Question

Formatted question description: https://leetcode.ca/all/200.html

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'.

Algorithm

Need to use DFS to solve, create a visited array to record whether a certain location has been visited.

For a location that is ‘1’ and has not been visited, recursively enter the number of ‘1’ in its up, down, left, and right positions, assign its visited corresponding value to true, and continue to enter all its adjacent neighbors, so that this Find all the numbers in the connected area, and assign true to the value in the visited.

After finding the adjacent area, increment the result by 1, and then continue to find the next position that is ‘1’ and has not been visited, and so on until the entire original array is traversed.

Or, skipping the visited array to record, just set all visited ‘1’ to be ‘0’. (Assuming we can alter the original array) grid[i][j] = '0';

Code

• 
public class Number_of_Islands {

public static void main (String[] args) {

Number_of_Islands out = new Number_of_Islands();
Solution s = out.new Solution();

System.out.println(s.numIslands(new char[][]{ {'1', '1', '0', '0', '0'}, {'1', '1', '0', '0', '0'}, {'0', '0', '1', '0', '0'}, {'0', '0', '0', '1', '1'} }));
}

// from online, directly overrite visited 1s
public class Solution {

private int n;
private int m;

public int numIslands(char[][] grid) {
int count = 0;
n = grid.length;
if (n == 0) return 0;
m = grid[0].length;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++)
if (grid[i][j] == '1') {
dfs(grid, i, j);
++count;
}
}
return count;
}

private void dfs(char[][] grid, int i, int j) {
// @note: grid[i][j] != '1', should last of if(), or else index out of boundary
if (i < 0 || j < 0 || i >= n || j >= m || grid[i][j] != '1') {
return;
}

grid[i][j] = '0';
dfs(grid, i + 1, j);
dfs(grid, i - 1, j);
dfs(grid, i, j + 1);
dfs(grid, i, j - 1);
}
}

class Solution_bfs {
public int numIslands(char[][] grid) {
int h = grid.length; // height
if (h == 0)
return 0;
int w = grid[0].length; // width
int count = 0;

boolean[][] visited = new boolean[h][w];

for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
if (!visited[i][j] && grid[i][j] == '1') {
// No need to clear queue, since previous bfs stops only when queue is empty
BFS(queue, grid, visited);
count++;
}
}
}

return count;
}

private void BFS(Queue<Integer[]> queue, char[][] islandGrid, boolean[][] visited) {

int H = islandGrid.length;
int L = islandGrid[0].length;

while (!queue.isEmpty()) {

Integer[] x = queue.remove();
int row = x[0];
int col = x[1];

if (row < 0 || col < 0 || row >= H || col >= L || visited[row][col] || islandGrid[row][col] != '1')
continue;

visited[row][col] = true;
}
}
}

// union find
class Solution_uf {

final int d[][] = { // direction
{0, 1},
{1, 0},
{0, -1},
{-1, 0}
};

public int numIslands(char[][] grid) {
int h = grid.length;
if (h == 0)
return 0;
int w = grid[0].length;

int N = 0; // id for each point, prepare for UF
List<Point> ptlist = new ArrayList<Point>(); // each position, Point list
Map<String, Integer> map = new HashMap<String, Integer>(); // (i,j) => count

for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
if (grid[i][j] == '1') {
Point pt = new Point(i,j);
map.put(i+","+j, N++);
}
}
}

UF uf = new UF(N);
for (int i = 0; i < N; i++) {
Point curr = ptlist.get(i);
int row = curr.x;
int col = curr.y;
for (int k = 0; k < 4; k++) {
int x = row + d[k][0];
int y = col + d[k][1];
if (isValid(grid, h, w, x, y)) {
uf.union(i, map.get(x + "," + y));
}
}
}

return uf.count();
}

private boolean isValid(char[][] grid, int width, int height, int row, int col) {
return (row >= 0) && (row < width) && (col >= 0) && (col < height) && grid[row][col] == '1';
}

class Point {
int x;
int y;

public Point(int x, int y) {
this.x = x;
this.y = y;
}
}
}

class UF {
private int[] parent;
private byte[] rank;
private int count;

public UF(int n) {
if (n < 0) throw new IllegalArgumentException();
count = n;
parent = new int[n];
rank = new byte[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
rank[i] = 0;
}
}

public int find(int p) {
validate(p);
while (p != parent[p]) {
parent[p] = parent[parent[p]];
p = parent[p];
}
return p;
}

public int count() {
return count;
}

public void union(int p, int q) {
int rootP = find(p);
int rootQ = find(q);
if (rootP == rootQ) return;

if      (rank[rootP] < rank[rootQ]) parent[rootP] = rootQ;
else if (rank[rootP] > rank[rootQ]) parent[rootQ] = rootP;
else {
parent[rootQ] = rootP;
rank[rootP]++;
}
count--;
}

private void validate(int p) {
int n = parent.length;
if (p < 0 || p >= n) {
throw new IllegalArgumentException("index " + p + " is not between 0 and " + (n-1));
}
}
}

}

############

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);
}
}
}
}

• // OJ: https://leetcode.com/problems/number-of-islands/
// Time: O(MN)
// Space: O(MN)
class UnionFind {
vector<int> id;
public:
UnionFind(int N) : id(N) {
iota(begin(id), end(id), 0);
}
void connect(int x, int y) {
int a = find(x), b = find(y);
if (a == b) return;
id[a] = b;
}
int find(int x) {
return id[x] == x ? x : (id[x] = find(id[x]));
}
};
class Solution {
int M, N;
int key(int x, int y) { return x * N + y; }
int dirs[4][2] = { {0,1},{0,-1},{1,0},{-1,0} };
public:
int numIslands(vector<vector<char>>& A) {
if (A.empty() || A[0].empty()) return 0;
M = A.size(), N = A[0].size();
UnionFind uf(M * N);
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (A[i][j] == '0') continue;
for (auto &dir : dirs) {
int x = i + dir[0], y = j + dir[1];
if (x < 0 || y < 0 || x >= M || y >= N || A[x][y] == '0') continue;
uf.connect(key(i, j), key(x, y));
}
}
}
unordered_set<int> s;
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (A[i][j] == '0') continue;
s.insert(uf.find(key(i, j)));
}
}
return s.size();
}
};

• # dfs
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

# bfs
from collections import deque

class Solution:
def numIslands(self, grid: List[List[str]]) -> int:
if not grid:
return 0

m, n = len(grid), len(grid[0])
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
islands = 0

for i in range(m):
for j in range(n):
if grid[i][j] == "1":
islands += 1
q = deque([(i, j)]) # no need to reset q, q already drained from previous bfs
while q:
x, y = q.popleft()
for dx, dy in directions:
nx, ny = x + dx, y + dy
if 0 <= nx < m and 0 <= ny < n and grid[nx][ny] == "1":
q.append((nx, ny))
grid[nx][ny] = "0"  # mark as visited
return islands

# union find
from typing import List, Tuple

class Solution:

def numIslands(self, grid: List[List[str]]) -> int:
h = len(grid)
if h == 0:
return 0
w = len(grid[0])

N = 0  # id for each point, prepare for UF
ptlist = []  # each position, Point list
count_map = {}  # (i,j) => count

for i in range(h):
for j in range(w):
if grid[i][j] == '1':
pt = (i, j)
ptlist.append(pt)
count_map[i, j] = N
N += 1

uf = UF(N)
for i in range(N):
curr = ptlist[i]
row, col = curr[0], curr[1]
for k in range(4):
x, y = row + Solution.d[k][0], col + Solution.d[k][1]
if self.isValid(grid, h, w, x, y):
uf.union(i, count_map[x, y])

return uf.count

def isValid(self, grid: List[List[str]], height: int, width: int, row: int, col: int) -> bool:
return (row >= 0) and (row < height) and (col >= 0) and (col < width) and grid[row][col] == '1'

d = [  # direction
(0, 1),
(1, 0),
(0, -1),
(-1, 0)
]

class UF:
def __init__(self, n: int):
if n < 0:
raise ValueError("n must be non-negative")
self.parent = list(range(n))
self.rank = [0] * n
self.count = n

def find(self, p: int) -> int:
self.validate(p)
while p != self.parent[p]:
self.parent[p] = self.parent[self.parent[p]]
p = self.parent[p]
return p

def count(self) -> int:
return self.count

def union(self, p: int, q: int):
rootP = self.find(p)
rootQ = self.find(q)
if rootP == rootQ:
return

if self.rank[rootP] < self.rank[rootQ]:
self.parent[rootP] = rootQ
elif self.rank[rootP] > self.rank[rootQ]:
self.parent[rootQ] = rootP
else:
self.parent[rootQ] = rootP
self.rank[rootP] += 1
self.count -= 1

def validate(self, p: int):
n = len(self.parent)
if p < 0 or p >= n:
raise ValueError(f"index {p} is not between 0 and {n-1}")

############

'''
>>> a = set()
>>>
>>> a |= {(1, 1)}
>>> a
{(1, 1)}
>>>
>>> a |= {(2, 2)}
>>> a
{(1, 1), (2, 2)}
>>>
>>> a
{(1, 1), (3, 3), (2, 2)}
>>>
>>> (1,1) in a
True
>>> (10,10) in a
False
'''
class Solution(object):
def numIslands(self, grid):
"""
:type grid: List[List[str]]
:rtype: int
"""
visited = set()
ans = 0

def dfs(grid, i, j, visited):
if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] == "0" or (i, j) in visited:
return False
visited |= {(i, j)}
for di, dj in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
newi, newj = i + di, j + dj
dfs(grid, newi, newj, visited)
return True

for i in range(0, len(grid)):
for j in range(0, len(grid[0])):
if dfs(grid, i, j, visited):
ans += 1
return ans


• 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
}

• function numIslands(grid: string[][]): number {
const m = grid.length;
const n = grid[0].length;
let ans = 0;
function dfs(i, j) {
grid[i][j] = '0';
const dirs = [-1, 0, 1, 0, -1];
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') {
dfs(x, y);
}
}
}
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;
}


• 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;
}
}

• 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
}
}