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 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 and Description
Solution()

Method Summary

Modifier and Type	Method and 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 Detail

Solution

public Solution()

Method Detail

numberOfPaths

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