LeetCode #3619 — MEDIUM

Count Islands With Total Value Divisible by K

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an m x n matrix grid and a positive integer k. An island is a group of positive integers (representing land) that are 4-directionally connected (horizontally or vertically).

The total value of an island is the sum of the values of all cells in the island.

Return the number of islands with a total value divisible by k.

Example 1:

Input: grid = [[0,2,1,0,0],[0,5,0,0,5],[0,0,1,0,0],[0,1,4,7,0],[0,2,0,0,8]], k = 5

Output: 2

Explanation:

The grid contains four islands. The islands highlighted in blue have a total value that is divisible by 5, while the islands highlighted in red do not.

Example 2:

Input: grid = [[3,0,3,0], [0,3,0,3], [3,0,3,0]], k = 3

Output: 6

Explanation:

The grid contains six islands, each with a total value that is divisible by 3.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 1000
1 <= m * n <= 10⁵
0 <= grid[i][j] <= 10⁶
1 <= k <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given an m x n matrix grid and a positive integer k. An island is a group of positive integers (representing land) that are 4-directionally connected (horizontally or vertically). The total value of an island is the sum of the values of all cells in the island. Return the number of islands with a total value divisible by k.

Baseline thinking

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

Pattern signal: Array · Union-Find

Example 1

[[0,2,1,0,0],[0,5,0,0,5],[0,0,1,0,0],[0,1,4,7,0],[0,2,0,0,8]]
5

Example 2

[[3,0,3,0],[0,3,0,3],[3,0,3,0]]
3

Core Insight

What unlocks the optimal approach

Use a BFS/DFS to find connected components.

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

Upper-end input sizes

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 #3619: Count Islands With Total Value Divisible by K
class Solution {
    private int m;
    private int n;
    private int[][] grid;
    private final int[] dirs = {-1, 0, 1, 0, -1};

    public int countIslands(int[][] grid, int k) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] > 0 && dfs(i, j) % k == 0) {
                    ++ans;
                }
            }
        }
        return ans;
    }

    private long dfs(int i, int j) {
        long s = grid[i][j];
        grid[i][j] = 0;
        for (int d = 0; d < 4; ++d) {
            int x = i + dirs[d], y = j + dirs[d + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] > 0) {
                s += dfs(x, y);
            }
        }
        return s;
    }
}

// Accepted solution for LeetCode #3619: Count Islands With Total Value Divisible by K
func countIslands(grid [][]int, k int) (ans int) {
	m, n := len(grid), len(grid[0])
	dirs := []int{-1, 0, 1, 0, -1}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		s := grid[i][j]
		grid[i][j] = 0
		for d := 0; d < 4; d++ {
			x, y := i+dirs[d], j+dirs[d+1]
			if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] > 0 {
				s += dfs(x, y)
			}
		}
		return s
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] > 0 && dfs(i, j)%k == 0 {
				ans++
			}
		}
	}
	return
}

# Accepted solution for LeetCode #3619: Count Islands With Total Value Divisible by K
class Solution:
    def countIslands(self, grid: List[List[int]], k: int) -> int:
        def dfs(i: int, j: int) -> int:
            s = grid[i][j]
            grid[i][j] = 0
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and grid[x][y]:
                    s += dfs(x, y)
            return s

        m, n = len(grid), len(grid[0])
        dirs = (-1, 0, 1, 0, -1)
        ans = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] and dfs(i, j) % k == 0:
                    ans += 1
        return ans

// Accepted solution for LeetCode #3619: Count Islands With Total Value Divisible by K
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3619: Count Islands With Total Value Divisible by K
// class Solution {
//     private int m;
//     private int n;
//     private int[][] grid;
//     private final int[] dirs = {-1, 0, 1, 0, -1};
// 
//     public int countIslands(int[][] grid, int k) {
//         m = grid.length;
//         n = grid[0].length;
//         this.grid = grid;
//         int ans = 0;
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 if (grid[i][j] > 0 && dfs(i, j) % k == 0) {
//                     ++ans;
//                 }
//             }
//         }
//         return ans;
//     }
// 
//     private long dfs(int i, int j) {
//         long s = grid[i][j];
//         grid[i][j] = 0;
//         for (int d = 0; d < 4; ++d) {
//             int x = i + dirs[d], y = j + dirs[d + 1];
//             if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] > 0) {
//                 s += dfs(x, y);
//             }
//         }
//         return s;
//     }
// }

// Accepted solution for LeetCode #3619: Count Islands With Total Value Divisible by K
function countIslands(grid: number[][], k: number): number {
    const m = grid.length,
        n = grid[0].length;
    const dirs = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number): number => {
        let s = grid[i][j];
        grid[i][j] = 0;
        for (let d = 0; d < 4; d++) {
            const x = i + dirs[d],
                y = j + dirs[d + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] > 0) {
                s += dfs(x, y);
            }
        }
        return s;
    };

    let ans = 0;
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (grid[i][j] > 0 && dfs(i, j) % k === 0) {
                ans++;
            }
        }
    }

    return ans;
}

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)

Space

O(m × n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

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.