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-rw-r--r--other/num_increasing_path.cpp54
1 files changed, 0 insertions, 54 deletions
diff --git a/other/num_increasing_path.cpp b/other/num_increasing_path.cpp
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--- a/other/num_increasing_path.cpp
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-// 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
-}