Package g2401_2500.s2435_paths_in_matrix_whose_sum_is_divisible_by_k

Class Solution

java.lang.Object
g2401_2500.s2435_paths_in_matrix_whose_sum_is_divisible_by_k.Solution
public class Solution extends java.lang.Object
2435 - Paths in Matrix Whose Sum Is Divisible by K.

Hard

You are given a 0-indexed m x n integer matrix grid and an integer k. You are currently at position (0, 0) and you want to reach position (m - 1, n - 1) moving only down or right.

Return the number of paths where the sum of the elements on the path is divisible by k. Since the answer may be very large, return it modulo 109 + 7.

Example 1:

Input: grid = [[5,2,4],[3,0,5],[0,7,2]], k = 3

Output: 2

Explanation: There are two paths where the sum of the elements on the path is divisible by k. The first path highlighted in red has a sum of 5 + 2 + 4 + 5 + 2 = 18 which is divisible by 3. The second path highlighted in blue has a sum of 5 + 3 + 0 + 5 + 2 = 15 which is divisible by 3.

Example 2:

Input: grid = [[0,0]], k = 5

Output: 1

Explanation: The path highlighted in red has a sum of 0 + 0 = 0 which is divisible by 5.

Example 3:

Input: grid = [[7,3,4,9],[2,3,6,2],[2,3,7,0]], k = 1

Output: 10

Explanation: Every integer is divisible by 1 so the sum of the elements on every possible path is divisible by k.

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 5 * 104
  • 1 <= m * n <= 5 * 104
  • 0 <= grid[i][j] <= 100
  • 1 <= k <= 50

  • Constructor Summary

    Constructors
    Constructor
    Description
    Solution()
     

  • Method Summary

    Modifier and Type
    Method
    Description
    int
    numberOfPaths(int[][] grid, int k)
     

    Methods inherited from class java.lang.Object

    clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • numberOfPaths

      public int numberOfPaths(int[][] grid, int k)