feat: add typescript solution to lc problem: No.2435

No.2435.Paths in Matrix Whose Sum Is Divisible by K

Author: zhaocchen

solution/2400-2499/2435.Paths in Matrix Whose Sum Is Divisible by K/README.md

@@ -324,7 +324,21 @@ func numberOfPaths(grid [][]int, k int) int {
 ### **TypeScript**
 
 ```ts
-
+function numberOfPaths(grid: number[][], k: number): number {
+  const MOD = 10 ** 9 + 7;
+  const m = grid.length, n = grid[0].length;
+  let ans = Array.from({ length: m + 1 }, () => Array.from({ length: n + 1 }, () => new Array(k).fill(0)));
+  ans[0][1][0] = 1;
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      for (let v = 0; v < k; v++) {
+        let key = (grid[i][j] + v) % k;
+        ans[i + 1][j + 1][key] = (ans[i][j + 1][v] + ans[i + 1][j][v] + ans[i + 1][j + 1][key]) % MOD;
+      }
+    }
+  }
+  return ans[m][n][0];
+};
 ```
 
 ### **...**

solution/2400-2499/2435.Paths in Matrix Whose Sum Is Divisible by K/README_EN.md

@@ -285,7 +285,21 @@ func numberOfPaths(grid [][]int, k int) int {
 ### **TypeScript**
 
 ```ts
-
+function numberOfPaths(grid: number[][], k: number): number {
+  const MOD = 10 ** 9 + 7;
+  const m = grid.length, n = grid[0].length;
+  let ans = Array.from({ length: m + 1 }, () => Array.from({ length: n + 1 }, () => new Array(k).fill(0)));
+  ans[0][1][0] = 1;
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      for (let v = 0; v < k; v++) {
+        let key = (grid[i][j] + v) % k;
+        ans[i + 1][j + 1][key] = (ans[i][j + 1][v] + ans[i + 1][j][v] + ans[i + 1][j + 1][key]) % MOD;
+      }
+    }
+  }
+  return ans[m][n][0];
+};
 ```
 
 ### **...**

solution/2400-2499/2435.Paths in Matrix Whose Sum Is Divisible by K/Solution.ts

@@ -0,0 +1,15 @@
+function numberOfPaths(grid: number[][], k: number): number {
+    const MOD = 10 ** 9 + 7;
+    const m = grid.length, n = grid[0].length;
+    let ans = Array.from({ length: m + 1 }, () => Array.from({ length: n + 1 }, () => new Array(k).fill(0)));
+    ans[0][1][0] = 1;
+    for (let i = 0; i < m; i++) {
+      for (let j = 0; j < n; j++) {
+        for (let v = 0; v < k; v++) {
+          let key = (grid[i][j] + v) % k;
+          ans[i + 1][j + 1][key] = (ans[i][j + 1][v] + ans[i + 1][j][v] + ans[i + 1][j + 1][key]) % MOD;
+        }
+      }
+    }
+    return ans[m][n][0];
+  };