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.lengthn == grid1[i].length == grid2[i].length1 <= m, n <= 500grid1[i][j]andgrid2[i][j]are either0or1.
BFS, DFS or Union find.
BFS:
class Solution:
def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
def bfs(i, j):
q = deque([(i, j)])
grid2[i][j] = 0
ans = True
while q:
i, j = q.popleft()
if grid1[i][j] == 0:
ans = False
for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and grid2[x][y]:
q.append((x, y))
grid2[x][y] = 0
return ans
m, n = len(grid1), len(grid1[0])
return sum(grid2[i][j] and bfs(i, j) for i in range(m) for j in range(n))DFS:
class Solution:
def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
def dfs(i, j):
ans = grid1[i][j] == 1
grid2[i][j] = 0
for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and grid2[x][y] == 1 and not dfs(x, y):
ans = False
return ans
m, n = len(grid1), len(grid1[0])
return sum(grid2[i][j] == 1 and dfs(i, j) for i in range(m) for j in range(n))Union find:
class Solution:
def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
m, n = len(grid1), len(grid1[0])
p = list(range(m * n))
for i in range(m):
for j in range(n):
if grid2[i][j]:
for a, b in [[0, 1], [1, 0]]:
x, y = i + a, j + b
if x < m and y < n and grid2[x][y]:
p[find(x * n + y)] = find(i * n + j)
s = [0] * (m * n)
for i in range(m):
for j in range(n):
if grid2[i][j]:
s[find(i * n + j)] = 1
for i in range(m):
for j in range(n):
root = find(i * n + j)
if s[root] and grid1[i][j] == 0:
s[root] = 0
return sum(s)DFS:
class Solution {
public int countSubIslands(int[][] grid1, int[][] grid2) {
int m = grid1.length;
int n = grid1[0].length;
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid2[i][j] == 1 && dfs(i, j, m, n, grid1, grid2)) {
++ans;
}
}
}
return ans;
}
private boolean dfs(int i, int j, int m, int n, int[][] grid1, int[][] grid2) {
boolean ans = grid1[i][j] == 1;
grid2[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 && grid2[x][y] == 1 && !dfs(x, y, m, n, grid1, grid2)) {
ans = false;
}
}
return ans;
}
}Union find:
class Solution {
private int[] p;
public int countSubIslands(int[][] grid1, int[][] grid2) {
int m = grid1.length;
int n = grid1[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 (grid2[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 && grid2[x][y] == 1) {
p[find(x * n + y)] = find(i * n + j);
}
}
}
}
}
int[] s = new int[m * n];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid2[i][j] == 1) {
s[find(i * n + j)] = 1;
}
}
}
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int root = find(i * n + j);
if (s[root] == 1 && grid1[i][j] == 0) {
s[root] = 0;
}
}
}
int ans = 0;
for (int i = 0; i < s.length; ++i) {
ans += s[i];
}
return ans;
}
private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}function countSubIslands(grid1: number[][], grid2: number[][]): number {
let m = grid1.length,
n = grid1[0].length;
let ans = 0;
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
if (grid2[i][j] == 1 && dfs(grid1, grid2, i, j)) {
++ans;
}
}
}
return ans;
}
function dfs(
grid1: number[][],
grid2: number[][],
i: number,
j: number,
): boolean {
let m = grid1.length,
n = grid1[0].length;
let ans = true;
if (grid1[i][j] == 0) {
ans = false;
}
grid2[i][j] = 0;
for (let [dx, dy] of [
[0, 1],
[0, -1],
[1, 0],
[-1, 0],
]) {
let x = i + dx,
y = j + dy;
if (x < 0 || x > m - 1 || y < 0 || y > n - 1 || grid2[x][y] == 0) {
continue;
}
if (!dfs(grid1, grid2, x, y)) {
ans = false;
}
}
return ans;
}BFS:
class Solution {
public:
int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) {
int m = grid1.size();
int n = grid1[0].size();
int ans = 0;
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
ans += (grid2[i][j] == 1 && bfs(i, j, m, n, grid1, grid2));
return ans;
}
bool bfs(int i, int j, int m, int n, vector<vector<int>>& grid1, vector<vector<int>>& grid2) {
queue<pair<int, int>> q;
q.push({i, j});
grid2[i][j] = 0;
bool ans = true;
vector<int> dirs = {-1, 0, 1, 0, -1};
while (!q.empty()) {
auto p = q.front();
q.pop();
i = p.first;
j = p.second;
if (grid1[i][j] == 0) ans = false;
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]) {
q.push({x, y});
grid2[x][y] = 0;
}
}
}
return ans;
}
};DFS:
class Solution {
public:
int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) {
int m = grid1.size();
int n = grid1[0].size();
int ans = 0;
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
if (grid2[i][j] == 1 && dfs(i, j, m, n, grid1, grid2))
++ans;
return ans;
}
bool dfs(int i, int j, int m, int n, vector<vector<int>>& grid1, vector<vector<int>>& grid2) {
bool ans = grid1[i][j];
grid2[i][j] = 0;
vector<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 && grid2[x][y] && !dfs(x, y, m, n, grid1, grid2))
ans = false;
}
return ans;
}
};BFS:
func countSubIslands(grid1 [][]int, grid2 [][]int) int {
m, n := len(grid1), len(grid1[0])
ans := 0
bfs := func(i, j int) bool {
q := [][]int{{i, j}}
grid2[i][j] = 0
res := true
dirs := []int{-1, 0, 1, 0, -1}
for len(q) > 0 {
i, j = q[0][0], q[0][1]
if grid1[i][j] == 0 {
res = false
}
q = q[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 && grid2[x][y] == 1 {
q = append(q, []int{x, y})
grid2[x][y] = 0
}
}
}
return res
}
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if grid2[i][j] == 1 && bfs(i, j) {
ans++
}
}
}
return ans
}DFS:
func countSubIslands(grid1 [][]int, grid2 [][]int) int {
m, n := len(grid1), len(grid1[0])
ans := 0
var dfs func(i, j int) bool
dfs = func(i, j int) bool {
res := grid1[i][j] == 1
grid2[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 && grid2[x][y] == 1 && !dfs(x, y) {
res = false
}
}
return res
}
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if grid2[i][j] == 1 && dfs(i, j) {
ans++
}
}
}
return ans
}