LeetCode #1254 — MEDIUM

Number of Closed Islands

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

The Problem

Problem Statement

Given a 2D grid consists of 0s (land) and 1s (water). An island is a maximal 4-directionally connected group of 0s and a closed island is an island totally (all left, top, right, bottom) surrounded by 1s.

Return the number of closed islands.

Example 1:

Input: grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
Output: 2
Explanation: 
Islands in gray are closed because they are completely surrounded by water (group of 1s).

Example 2:

Input: grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
Output: 1

Example 3:

Input: grid = [[1,1,1,1,1,1,1],
               [1,0,0,0,0,0,1],
               [1,0,1,1,1,0,1],
               [1,0,1,0,1,0,1],
               [1,0,1,1,1,0,1],
               [1,0,0,0,0,0,1],
               [1,1,1,1,1,1,1]]
Output: 2

Constraints:

1 <= grid.length, grid[0].length <= 100
0 <= grid[i][j] <=1

Step 01

Brute Force Baseline

Problem summary: Given a 2D grid consists of 0s (land) and 1s (water). An island is a maximal 4-directionally connected group of 0s and a closed island is an island totally (all left, top, right, bottom) surrounded by 1s. Return the number of closed islands.

Baseline thinking

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

Pattern signal: Array · Union-Find

Example 1

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

Example 2

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

Example 3

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

Step 02

Core Insight

What unlocks the optimal approach

Exclude connected group of 0s on the corners because they are not closed island.
Return number of connected component of 0s on the grid.

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 #1254: Number of Closed Islands
class Solution {
    private int m;
    private int n;
    private int[][] grid;

    public int closedIsland(int[][] grid) {
        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) {
                    ans += dfs(i, j);
                }
            }
        }
        return ans;
    }

    private int dfs(int i, int j) {
        int res = i > 0 && i < m - 1 && j > 0 && j < n - 1 ? 1 : 0;
        grid[i][j] = 1;
        int[] dirs = {-1, 0, 1, 0, -1};
        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 && grid[x][y] == 0) {
                res &= dfs(x, y);
            }
        }
        return res;
    }
}

// Accepted solution for LeetCode #1254: Number of Closed Islands
func closedIsland(grid [][]int) (ans int) {
	m, n := len(grid), len(grid[0])
	dirs := [5]int{-1, 0, 1, 0, -1}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		res := 1
		if i == 0 || i == m-1 || j == 0 || j == n-1 {
			res = 0
		}
		grid[i][j] = 1
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 0 {
				res &= dfs(x, y)
			}
		}
		return res
	}
	for i, row := range grid {
		for j, v := range row {
			if v == 0 {
				ans += dfs(i, j)
			}
		}
	}
	return
}

# Accepted solution for LeetCode #1254: Number of Closed Islands
class Solution:
    def closedIsland(self, grid: List[List[int]]) -> int:
        def dfs(i: int, j: int) -> int:
            res = int(0 < i < m - 1 and 0 < j < n - 1)
            grid[i][j] = 1
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and grid[x][y] == 0:
                    res &= dfs(x, y)
            return res

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

// Accepted solution for LeetCode #1254: Number of Closed Islands
struct Solution;

impl Solution {
    fn closed_island(mut grid: Vec<Vec<i32>>) -> i32 {
        let mut res = 0;
        let n = grid.len();
        let m = grid[0].len();
        for i in 0..n {
            for j in 0..m {
                if grid[i][j] == 0 && Self::dfs(i, j, &mut grid, n, m) {
                    res += 1;
                }
            }
        }
        res
    }

    fn dfs(i: usize, j: usize, grid: &mut Vec<Vec<i32>>, n: usize, m: usize) -> bool {
        grid[i][j] = 1;
        let top = if i > 0 {
            if grid[i - 1][j] == 1 {
                true
            } else {
                Self::dfs(i - 1, j, grid, n, m)
            }
        } else {
            false
        };
        let left = if j > 0 {
            if grid[i][j - 1] == 1 {
                true
            } else {
                Self::dfs(i, j - 1, grid, n, m)
            }
        } else {
            false
        };
        let bottom = if i + 1 < n {
            if grid[i + 1][j] == 1 {
                true
            } else {
                Self::dfs(i + 1, j, grid, n, m)
            }
        } else {
            false
        };
        let right = if j + 1 < m {
            if grid[i][j + 1] == 1 {
                true
            } else {
                Self::dfs(i, j + 1, grid, n, m)
            }
        } else {
            false
        };
        top && left && bottom && right
    }
}

#[test]
fn test() {
    let grid = vec_vec_i32![
        [1, 1, 1, 1, 1, 1, 1, 0],
        [1, 0, 0, 0, 0, 1, 1, 0],
        [1, 0, 1, 0, 1, 1, 1, 0],
        [1, 0, 0, 0, 0, 1, 0, 1],
        [1, 1, 1, 1, 1, 1, 1, 0]
    ];
    let res = 2;
    assert_eq!(Solution::closed_island(grid), res);
    let grid = vec_vec_i32![[0, 0, 1, 0, 0], [0, 1, 0, 1, 0], [0, 1, 1, 1, 0]];
    let res = 1;
    assert_eq!(Solution::closed_island(grid), res);
}

// Accepted solution for LeetCode #1254: Number of Closed Islands
function closedIsland(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    const dirs = [-1, 0, 1, 0, -1];
    const dfs = (i: number, j: number): number => {
        let res = i > 0 && j > 0 && i < m - 1 && j < n - 1 ? 1 : 0;
        grid[i][j] = 1;
        for (let k = 0; k < 4; ++k) {
            const [x, y] = [i + dirs[k], j + dirs[k + 1]];
            if (x >= 0 && y >= 0 && x < m && y < n && grid[x][y] === 0) {
                res &= dfs(x, y);
            }
        }
        return res;
    };
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; j++) {
            if (grid[i][j] === 0) {
                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.