LeetCode #2435 — HARD

Paths in Matrix Whose Sum Is Divisible by K

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[[5,2,4],[3,0,5],[0,7,2]]
3

Example 2

[[0,0]]
5

Example 3

[[7,3,4,9],[2,3,6,2],[2,3,7,0]]
1

Core Insight

What unlocks the optimal approach

The actual numbers in grid do not matter. What matters are the remainders you get when you divide the numbers by k.
We can use dynamic programming to solve this problem. What can we use as states?
Let dp[i][j][value] represent the number of paths where the sum of the elements on the path has a remainder of value when divided by k.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2435: Paths in Matrix Whose Sum Is Divisible by K
class Solution {
    public int numberOfPaths(int[][] grid, int K) {
        final int mod = (int) 1e9 + 7;
        int m = grid.length, n = grid[0].length;
        int[][][] f = new int[m][n][K];
        f[0][0][grid[0][0] % K] = 1;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                for (int k = 0; k < K; ++k) {
                    int k0 = ((k - grid[i][j] % K) + K) % K;
                    if (i > 0) {
                        f[i][j][k] = (f[i][j][k] + f[i - 1][j][k0]) % mod;
                    }
                    if (j > 0) {
                        f[i][j][k] = (f[i][j][k] + f[i][j - 1][k0]) % mod;
                    }
                }
            }
        }
        return f[m - 1][n - 1][0];
    }
}

// Accepted solution for LeetCode #2435: Paths in Matrix Whose Sum Is Divisible by K
func numberOfPaths(grid [][]int, K int) int {
	const mod = 1e9 + 7
	m, n := len(grid), len(grid[0])
	f := make([][][]int, m)
	for i := range f {
		f[i] = make([][]int, n)
		for j := range f[i] {
			f[i][j] = make([]int, K)
		}
	}
	f[0][0][grid[0][0]%K] = 1
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			for k := 0; k < K; k++ {
				k0 := ((k - grid[i][j]%K) + K) % K
				if i > 0 {
					f[i][j][k] = (f[i][j][k] + f[i-1][j][k0]) % mod
				}
				if j > 0 {
					f[i][j][k] = (f[i][j][k] + f[i][j-1][k0]) % mod
				}
			}
		}
	}
	return f[m-1][n-1][0]
}

# Accepted solution for LeetCode #2435: Paths in Matrix Whose Sum Is Divisible by K
class Solution:
    def numberOfPaths(self, grid: List[List[int]], K: int) -> int:
        mod = 10**9 + 7
        m, n = len(grid), len(grid[0])
        f = [[[0] * K for _ in range(n)] for _ in range(m)]
        f[0][0][grid[0][0] % K] = 1
        for i in range(m):
            for j in range(n):
                for k in range(K):
                    k0 = ((k - grid[i][j] % K) + K) % K
                    if i:
                        f[i][j][k] += f[i - 1][j][k0]
                    if j:
                        f[i][j][k] += f[i][j - 1][k0]
                    f[i][j][k] %= mod
        return f[m - 1][n - 1][0]

// Accepted solution for LeetCode #2435: Paths in Matrix Whose Sum Is Divisible by K
impl Solution {
    pub fn number_of_paths(grid: Vec<Vec<i32>>, K: i32) -> i32 {
        const MOD: i32 = 1_000_000_007;
        let m = grid.len();
        let n = grid[0].len();
        let K = K as usize;
        let mut f = vec![vec![vec![0; K]; n]; m];
        f[0][0][grid[0][0] as usize % K] = 1;
        for i in 0..m {
            for j in 0..n {
                for k in 0..K {
                    let k0 = ((k + K - grid[i][j] as usize % K) % K) as usize;
                    if i > 0 {
                        f[i][j][k] = (f[i][j][k] + f[i - 1][j][k0]) % MOD;
                    }
                    if j > 0 {
                        f[i][j][k] = (f[i][j][k] + f[i][j - 1][k0]) % MOD;
                    }
                }
            }
        }
        f[m - 1][n - 1][0]
    }
}

// Accepted solution for LeetCode #2435: Paths in Matrix Whose Sum Is Divisible by K
function numberOfPaths(grid: number[][], K: number): number {
    const mod = 1e9 + 7;
    const m = grid.length;
    const n = grid[0].length;
    const f: number[][][] = Array.from({ length: m }, () =>
        Array.from({ length: n }, () => Array(K).fill(0)),
    );
    f[0][0][grid[0][0] % K] = 1;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            for (let k = 0; k < K; ++k) {
                const k0 = (k - (grid[i][j] % K) + K) % K;
                if (i > 0) {
                    f[i][j][k] = (f[i][j][k] + f[i - 1][j][k0]) % mod;
                }
                if (j > 0) {
                    f[i][j][k] = (f[i][j][k] + f[i][j - 1][k0]) % mod;
                }
            }
        }
    }
    return f[m - 1][n - 1][0];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n × K)

Space

O(m × n × K)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.