1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
|
// 2328. Number of Increasing Paths in a Grid
#include <stdio.h>
#include <vector>
using namespace std;
/*
* given an 'm x n' integer matrix 'grid' where you can move from a cell to any
* adjacent cell in all 4 directions. return th 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.
*/
class Solution {
int mod = 1e9 + 7;
int dfs(vector<vector<int>> &grid, int i, int j, vector<vector<int>> &dp,
int prev) {
if (i < 0 || j < 0 || i >= grid.size() || j >= grid[0].size() ||
prev >= grid[i][j])
return 0;
if (dp[i][j] != -1)
return dp[i][j];
int top = dfs(grid, i - 1, j, dp, grid[i][j]);
int bottom = dfs(grid, i + 1, j, dp, grid[i][j]);
int right = dfs(grid, i, j + 1, dp, grid[i][j]);
int left = dfs(grid, i, j - 1, dp, grid[i][j]);
dp[i][j] = (1 + top + bottom + right + left) % mod;
return dp[i][j];
}
public:
int countPaths(vector<vector<int>> &grid) {
int m = grid.size(), n = grid[0].size(), ans = 0;
if (m == 1 && n == 1)
return 1;
vector<vector<int>> dp(m, vector<int>(n, -1));
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++) {
if (dp[i][j] == -1)
ans = (ans + dfs(grid, i, j, dp, -1)) % mod;
else
ans = (ans + dp[i][j]) % mod;
}
return ans;
}
};
int main() {
Solution obj;
vector<vector<int>> g1 = {{1, 1}, {3, 4}}, g2 = {{1}, {2}};
printf("%d\n", obj.countPaths(g1)); // expect: 8
printf("%d\n", obj.countPaths(g2)); // expect: 3
}
|