
	}
	return
}

# Accepted solution for LeetCode #2328: Number of Increasing Paths in a Grid
class Solution:
    def countPaths(self, grid: List[List[int]]) -> int:
        @cache
        def dfs(i: int, j: int) -> int:
            ans = 1
            for a, b in pairwise((-1, 0, 1, 0, -1)):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and grid[i][j] < grid[x][y]:
                    ans = (ans + dfs(x, y)) % mod
            return ans

        mod = 10**9 + 7
        m, n = len(grid), len(grid[0])
        return sum(dfs(i, j) for i in range(m) for j in range(n)) % mod

// Accepted solution for LeetCode #2328: Number of Increasing Paths in a Grid
/**
 * [2328] Number of Increasing Paths in a Grid
 *
 * You are given an m x n integer matrix grid, where you can move from a cell to any adjacent cell in all 4 directions.
 * Return the number of strictly increasing paths in the grid such that you can start from any cell and end at any cell. Since the answer may be very large, return it modulo 10^9 + 7.
 * Two paths are considered different if they do not have exactly the same sequence of visited cells.
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2022/05/10/griddrawio-4.png" style="width: 181px; height: 121px;" />
 * Input: grid = [[1,1],[3,4]]
 * Output: 8
 * Explanation: The strictly increasing paths are:
 * - Paths with length 1: [1], [1], [3], [4].
 * - Paths with length 2: [1 -> 3], [1 -> 4], [3 -> 4].
 * - Paths with length 3: [1 -> 3 -> 4].
 * The total number of paths is 4 + 3 + 1 = 8.
 *
 * Example 2:
 *
 * Input: grid = [[1],[2]]
 * Output: 3
 * Explanation: The strictly increasing paths are:
 * - Paths with length 1: [1], [2].
 * - Paths with length 2: [1 -> 2].
 * The total number of paths is 2 + 1 = 3.
 *
 *  
 * Constraints:
 *
 * 	m == grid.length
 * 	n == grid[i].length
 * 	1 <= m, n <= 1000
 * 	1 <= m * n <= 10^5
 * 	1 <= grid[i][j] <= 10^5
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/number-of-increasing-paths-in-a-grid/
// discuss: https://leetcode.com/problems/number-of-increasing-paths-in-a-grid/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn count_paths(grid: Vec<Vec<i32>>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2328_example_1() {
        let grid = vec![vec![1, 1], vec![3, 4]];

        let result = 8;

        assert_eq!(Solution::count_paths(grid), result);
    }

    #[test]
    #[ignore]
    fn test_2328_example_2() {
        let grid = vec![vec![1], vec![2]];

        let result = 3;

        assert_eq!(Solution::count_paths(grid), result);
    }
}

// Accepted solution for LeetCode #2328: Number of Increasing Paths in a Grid
function countPaths(grid: number[][]): number {
    const mod = 1e9 + 7;
    const m = grid.length;
    const n = grid[0].length;
    const f = new Array(m).fill(0).map(() => new Array(n).fill(0));
    const dfs = (i: number, j: number): number => {
        if (f[i][j]) {
            return f[i][j];
        }
        let ans = 1;
        const dirs: number[] = [-1, 0, 1, 0, -1];
        for (let k = 0; k < 4; ++k) {
            const x = i + dirs[k];
            const y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[i][j] < grid[x][y]) {
                ans = (ans + dfs(x, y)) % mod;
            }
        }
        return (f[i][j] = ans);
    };
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            ans = (ans + dfs(i, j)) % mod;
        }
    }
    return ans;
}