LeetCode in Kotlin

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

Solution

class Solution {
    private val mod = (1e9 + 7).toInt()
    private var row = 0
    private var col = 0
    private lateinit var cache: Array<Array<IntArray>>

    fun numberOfPaths(grid: Array<IntArray>, k: Int): Int {
        row = grid.size
        col = grid[0].size
        cache = Array(row) { Array(col) { IntArray(k) { -1 } } }

        return numberOfPaths(grid, 0, 0, k, 0)
    }

    // return the number of path with <Sum([r][c] ~ [ROW][COL]) % k == remainder>
    private fun numberOfPaths(grid: Array<IntArray>, r: Int, c: Int, k: Int, remainder: Int): Int {
        if (r to c !in grid) return 0
        if (cache[r][c][remainder] != -1) return cache[r][c][remainder]
        if (r == row - 1 && c == col - 1)
            return if (grid[r][c] % k == remainder) 1 else 0

        return ((remainder - grid[r][c] + 100 * k) % k).let {
            (numberOfPaths(grid, r + 1, c, k, it) + numberOfPaths(grid, r, c + 1, k, it)) % mod
        }.also {
            cache[r][c][remainder] = it
        }
    }

    private operator fun Array<IntArray>.contains(pair: Pair<Int, Int>): Boolean {
        return (0 <= pair.first && pair.first < this.size) &&
            (0 <= pair.second && pair.second < this[0].size)
    }
}

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