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200. Number of Islands

Description

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

 

Example 1:

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

Example 2:

Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3

 

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 300
  • grid[i][j] is '0' or '1'.

Solutions

Solution 1

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class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        def dfs(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] == '1':
                    dfs(x, y)

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

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class Solution {
    private char[][] grid;
    private int m;
    private int n;

    public int numIslands(char[][] 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] == '1') {
                    dfs(i, j);
                    ++ans;
                }
            }
        }
        return ans;
    }

    private void dfs(int i, int j) {
        grid[i][j] = '0';
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k];
            int y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1') {
                dfs(x, y);
            }
        }
    }
}

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class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int ans = 0;
        int dirs[5] = {-1, 0, 1, 0, -1};
        function<void(int, int)> dfs = [&](int i, int j) {
            grid[i][j] = '0';
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < grid.size() && y >= 0 && y < grid[0].size() && grid[x][y] == '1') {
                    dfs(x, y);
                }
            }
        };
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '1') {
                    dfs(i, j);
                    ++ans;
                }
            }
        }
        return ans;
    }
};

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func numIslands(grid [][]byte) int {
    m, n := len(grid), len(grid[0])
    var dfs func(i, j int)
    dfs = func(i, j int) {
        grid[i][j] = '0'
        dirs := []int{-1, 0, 1, 0, -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] == '1' {
                dfs(x, y)
            }
        }
    }
    ans := 0
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            if grid[i][j] == '1' {
                dfs(i, j)
                ans++
            }
        }
    }
    return ans
}

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function numIslands(grid: string[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let ans = 0;
    const dfs = (i: number, j: number) => {
        if (grid[i]?.[j] !== '1') {
            return;
        }
        grid[i][j] = '0';
        dfs(i + 1, j);
        dfs(i - 1, j);
        dfs(i, j + 1);
        dfs(i, j - 1);
    };
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] === '1') {
                dfs(i, j);
                ++ans;
            }
        }
    }
    return ans;
}

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const DIRS: [i32; 5] = [-1, 0, 1, 0, -1];

impl Solution {
    pub fn num_islands(grid: Vec<Vec<char>>) -> i32 {
        fn dfs(grid: &mut Vec<Vec<char>>, i: usize, j: usize) {
            grid[i][j] = '0';
            for k in 0..4 {
                let x = (i as i32) + DIRS[k];
                let y = (j as i32) + DIRS[k + 1];
                if x >= 0
                    && (x as usize) < grid.len()
                    && y >= 0
                    && (y as usize) < grid[0].len()
                    && grid[x as usize][y as usize] == '1'
                {
                    dfs(grid, x as usize, y as usize);
                }
            }
        }

        let mut grid = grid;
        let mut ans = 0;
        for i in 0..grid.len() {
            for j in 0..grid[0].len() {
                if grid[i][j] == '1' {
                    dfs(&mut grid, i, j);
                    ans += 1;
                }
            }
        }
        ans
    }
}

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using System;
using System.Collections.Generic;
using System.Linq;

public class Solution {
    public int NumIslands(char[][] grid)
    {
        var queue = new Queue<Tuple<int, int>>();
        var lenI = grid.Length;
        var lenJ = lenI == 0 ? 0 : grid[0].Length;
        var paths = new int[,] { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } };
        var result = 0;
        for (var i = 0; i < lenI; ++i)
        {
            for (var j = 0; j < lenJ; ++j)
            {
                if (grid[i][j] == '1')
                {
                    ++result;
                    grid[i][j] = '0';
                    queue.Enqueue(Tuple.Create(i, j));
                    while (queue.Any())
                    {
                        var position = queue.Dequeue();
                        for (var k = 0; k < 4; ++k)
                        {
                            var next = Tuple.Create(position.Item1 + paths[k, 0], position.Item2 + paths[k, 1]);
                            if (next.Item1 >= 0 && next.Item1 < lenI && next.Item2 >= 0 && next.Item2 < lenJ && grid[next.Item1][next.Item2] == '1')
                            {
                                grid[next.Item1][next.Item2] = '0';
                                queue.Enqueue(next);
                            }
                        }
                    }
                }
            }
        }
        return result;
    }
}

Solution 2

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class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        def bfs(i, j):
            grid[i][j] = '0'
            q = deque([(i, j)])
            while q:
                i, j = q.popleft()
                for a, b in pairwise(dirs):
                    x, y = i + a, j + b
                    if 0 <= x < m and 0 <= y < n and grid[x][y] == '1':
                        q.append((x, y))
                        grid[x][y] = 0

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

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class Solution {
    private char[][] grid;
    private int m;
    private int n;

    public int numIslands(char[][] 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] == '1') {
                    bfs(i, j);
                    ++ans;
                }
            }
        }
        return ans;
    }

    private void bfs(int i, int j) {
        grid[i][j] = '0';
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {i, j});
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!q.isEmpty()) {
            int[] p = q.poll();
            for (int k = 0; k < 4; ++k) {
                int x = p[0] + dirs[k];
                int y = p[1] + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1') {
                    q.offer(new int[] {x, y});
                    grid[x][y] = '0';
                }
            }
        }
    }
}

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class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int ans = 0;
        int dirs[5] = {-1, 0, 1, 0, -1};
        function<void(int, int)> bfs = [&](int i, int j) {
            grid[i][j] = '0';
            queue<pair<int, int>> q;
            q.push({i, j});
            vector<int> dirs = {-1, 0, 1, 0, -1};
            while (!q.empty()) {
                auto [a, b] = q.front();
                q.pop();
                for (int k = 0; k < 4; ++k) {
                    int x = a + dirs[k];
                    int y = b + dirs[k + 1];
                    if (x >= 0 && x < grid.size() && y >= 0 && y < grid[0].size() && grid[x][y] == '1') {
                        q.push({x, y});
                        grid[x][y] = '0';
                    }
                }
            }
        };
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '1') {
                    bfs(i, j);
                    ++ans;
                }
            }
        }
        return ans;
    }
};

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func numIslands(grid [][]byte) int {
    m, n := len(grid), len(grid[0])
    bfs := func(i, j int) {
        grid[i][j] = '0'
        q := [][]int{[]int{i, j}}
        dirs := []int{-1, 0, 1, 0, -1}
        for len(q) > 0 {
            p := q[0]
            q = q[1:]
            for k := 0; k < 4; k++ {
                x, y := p[0]+dirs[k], p[1]+dirs[k+1]
                if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1' {
                    q = append(q, []int{x, y})
                    grid[x][y] = '0'
                }
            }
        }
    }
    ans := 0
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            if grid[i][j] == '1' {
                bfs(i, j)
                ans++
            }
        }
    }
    return ans
}

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function numIslands(grid: string[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let ans = 0;
    function bfs(i, j) {
        grid[i][j] = '0';
        let q = [[i, j]];
        const dirs = [-1, 0, 1, 0, -1];
        while (q.length) {
            [i, j] = q.shift();
            for (let k = 0; k < 4; ++k) {
                const x = i + dirs[k];
                const y = j + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1') {
                    q.push([x, y]);
                    grid[x][y] = '0';
                }
            }
        }
    }
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] == '1') {
                bfs(i, j);
                ++ans;
            }
        }
    }
    return ans;
}

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use std::collections::VecDeque;

const DIRS: [i32; 5] = [-1, 0, 1, 0, -1];

impl Solution {
    pub fn num_islands(grid: Vec<Vec<char>>) -> i32 {
        fn bfs(grid: &mut Vec<Vec<char>>, i: usize, j: usize) {
            grid[i][j] = '0';
            let mut queue = VecDeque::from([(i, j)]);
            while !queue.is_empty() {
                let (i, j) = queue.pop_front().unwrap();
                for k in 0..4 {
                    let x = (i as i32) + DIRS[k];
                    let y = (j as i32) + DIRS[k + 1];
                    if x >= 0
                        && (x as usize) < grid.len()
                        && y >= 0
                        && (y as usize) < grid[0].len()
                        && grid[x as usize][y as usize] == '1'
                    {
                        grid[x as usize][y as usize] = '0';
                        queue.push_back((x as usize, y as usize));
                    }
                }
            }
        }

        let mut grid = grid;
        let mut ans = 0;
        for i in 0..grid.len() {
            for j in 0..grid[0].len() {
                if grid[i][j] == '1' {
                    bfs(&mut grid, i, j);
                    ans += 1;
                }
            }
        }
        ans
    }
}

Solution 3

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class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        dirs = (0, 1, 0)
        m, n = len(grid), len(grid[0])
        p = list(range(m * n))
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '1':
                    for a, b in pairwise(dirs):
                        x, y = i + a, j + b
                        if x < m and y < n and grid[x][y] == '1':
                            p[find(i * n + j)] = find(x * n + y)
        return sum(
            grid[i][j] == '1' and i * n + j == find(i * n + j)
            for i in range(m)
            for j in range(n)
        )

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class Solution {
    private int[] p;

    public int numIslands(char[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        p = new int[m * n];
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
        }
        int[] dirs = {1, 0, 1};
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '1') {
                    for (int k = 0; k < 2; ++k) {
                        int x = i + dirs[k];
                        int y = j + dirs[k + 1];
                        if (x < m && y < n && grid[x][y] == '1') {
                            p[find(x * n + y)] = find(i * n + j);
                        }
                    }
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '1' && i * n + j == find(i * n + j)) {
                    ++ans;
                }
            }
        }
        return ans;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

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class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vector<int> p(m * n);
        iota(p.begin(), p.end(), 0);
        function<int(int)> find = [&](int x) -> int {
            if (p[x] != x) {
                p[x] = find(p[x]);
            }
            return p[x];
        };
        int dirs[3] = {1, 0, 1};
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '1') {
                    for (int k = 0; k < 2; ++k) {
                        int x = i + dirs[k];
                        int y = j + dirs[k + 1];
                        if (x < m && y < n && grid[x][y] == '1') {
                            p[find(x * n + y)] = find(i * n + j);
                        }
                    }
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans += grid[i][j] == '1' && i * n + j == find(i * n + j);
            }
        }
        return ans;
    }
};

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func numIslands(grid [][]byte) int {
    m, n := len(grid), len(grid[0])
    p := make([]int, m*n)
    for i := range p {
        p[i] = i
    }
    var find func(x int) int
    find = func(x int) int {
        if p[x] != x {
            p[x] = find(p[x])
        }
        return p[x]
    }
    dirs := []int{1, 0, 1}
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            if grid[i][j] == '1' {
                for k := 0; k < 2; k++ {
                    x, y := i+dirs[k], j+dirs[k+1]
                    if x < m && y < n && grid[x][y] == '1' {
                        p[find(x*n+y)] = find(i*n + j)
                    }
                }
            }
        }
    }
    ans := 0
    for i := 0; i < m; i++ {
        for j := 0; j < n; j++ {
            if grid[i][j] == '1' && i*n+j == find(i*n+j) {
                ans++
            }
        }
    }
    return ans
}

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function numIslands(grid: string[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let p = [];
    for (let i = 0; i < m * n; ++i) {
        p.push(i);
    }
    function find(x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
    const dirs = [1, 0, 1];
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] == '1') {
                for (let k = 0; k < 2; ++k) {
                    const x = i + dirs[k];
                    const y = j + dirs[k + 1];
                    if (x < m && y < n && grid[x][y] == '1') {
                        p[find(i * n + j)] = find(x * n + y);
                    }
                }
            }
        }
    }
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] == '1' && i * n + j == find(i * n + j)) {
                ++ans;
            }
        }
    }
    return ans;
}

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const DIRS: [usize; 3] = [1, 0, 1];

impl Solution {
    pub fn num_islands(grid: Vec<Vec<char>>) -> i32 {
        let m = grid.len();
        let n = grid[0].len();
        let mut p: Vec<i32> = (0..(m * n) as i32).collect();

        fn find(p: &mut Vec<i32>, x: usize) -> i32 {
            if p[x] != (x as i32) {
                p[x] = find(p, p[x] as usize);
            }
            p[x]
        }

        for i in 0..m {
            for j in 0..n {
                if grid[i][j] == '1' {
                    for k in 0..2 {
                        let x = i + DIRS[k];
                        let y = j + DIRS[k + 1];
                        if x < m && y < n && grid[x][y] == '1' {
                            let f1 = find(&mut p, x * n + y);
                            let f2 = find(&mut p, i * n + j);
                            p[f1 as usize] = f2;
                        }
                    }
                }
            }
        }

        let mut ans = 0;
        for i in 0..m {
            for j in 0..n {
                if grid[i][j] == '1' && p[i * n + j] == ((i * n + j) as i32) {
                    ans += 1;
                }
            }
        }
        ans
    }
}

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