200. Number of Islands - Explanation

Description

Given a 2D grid grid where '1' represents land and '0' represents water, count and return the number of islands.

An island is formed by connecting adjacent lands horizontally or vertically and is surrounded by water. You may assume water is surrounding the grid (i.e., all the edges are water).

Example 1:

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

Example 2:

Input: grid = [
    ["1","1","0","0","1"],
    ["1","1","0","0","1"],
    ["0","0","1","0","0"],
    ["0","0","0","1","1"]
  ]
Output: 4

Constraints:

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

Topics

Recommended Time & Space Complexity

You should aim for a solution with O(m * n) time and O(m * n) space, where m is the number of rows and n is the number of columns in the grid.

Hint 1

An island is a group of 1's in which every 1 is reachable from any other 1 in that group. Can you think of an algorithm that can find the number of groups by visiting a group only once? Maybe there is a recursive way of doing it.

Hint 2

We can use the Depth First Search (DFS) algorithm to traverse each group independently. We iterate through each cell of the grid. When we encounter a 1, we perform a DFS starting at that cell and recursively visit every other 1 that is reachable. During this process, we mark the visited 1's as 0 to ensure we don't revisit them, as they belong to the same group. The number of groups corresponds to the number of islands.

Company Tags

Prerequisites

Before attempting this problem, you should be comfortable with:

2D Arrays/Matrices - Navigating and manipulating grid-based data structures
Depth First Search (DFS) - Recursive traversal to explore all connected components
Breadth First Search (BFS) - Level-by-level traversal using a queue
Union-Find (Disjoint Set Union) - Data structure for efficiently tracking connected components

1. Depth First Search

Intuition

Think of the grid as a map where '1' is land and '0' is water.
An island is a group of connected land cells (up, down, left, right).
Whenever we find a land cell that hasn’t been visited, we start a DFS to sink the entire island by marking all its connected land as water. Each DFS call corresponds to one island.

Algorithm

Iterate through every cell in the grid.
When a cell with value '1' is found:
- Increment the island count.
- Run DFS from that cell.
In DFS:
- If the cell is out of bounds or is '0', return.
- Mark the current cell as '0' (visited).
- Recursively explore all 4 directions (up, down, left, right).
Continue until all cells are processed.
Return the total island count.

Example - Dry Run

Input Grid:

grid = [
  ["1","1","0","0"],
  ["1","1","0","0"],
  ["0","0","1","0"],
  ["0","0","0","1"]
]

Visual Representation:

Input Grid (1 = land, 0 = water):

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ █ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ █ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘

Legend: █ = land ('1'), ░ = water ('0'), ✓ = visited

Step-by-Step Walkthrough:


Step 1: Found land at (0,0), start DFS
        Island count: 0 → 1

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ █ │ ░ │ ░ │  ← Starting DFS here
  ├───┼───┼───┼───┤
1 │ █ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘



Step 2: DFS explores neighbors of (0,0)
        Visit (1,0) - it's land, mark visited

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ █ │ ░ │ ░ │  ← Visited from (0,0)
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘



Step 3: DFS explores neighbors of (1,0)
        Visit (1,1) - it's land, mark visited

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │  ← Visited from (1,0)
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘



Step 4: DFS explores neighbors of (1,1)
        Visit (0,1) - it's land, mark visited
        All other neighbors are water or visited

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │  ← Visited from (1,1)
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘

        DFS complete for first island.
        Island count: 1



Step 5: Continue scanning, find land at (2,2)
        Start DFS, island count: 1 → 2

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ ✓ │ ░ │  ← New island found!
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘

        No land neighbors, DFS complete.
        Island count: 2



Step 6: Continue scanning, find land at (3,3)
        Start DFS, island count: 2 → 3

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ ✓ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ ✓ │  ← New island found!
  └───┴───┴───┴───┘

        No land neighbors, DFS complete.
        Island count: 3

Final Result: 3 islands

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        directions = [[1, 0], [-1, 0], [0, 1], [0, -1]]
        ROWS, COLS = len(grid), len(grid[0])
        islands = 0

        def dfs(r, c):
            if (r < 0 or c < 0 or r >= ROWS or
                c >= COLS or grid[r][c] == "0"
            ):
                return

            grid[r][c] = "0"
            for dr, dc in directions:
                dfs(r + dr, c + dc)

        for r in range(ROWS):
            for c in range(COLS):
                if grid[r][c] == "1":
                    dfs(r, c)
                    islands += 1

        return islands

public class Solution {
    private static final int[][] directions = {{1, 0}, {-1, 0},
                                               {0, 1}, {0, -1}};

    public int numIslands(char[][] grid) {
        int ROWS = grid.length, COLS = grid[0].length;
        int islands = 0;

        for (int r = 0; r < ROWS; r++) {
            for (int c = 0; c < COLS; c++) {
                if (grid[r][c] == '1') {
                    dfs(grid, r, c);
                    islands++;
                }
            }
        }

        return islands;
    }

    private void dfs(char[][] grid, int r, int c) {
        if (r < 0 || c < 0 || r >= grid.length ||
            c >= grid[0].length || grid[r][c] == '0') {
            return;
        }

        grid[r][c] = '0';
        for (int[] dir : directions) {
            dfs(grid, r + dir[0], c + dir[1]);
        }
    }
}

class Solution {
    /**
     * @param {character[][]} grid
     * @return {number}
     */
    numIslands(grid) {
        const directions = [
            [1, 0],
            [-1, 0],
            [0, 1],
            [0, -1],
        ];
        const ROWS = grid.length,
            COLS = grid[0].length;
        let islands = 0;

        const dfs = (r, c) => {
            if (r < 0 || c < 0 || r >= ROWS || c >= COLS || grid[r][c] === '0')
                return;

            grid[r][c] = '0';
            for (const [dr, dc] of directions) {
                dfs(r + dr, c + dc);
            }
        };

        for (let r = 0; r < ROWS; r++) {
            for (let c = 0; c < COLS; c++) {
                if (grid[r][c] === '1') {
                    dfs(r, c);
                    islands++;
                }
            }
        }

        return islands;
    }
}

class Solution {
    fun numIslands(grid: Array<CharArray>): Int {
        val directions = arrayOf(intArrayOf(1, 0),
                                 intArrayOf(-1, 0),
                                 intArrayOf(0, 1),
                                 intArrayOf(0, -1))
        val rows = grid.size
        val cols = grid[0].size
        var islands = 0

        fun dfs(r: Int, c: Int) {
            if (r < 0 || c < 0 || r >= rows ||
                c >= cols || grid[r][c] == '0') {
                return
            }
            grid[r][c] = '0'
            for (dir in directions) {
                dfs(r + dir[0], c + dir[1])
            }
        }

        for (r in 0 until rows) {
            for (c in 0 until cols) {
                if (grid[r][c] == '1') {
                    dfs(r, c)
                    islands++
                }
            }
        }

        return islands
    }
}

Time & Space Complexity

Time complexity: $O(m * n)$
Space complexity: $O(m * n)$

Where $m$ is the number of rows and $n$ is the number of columns in the $grid$ .

2. Breadth First Search

Intuition

Treat the grid like a map where '1' represents land and '0' represents water.
Each island is a group of connected land cells.
When we encounter a land cell, we use BFS to visit all connected land cells and mark them as water, ensuring the same island is not counted again.

Algorithm

Traverse every cell in the grid.
When a '1' (land) cell is found:
- Increment the island count.
- Start BFS from that cell.
In BFS:
- Push the starting cell into a queue and mark it as '0'.
- While the queue is not empty:
  - Pop a cell.
  - Explore its 4 neighbors (up, down, left, right).
  - If a neighbor is land, mark it as '0' and add it to the queue.
Continue scanning the grid.
Return the total number of islands.

Example - Dry Run

Input Grid:

grid = [
  ["1","1","0","0"],
  ["1","1","0","0"],
  ["0","0","1","0"],
  ["0","0","0","1"]
]

Visual Representation:

Input Grid (1 = land, 0 = water):

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ █ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ █ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘

Legend: █ = land ('1'), ░ = water ('0'), ✓ = visited

Step-by-Step Walkthrough:


Step 1: Found land at (0,0), start BFS
        Add (0,0) to queue, mark visited
        Island count: 0 → 1
        Queue: [(0,0)]

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ █ │ ░ │ ░ │  ← Starting BFS here
  ├───┼───┼───┼───┤
1 │ █ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘



Step 2: Dequeue (0,0), check neighbors
        (0,1) is land → add to queue, mark visited
        (1,0) is land → add to queue, mark visited
        Queue: [(0,1), (1,0)]

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │  ← Added (0,1)
  ├───┼───┼───┼───┤
1 │ ✓ │ █ │ ░ │ ░ │  ← Added (1,0)
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘



Step 3: Dequeue (0,1), check neighbors
        (0,0) already visited
        (1,1) is land → add to queue, mark visited
        Queue: [(1,0), (1,1)]

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │  ← Added (1,1)
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘



Step 4: Dequeue (1,0), check neighbors
        All neighbors are water or visited
        Queue: [(1,1)]

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘



Step 5: Dequeue (1,1), check neighbors
        All neighbors are water or visited
        Queue: [] (empty)

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘

        BFS complete for first island.
        Island count: 1



Step 6: Continue scanning, find land at (2,2)
        Start BFS, add (2,2) to queue, mark visited
        Island count: 1 → 2
        Queue: [(2,2)]

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ ✓ │ ░ │  ← New island found!
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘



Step 7: Dequeue (2,2), no land neighbors
        Queue: [] (empty)

        BFS complete.
        Island count: 2



Step 8: Continue scanning, find land at (3,3)
        Start BFS, add (3,3) to queue, mark visited
        Island count: 2 → 3
        Queue: [(3,3)]

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ ✓ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ ✓ │  ← New island found!
  └───┴───┴───┴───┘



Step 9: Dequeue (3,3), no land neighbors
        Queue: [] (empty)

        BFS complete.
        Island count: 3

Final Result: 3 islands

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        directions = [[1, 0], [-1, 0], [0, 1], [0, -1]]
        ROWS, COLS = len(grid), len(grid[0])
        islands = 0

        def bfs(r, c):
            q = deque()
            grid[r][c] = "0"
            q.append((r, c))

            while q:
                row, col = q.popleft()
                for dr, dc in directions:
                    nr, nc = dr + row, dc + col
                    if (nr < 0 or nc < 0 or nr >= ROWS or
                        nc >= COLS or grid[nr][nc] == "0"
                    ):
                        continue
                    q.append((nr, nc))
                    grid[nr][nc] = "0"

        for r in range(ROWS):
            for c in range(COLS):
                if grid[r][c] == "1":
                    bfs(r, c)
                    islands += 1

        return islands

public class Solution {
    private static final int[][] directions = {{1, 0}, {-1, 0},
                                               {0, 1}, {0, -1}};

    public int numIslands(char[][] grid) {
        int ROWS = grid.length, COLS = grid[0].length;
        int islands = 0;

        for (int r = 0; r < ROWS; r++) {
            for (int c = 0; c < COLS; c++) {
                if (grid[r][c] == '1') {
                    bfs(grid, r, c);
                    islands++;
                }
            }
        }

        return islands;
    }

    private void bfs(char[][] grid, int r, int c) {
        Queue<int[]> q = new LinkedList<>();
        grid[r][c] = '0';
        q.add(new int[]{r, c});

        while (!q.isEmpty()) {
            int[] node = q.poll();
            int row = node[0], col = node[1];

            for (int[] dir : directions) {
                int nr = row + dir[0], nc = col + dir[1];
                if (nr >= 0 && nc >= 0 && nr < grid.length &&
                    nc < grid[0].length && grid[nr][nc] == '1') {
                    q.add(new int[]{nr, nc});
                    grid[nr][nc] = '0';
                }
            }
        }
    }
}

class Solution {
    /**
     * @param {character[][]} grid
     * @return {number}
     */
    numIslands(grid) {
        const directions = [
            [1, 0],
            [-1, 0],
            [0, 1],
            [0, -1],
        ];
        const ROWS = grid.length,
            COLS = grid[0].length;
        let islands = 0;

        const bfs = (r, c) => {
            const q = new Queue();
            q.push([r, c]);
            grid[r][c] = '0';

            while (!q.isEmpty()) {
                const [row, col] = q.pop();
                for (const [dr, dc] of directions) {
                    const nr = row + dr,
                        nc = col + dc;
                    if (
                        nr >= 0 &&
                        nc >= 0 &&
                        nr < ROWS &&
                        nc < COLS &&
                        grid[nr][nc] === '1'
                    ) {
                        q.push([nr, nc]);
                        grid[nr][nc] = '0';
                    }
                }
            }
        };

        for (let r = 0; r < ROWS; r++) {
            for (let c = 0; c < COLS; c++) {
                if (grid[r][c] === '1') {
                    bfs(r, c);
                    islands++;
                }
            }
        }

        return islands;
    }
}

class Solution {
    fun numIslands(grid: Array<CharArray>): Int {
        val directions = arrayOf(intArrayOf(1, 0),
                                 intArrayOf(-1, 0),
                                 intArrayOf(0, 1),
                                 intArrayOf(0, -1))
        val rows = grid.size
        val cols = grid[0].size
        var islands = 0

        fun bfs(r: Int, c: Int) {
            val q: Queue<IntArray> = LinkedList()
            grid[r][c] = '0'
            q.add(intArrayOf(r, c))

            while (q.isNotEmpty()) {
                val (row, col) = q.poll()
                for (dir in directions) {
                    val nr = row + dir[0]
                    val nc = col + dir[1]
                    if (nr < 0 || nc < 0 || nr >= rows ||
                        nc >= cols || grid[nr][nc] == '0') {
                        continue
                    }
                    q.add(intArrayOf(nr, nc))
                    grid[nr][nc] = '0'
                }
            }
        }

        for (r in 0 until rows) {
            for (c in 0 until cols) {
                if (grid[r][c] == '1') {
                    bfs(r, c)
                    islands++
                }
            }
        }

        return islands
    }
}

Time & Space Complexity

Time complexity: $O(m * n)$
Space complexity: $O(m * n)$

Where $m$ is the number of rows and $n$ is the number of columns in the $grid$ .

3. Disjoint Set Union

Intuition

Think of every land cell ('1') as its own separate island initially.
When two land cells are adjacent (up, down, left, right), they actually belong to the same island, so we should merge them.

Disjoint Set Union (Union-Find) helps us:

Quickly connect adjacent land cells
Avoid counting the same island multiple times

Each successful merge reduces the total island count by 1.

Algorithm

Treat each cell as a node and map (row, col) to a unique index.
Initialize DSU for all cells.
Traverse the grid:
- If a cell is land ('1'), increment island count.
- Check its 4 neighbors.
- If a neighbor is also land:
  - Union the two cells.
  - If a union actually happens, decrement island count.
After processing all cells, the remaining count is the number of islands.
Return the island count.

Example - Dry Run

Input Grid:

grid = [
  ["1","1","0","0"],
  ["1","1","0","0"],
  ["0","0","1","0"],
  ["0","0","0","1"]
]

Visual Representation:

Input Grid (1 = land, 0 = water):

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ █ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ █ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘

Cell Indices (row * 4 + col):

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ 0 │ 1 │ 2 │ 3 │
  ├───┼───┼───┼───┤
1 │ 4 │ 5 │ 6 │ 7 │
  ├───┼───┼───┼───┤
2 │ 8 │ 9 │10 │11 │
  ├───┼───┼───┼───┤
3 │12 │13 │14 │15 │
  └───┴───┴───┴───┘

Legend: █ = land ('1'), ░ = water ('0'), ✓ = visited

Step-by-Step Walkthrough:


Initial State:
  Parent: [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]
  Each cell is its own parent (separate component)



Step 1: Process cell (0,0) - index 0, it's land
        Island count: 0 → 1

        Check neighbor (1,0) - index 4, it's land
        Union(0, 4): different parents, merge!
        Island count: 1 → 0

        Check neighbor (0,1) - index 1, it's land
        Union(0, 1): different parents, merge!
        Island count: 0 → -1

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ █ │ ░ │ ░ │  ← Processing (0,0)
  ├───┼───┼───┼───┤
1 │ █ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘

        DSU: {0, 1, 4} are connected



Step 2: Process cell (0,1) - index 1, it's land
        Island count: -1 → 0

        Check neighbor (0,0) - index 0, it's land
        Union(1, 0): SAME parent (already connected)!
        No change

        Check neighbor (1,1) - index 5, it's land
        Union(1, 5): different parents, merge!
        Island count: 0 → -1

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │  ← Processing (0,1)
  ├───┼───┼───┼───┤
1 │ █ │ █ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘

        DSU: {0, 1, 4, 5} are connected



Step 3: Process cell (1,0) - index 4, it's land
        Island count: -1 → 0

        Check neighbor (0,0) - index 0
        Union(4, 0): SAME parent, no change

        Check neighbor (1,1) - index 5
        Union(4, 5): SAME parent, no change

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ █ │ ░ │ ░ │  ← Processing (1,0)
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘

        DSU: {0, 1, 4, 5} still connected



Step 4: Process cell (1,1) - index 5, it's land
        Island count: 0 → 1

        Check neighbor (0,1) - index 1
        Union(5, 1): SAME parent, no change

        Check neighbor (1,0) - index 4
        Union(5, 4): SAME parent, no change

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │  ← Processing (1,1)
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ █ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘

        First island complete!
        DSU: {0, 1, 4, 5} unified
        Island count: 1



Step 5: Process cell (2,2) - index 10, it's land
        Island count: 1 → 2

        No land neighbors to union with

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ ✓ │ ░ │  ← New island found!
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ █ │
  └───┴───┴───┴───┘

        Island count: 2



Step 6: Process cell (3,3) - index 15, it's land
        Island count: 2 → 3

        No land neighbors to union with

    0   1   2   3
  ┌───┬───┬───┬───┐
0 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
1 │ ✓ │ ✓ │ ░ │ ░ │
  ├───┼───┼───┼───┤
2 │ ░ │ ░ │ ✓ │ ░ │
  ├───┼───┼───┼───┤
3 │ ░ │ ░ │ ░ │ ✓ │  ← New island found!
  └───┴───┴───┴───┘

        Island count: 3

Final DSU State:

Components:
  - Component 1: {0, 1, 4, 5} (first island)
  - Component 2: {10} (second island)
  - Component 3: {15} (third island)

Final Result: 3 islands

class DSU:
    def __init__(self, n):
        self.Parent = list(range(n + 1))
        self.Size = [1] * (n + 1)

    def find(self, node):
        if self.Parent[node] != node:
            self.Parent[node] = self.find(self.Parent[node])
        return self.Parent[node]

    def union(self, u, v):
        pu = self.find(u)
        pv = self.find(v)
        if pu == pv:
            return False
        if self.Size[pu] >= self.Size[pv]:
            self.Size[pu] += self.Size[pv]
            self.Parent[pv] = pu
        else:
            self.Size[pv] += self.Size[pu]
            self.Parent[pu] = pv
        return True

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        ROWS, COLS = len(grid), len(grid[0])
        dsu = DSU(ROWS * COLS)

        def index(r, c):
            return r * COLS + c

        directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
        islands = 0

        for r in range(ROWS):
            for c in range(COLS):
                if grid[r][c] == '1':
                    islands += 1
                    for dr, dc in directions:
                        nr, nc = r + dr, c + dc
                        if (nr < 0 or nc < 0 or nr >= ROWS or
                            nc >= COLS or grid[nr][nc] == "0"
                        ):
                            continue

                        if dsu.union(index(r, c), index(nr, nc)):
                            islands -= 1

        return islands

class DSU {
    vector<int> Parent, Size;
public:
    DSU(int n) {
        Parent.resize(n + 1);
        Size.resize(n + 1);
        for (int i = 0; i <= n; i++) {
            Parent[i] = i;
            Size[i] = 1;
        }
    }

    int find(int node) {
        if (node != Parent[node]) {
            Parent[node] = find(Parent[node]);
        }
        return Parent[node];
    }

    bool unionBySize(int u, int v) {
        int pu = find(u);
        int pv = find(v);
        if (pu == pv) return false;
        if (Size[pu] >= Size[pv]) {
            Size[pu] += Size[pv];
            Parent[pv] = pu;
        } else {
            Size[pv] += Size[pu];
            Parent[pu] = pv;
        }
        return true;
    }
};

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int ROWS = grid.size();
        int COLS = grid[0].size();
        DSU dsu(ROWS * COLS);

        int directions[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        int islands = 0;

        auto index = [&](int r, int c) {
            return r * COLS + c;
        };

        for (int r = 0; r < ROWS; r++) {
            for (int c = 0; c < COLS; c++) {
                if (grid[r][c] == '1') {
                    islands++;
                    for (auto& d : directions) {
                        int nr = r + d[0];
                        int nc = c + d[1];
                        if (nr >= 0 && nc >= 0 && nr < ROWS &&
                            nc < COLS && grid[nr][nc] == '1') {
                            if (dsu.unionBySize(index(r, c), index(nr, nc))) {
                                islands--;
                            }
                        }
                    }
                }
            }
        }

        return islands;
    }
};

class DSU {
    constructor(n) {
        this.Parent = Array(n + 1)
            .fill(0)
            .map((_, i) => i);
        this.Size = Array(n + 1).fill(1);
    }

    /**
     * @param {number} node
     * @return {number}
     */
    find(node) {
        if (this.Parent[node] !== node) {
            this.Parent[node] = this.find(this.Parent[node]);
        }
        return this.Parent[node];
    }

    /**
     * @param {number} u
     * @param {number} v
     * @return {boolean}
     */
    union(u, v) {
        let pu = this.find(u);
        let pv = this.find(v);
        if (pu === pv) return false;
        if (this.Size[pu] >= this.Size[pv]) {
            this.Size[pu] += this.Size[pv];
            this.Parent[pv] = pu;
        } else {
            this.Size[pv] += this.Size[pu];
            this.Parent[pu] = pv;
        }
        return true;
    }
}

class Solution {
    /**
     * @param {character[][]} grid
     * @return {number}
     */
    numIslands(grid) {
        const ROWS = grid.length;
        const COLS = grid[0].length;
        const dsu = new DSU(ROWS * COLS);

        const directions = [
            [1, 0],
            [-1, 0],
            [0, 1],
            [0, -1],
        ];

        let islands = 0;

        const index = (r, c) => r * COLS + c;

        for (let r = 0; r < ROWS; r++) {
            for (let c = 0; c < COLS; c++) {
                if (grid[r][c] === '1') {
                    islands++;
                    for (let [dr, dc] of directions) {
                        let nr = r + dr,
                            nc = c + dc;
                        if (
                            nr >= 0 &&
                            nc >= 0 &&
                            nr < ROWS &&
                            nc < COLS &&
                            grid[nr][nc] === '1'
                        ) {
                            if (dsu.union(index(r, c), index(nr, nc))) {
                                islands--;
                            }
                        }
                    }
                }
            }
        }

        return islands;
    }
}

public class DSU {
    private int[] Parent, Size;

    public DSU(int n) {
        Parent = new int[n + 1];
        Size = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            Parent[i] = i;
            Size[i] = 1;
        }
    }

    public int Find(int node) {
        if (node != Parent[node]) {
            Parent[node] = Find(Parent[node]);
        }
        return Parent[node];
    }

    public bool Union(int u, int v) {
        int pu = Find(u);
        int pv = Find(v);
        if (pu == pv) return false;
        if (Size[pu] >= Size[pv]) {
            Size[pu] += Size[pv];
            Parent[pv] = pu;
        } else {
            Size[pv] += Size[pu];
            Parent[pu] = pv;
        }
        return true;
    }
}

public class Solution {
    public int NumIslands(char[][] grid) {
        int ROWS = grid.Length;
        int COLS = grid[0].Length;
        DSU dsu = new DSU(ROWS * COLS);

        int[][] directions = new int[][] {
            new int[] { 1, 0 }, new int[] { -1, 0 },
            new int[] { 0, 1 }, new int[] { 0, -1 }
        };
        int islands = 0;

        for (int r = 0; r < ROWS; r++) {
            for (int c = 0; c < COLS; c++) {
                if (grid[r][c] == '1') {
                    islands++;
                    foreach (var d in directions) {
                        int nr = r + d[0];
                        int nc = c + d[1];
                        if (nr >= 0 && nc >= 0 && nr < ROWS &&
                            nc < COLS && grid[nr][nc] == '1') {
                            if (dsu.Union(r * COLS + c, nr * COLS + nc)) {
                                islands--;
                            }
                        }
                    }
                }
            }
        }

        return islands;
    }
}

type DSU struct {
    Parent []int
    Size   []int
}

func NewDSU(n int) *DSU {
    dsu := &DSU{
        Parent: make([]int, n+1),
        Size:   make([]int, n+1),
    }
    for i := 0; i <= n; i++ {
        dsu.Parent[i] = i
        dsu.Size[i] = 1
    }
    return dsu
}

func (dsu *DSU) find(node int) int {
    if dsu.Parent[node] != node {
        dsu.Parent[node] = dsu.find(dsu.Parent[node])
    }
    return dsu.Parent[node]
}

func (dsu *DSU) union(u, v int) bool {
    pu := dsu.find(u)
    pv := dsu.find(v)
    if pu == pv {
        return false
    }
    if dsu.Size[pu] >= dsu.Size[pv] {
        dsu.Size[pu] += dsu.Size[pv]
        dsu.Parent[pv] = pu
    } else {
        dsu.Size[pv] += dsu.Size[pu]
        dsu.Parent[pu] = pv
    }
    return true
}

func numIslands(grid [][]byte) int {
    rows, cols := len(grid), len(grid[0])
    dsu := NewDSU(rows * cols)
    directions := [][]int{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}
    islands := 0

    index := func(r, c int) int {
        return r*cols + c
    }

    for r := 0; r < rows; r++ {
        for c := 0; c < cols; c++ {
            if grid[r][c] == '1' {
                islands++
                for _, dir := range directions {
                    nr, nc := r+dir[0], c+dir[1]
                    if nr < 0 || nc < 0 || nr >= rows ||
                       nc >= cols || grid[nr][nc] == '0' {
                        continue
                    }
                    if dsu.union(index(r, c), index(nr, nc)) {
                        islands--
                    }
                }
            }
        }
    }

    return islands
}

class DSU(n: Int) {
    val Parent = IntArray(n + 1) { it }
    val Size = IntArray(n + 1) { 1 }

    fun find(node: Int): Int {
        if (Parent[node] != node) {
            Parent[node] = find(Parent[node])
        }
        return Parent[node]
    }

    fun union(u: Int, v: Int): Boolean {
        val pu = find(u)
        val pv = find(v)
        if (pu == pv) return false
        if (Size[pu] >= Size[pv]) {
            Size[pu] += Size[pv]
            Parent[pv] = pu
        } else {
            Size[pv] += Size[pu]
            Parent[pu] = pv
        }
        return true
    }
}

class Solution {
    fun numIslands(grid: Array<CharArray>): Int {
        val rows = grid.size
        val cols = grid[0].size
        val dsu = DSU(rows * cols)
        val directions = arrayOf(intArrayOf(1, 0),
                                 intArrayOf(-1, 0),
                                 intArrayOf(0, 1),
                                 intArrayOf(0, -1))
        var islands = 0

        fun index(r: Int, c: Int): Int {
            return r * cols + c
        }

        for (r in 0 until rows) {
            for (c in 0 until cols) {
                if (grid[r][c] == '1') {
                    islands++
                    for (dir in directions) {
                        val nr = r + dir[0]
                        val nc = c + dir[1]
                        if (nr < 0 || nc < 0 || nr >= rows ||
                            nc >= cols || grid[nr][nc] == '0') {
                            continue
                        }
                        if (dsu.union(index(r, c), index(nr, nc))) {
                            islands--
                        }
                    }
                }
            }
        }

        return islands
    }
}

class DSU {
    private var parent: [Int]
    private var size: [Int]

    init(_ n: Int) {
        parent = Array(0...n)
        size = Array(repeating: 1, count: n + 1)
    }

    func find(_ node: Int) -> Int {
        if parent[node] != node {
            parent[node] = find(parent[node])
        }
        return parent[node]
    }

    func union(_ u: Int, _ v: Int) -> Bool {
        let pu = find(u)
        let pv = find(v)
        if pu == pv {
            return false
        }
        if size[pu] >= size[pv] {
            size[pu] += size[pv]
            parent[pv] = pu
        } else {
            size[pv] += size[pu]
            parent[pu] = pv
        }
        return true
    }
}

class Solution {
    func numIslands(_ grid: [[Character]]) -> Int {
        let ROWS = grid.count
        let COLS = grid[0].count
        let dsu = DSU(ROWS * COLS)

        func index(_ r: Int, _ c: Int) -> Int {
            return r * COLS + c
        }

        let directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
        var islands = 0
        var grid = grid

        for r in 0..<ROWS {
            for c in 0..<COLS {
                if grid[r][c] == "1" {
                    islands += 1
                    for dir in directions {
                        let nr = r + dir.0
                        let nc = c + dir.1
                        if nr < 0 || nc < 0 || nr >= ROWS || nc >= COLS || grid[nr][nc] == "0" {
                            continue
                        }
                        if dsu.union(index(r, c), index(nr, nc)) {
                            islands -= 1
                        }
                    }
                }
            }
        }

        return islands
    }
}

Time & Space Complexity

Time complexity: $O(m * n)$
Space complexity: $O(m * n)$

Where $m$ is the number of rows and $n$ is the number of columns in the $grid$ .

Common Pitfalls

Not Marking Cells as Visited

When exploring an island, you must mark cells as visited (either by changing them to '0' or using a visited set). Forgetting this causes infinite loops as the DFS/BFS keeps revisiting the same cells.

# Wrong: doesn't mark visited, infinite loop
if grid[r][c] == '1':
    dfs(r + 1, c)
# Correct: mark as visited
if grid[r][c] == '1':
    grid[r][c] = '0'  # or visited.add((r, c))
    dfs(r + 1, c)

Counting Every Land Cell Instead of Islands

Each island should be counted once when you first encounter it. A common mistake is incrementing the count for every '1' cell rather than only incrementing when starting a new DFS/BFS traversal.

Incorrect Boundary Checks

Always verify that row and column indices are within bounds before accessing the grid. Off-by-one errors or missing boundary checks cause index out of bounds errors.

# Wrong: checks after access
if grid[nr][nc] == '1' and 0 <= nr < rows:
# Correct: check bounds first
if 0 <= nr < rows and 0 <= nc < cols and grid[nr][nc] == '1':

Diagonal Connections

This problem only considers horizontal and vertical connections (4-directional). Including diagonal neighbors incorrectly merges separate islands and undercounts the total.