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-rw-r--r--other/num_increasing_path.cpp54
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diff --git a/other/num_increasing_path.cpp b/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
+}