Class Solution

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    public final class Solution
    
                        

    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 <code>10<sup>9</sup> + 7</code>.

    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

    • <code>1 <= m, n <= 5 * 10<sup>4</sup></code>

    • <code>1 <= m * n <= 5 * 10<sup>4</sup></code>

    • 0 &lt;= grid[i][j] &lt;= 100

    • 1 &lt;= k &lt;= 50

    • Nested Class Summary

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      Modifier and Type Class Description
    • Field Summary

      Fields 
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    • Constructor Summary

      Constructors 
      Constructor Description
      Solution()
    • Enum Constant Summary

      Enum Constants 
      Enum Constant Description
    • Method Summary

      Modifier and Type Method Description
      final Integer numberOfPaths(Array<IntArray> grid, Integer k)
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        clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait