LeetCode #1905 — MEDIUM

Count Sub Islands

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

The Problem

Problem Statement

You are given two m x n binary matrices grid1 and grid2 containing only 0's (representing water) and 1's (representing land). An island is a group of 1's connected 4-directionally (horizontal or vertical). Any cells outside of the grid are considered water cells.

An island in grid2 is considered a sub-island if there is an island in grid1 that contains all the cells that make up this island in grid2.

Return the number of islands in grid2 that are considered sub-islands.

Example 1:

Input: grid1 = [[1,1,1,0,0],[0,1,1,1,1],[0,0,0,0,0],[1,0,0,0,0],[1,1,0,1,1]], grid2 = [[1,1,1,0,0],[0,0,1,1,1],[0,1,0,0,0],[1,0,1,1,0],[0,1,0,1,0]]
Output: 3
Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are three sub-islands.

Example 2:

Input: grid1 = [[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]], grid2 = [[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]]
Output: 2 
Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are two sub-islands.

Constraints:

m == grid1.length == grid2.length
n == grid1[i].length == grid2[i].length
1 <= m, n <= 500
grid1[i][j] and grid2[i][j] are either 0 or 1.

Step 01

Brute Force Baseline

Problem summary: You are given two m x n binary matrices grid1 and grid2 containing only 0's (representing water) and 1's (representing land). An island is a group of 1's connected 4-directionally (horizontal or vertical). Any cells outside of the grid are considered water cells. An island in grid2 is considered a sub-island if there is an island in grid1 that contains all the cells that make up this island in grid2. Return the number of islands in grid2 that are considered sub-islands.

Baseline thinking

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

Pattern signal: Array · Union-Find

Example 1

[[1,1,1,0,0],[0,1,1,1,1],[0,0,0,0,0],[1,0,0,0,0],[1,1,0,1,1]]
[[1,1,1,0,0],[0,0,1,1,1],[0,1,0,0,0],[1,0,1,1,0],[0,1,0,1,0]]

Example 2

[[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]]
[[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]]

Core Insight

What unlocks the optimal approach

Let's use floodfill to iterate over the islands of the second grid
Let's note that if all the cells in an island in the second grid if they are represented by land in the first grid then they are connected hence making that island a sub-island

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 #1905: Count Sub Islands
class Solution {
    private final int[] dirs = {-1, 0, 1, 0, -1};
    private int[][] grid1;
    private int[][] grid2;
    private int m;
    private int n;

    public int countSubIslands(int[][] grid1, int[][] grid2) {
        m = grid1.length;
        n = grid1[0].length;
        this.grid1 = grid1;
        this.grid2 = grid2;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid2[i][j] == 1) {
                    ans += dfs(i, j);
                }
            }
        }
        return ans;
    }

    private int dfs(int i, int j) {
        int ok = grid1[i][j];
        grid2[i][j] = 0;
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid2[x][y] == 1) {
                ok &= dfs(x, y);
            }
        }
        return ok;
    }
}

// Accepted solution for LeetCode #1905: Count Sub Islands
func countSubIslands(grid1 [][]int, grid2 [][]int) (ans int) {
	m, n := len(grid1), len(grid1[0])
	dirs := [5]int{-1, 0, 1, 0, -1}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		ok := grid1[i][j]
		grid2[i][j] = 0
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < m && y >= 0 && y < n && grid2[x][y] == 1 && dfs(x, y) == 0 {
				ok = 0
			}
		}
		return ok
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if grid2[i][j] == 1 {
				ans += dfs(i, j)
			}
		}
	}
	return
}

# Accepted solution for LeetCode #1905: Count Sub Islands
class Solution:
    def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
        def dfs(i: int, j: int) -> int:
            ok = grid1[i][j]
            grid2[i][j] = 0
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and grid2[x][y] and not dfs(x, y):
                    ok = 0
            return ok

        m, n = len(grid1), len(grid1[0])
        dirs = (-1, 0, 1, 0, -1)
        return sum(dfs(i, j) for i in range(m) for j in range(n) if grid2[i][j])

// Accepted solution for LeetCode #1905: Count Sub Islands
/**
 * [1905] Count Sub Islands
 *
 * You are given two m x n binary matrices grid1 and grid2 containing only 0's (representing water) and 1's (representing land). An island is a group of 1's connected 4-directionally (horizontal or vertical). Any cells outside of the grid are considered water cells.
 * An island in grid2 is considered a sub-island if there is an island in grid1 that contains all the cells that make up this island in grid2.
 * Return the number of islands in grid2 that are considered sub-islands.
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/06/10/test1.png" style="width: 493px; height: 205px;" />
 * Input: grid1 = [[1,1,1,0,0],[0,1,1,1,1],[0,0,0,0,0],[1,0,0,0,0],[1,1,0,1,1]], grid2 = [[1,1,1,0,0],[0,0,1,1,1],[0,1,0,0,0],[1,0,1,1,0],[0,1,0,1,0]]
 * Output: 3
 * Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
 * The 1s colored red in grid2 are those considered to be part of a sub-island. There are three sub-islands.
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/06/03/testcasex2.png" style="width: 491px; height: 201px;" />
 * Input: grid1 = [[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]], grid2 = [[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]]
 * Output: 2
 * Explanation: In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
 * The 1s colored red in grid2 are those considered to be part of a sub-island. There are two sub-islands.
 *
 *  
 * Constraints:
 *
 * 	m == grid1.length == grid2.length
 * 	n == grid1[i].length == grid2[i].length
 * 	1 <= m, n <= 500
 * 	grid1[i][j] and grid2[i][j] are either 0 or 1.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/count-sub-islands/
// discuss: https://leetcode.com/problems/count-sub-islands/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn count_sub_islands(grid1: Vec<Vec<i32>>, grid2: Vec<Vec<i32>>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1905_example_1() {
        let grid1 = vec![
            vec![1, 1, 1, 0, 0],
            vec![0, 1, 1, 1, 1],
            vec![0, 0, 0, 0, 0],
            vec![1, 0, 0, 0, 0],
            vec![1, 1, 0, 1, 1],
        ];
        let grid2 = vec![
            vec![1, 1, 1, 0, 0],
            vec![0, 0, 1, 1, 1],
            vec![0, 1, 0, 0, 0],
            vec![1, 0, 1, 1, 0],
            vec![0, 1, 0, 1, 0],
        ];

        let result = 3;

        assert_eq!(Solution::count_sub_islands(grid1, grid2), result);
    }

    #[test]
    #[ignore]
    fn test_1905_example_2() {
        let grid1 = vec![
            vec![1, 0, 1, 0, 1],
            vec![1, 1, 1, 1, 1],
            vec![0, 0, 0, 0, 0],
            vec![1, 1, 1, 1, 1],
            vec![1, 0, 1, 0, 1],
        ];
        let grid2 = vec![
            vec![0, 0, 0, 0, 0],
            vec![1, 1, 1, 1, 1],
            vec![0, 1, 0, 1, 0],
            vec![0, 1, 0, 1, 0],
            vec![1, 0, 0, 0, 1],
        ];

        let result = 2;

        assert_eq!(Solution::count_sub_islands(grid1, grid2), result);
    }
}

// Accepted solution for LeetCode #1905: Count Sub Islands
function countSubIslands(grid1: number[][], grid2: number[][]): number {
    const [m, n] = [grid1.length, grid1[0].length];
    let ans = 0;
    const dirs: number[] = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number): number => {
        let ok = grid1[i][j];
        grid2[i][j] = 0;
        for (let k = 0; k < 4; ++k) {
            const [x, y] = [i + dirs[k], j + dirs[k + 1]];
            if (x >= 0 && x < m && y >= 0 && y < n && grid2[x][y]) {
                ok &= dfs(x, y);
            }
        }
        return ok;
    };
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; j++) {
            if (grid2[i][j]) {
                ans += dfs(i, j);
            }
        }
    }
    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.