# 2328. Number of Increasing Paths in a Grid

## Description

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 109 + 7.

Two paths are considered different if they do not have exactly the same sequence of visited cells.

Example 1:

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 <= 105
• 1 <= grid[i][j] <= 105

## Solutions

Solution 1: DFS + Memorization

We design a function $dfs(i, j)$, which represents the number of strictly increasing paths that can be reached from the grid graph starting at the $i$-th row and $j$-th column. Then the answer is $\sum_{i=0}^{m-1} \sum_{j=0}^{n-1} dfs(i, j)$. In the search process, we can use a two-dimensional array $f$ to record the calculated results to avoid repeated calculation.

The calculation process of the function $dfs(i, j)$ is as follows:

• If $f[i][j]$ is not $0$, it means that it has been calculated, and $f[i][j]$ is returned directly;
• Otherwise, we initialize $f[i][j] = 1$, and then enumerate the four directions of $(i, j)$. If the grid $(x, y)$ in a certain direction satisfies $0 \leq x \lt m$, $0 \leq y \lt n$, and $grid[i][j] \lt grid[x][y]$, we can start from the grid $(i, j)$ to the grid $(x, y)$, and the number on the path is strictly increasing, so $f[i][j] += dfs(x, y)$.

Finally, we return $f[i][j]$.

The answer is $\sum_{i=0}^{m-1} \sum_{j=0}^{n-1} dfs(i, j)$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns in the grid graph, respectively.

• 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;
}
}

• class Solution {
public:
int countPaths(vector<vector<int>>& grid) {
const int mod = 1e9 + 7;
int m = grid.size(), n = grid[0].size();
int f[m][n];
memset(f, 0, sizeof(f));
function<int(int, int)> dfs = [&](int i, int j) -> int {
if (f[i][j]) {
return f[i][j];
}
int ans = 1;
int dirs[5] = {-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;
};
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;
}
};

• 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


• 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
}

• 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;
}