# Algorithm Pattern – Island Programs LC 200 1254 1020 695 1905 694

#### Algorithm Pattern

In a traditional way, we can use a visited array to mark if a cell has been visited.

However, if it is allowed, we can also set the visited cell as 0 (ocean indicator) as well. And in some special cases, like if you need to know if a visited cell is valid or invalid(connected with boundary, etc), set visited cell as 0 may be a better choice.

DFS:

``````int[][] dirs = new int[][]{{-1,0}, {1,0}, {0,-1}, {0,1}};

void dfs(int[][] grid, int i, int j, boolean[][] visited) {
int m = grid.length, n = grid[0].length;
if (i < 0 || j < 0 || i >= m || j >= n) {
// invalid coordinate
return;
}
if (visited[i][j]) {
// visited (i, j)
return;
}

// mark (i, j) visited
visited[i][j] = true;
for (int[] d : dirs) {
int next_i = i + d[0];
int next_j = j + d[1];
dfs(grid, next_i, next_j, visited);
}
// leave (i, j)
}``````

BFS:

Check example 695

Example

1. Number of Islands

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

My Solution

``````class Solution {
private int count;
private int[][] visited;
public int numIslands(char[][] grid) {
visited = new int[grid.length][grid[0].length];
count = 0;

for(int r = 0; r < grid.length; r++){
for(int c = 0; c < grid[0].length; c++){
if(grid[r][c] == '1' && visited[r][c] == 0){
search(grid, r, c);
count++;
}
}
}

return count;
}

public void search(char[][] grid, int r, int c){
// invalid
if(r < 0 || r > grid.length - 1 || c < 0 || c > grid[r].length - 1 || visited[r][c] == 1 || grid[r][c] == '0'){
return;
}

visited[r][c] = 1; //  or we can set current cell as 0 to avoid using visited array

int[][] dir = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

for(int[] d : dir){
int n_r = r + d[0];
int n_c = c + d[1];
search(grid, n_r, n_c);
}
}
}``````
1. Number of Closed Islands

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

Solution:

``````class Solution {
// 主函数：计算封闭岛屿的数量
int closedIsland(int[][] grid) {
int m = grid.length, n = grid[0].length;
for (int j = 0; j < n; j++) {
// 把靠上边的岛屿淹掉
dfs(grid, 0, j);
// 把靠下边的岛屿淹掉
dfs(grid, m - 1, j);
}
for (int i = 0; i < m; i++) {
// 把靠左边的岛屿淹掉
dfs(grid, i, 0);
// 把靠右边的岛屿淹掉
dfs(grid, i, n - 1);
}
// 遍历 grid，剩下的岛屿都是封闭岛屿
int res = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 0) {
res++;
dfs(grid, i, j);
}
}
}
return res;
}

// 从 (i, j) 开始，将与之相邻的陆地都变成海水
void dfs(int[][] grid, int i, int j) {
int m = grid.length, n = grid[0].length;
if (i < 0 || j < 0 || i >= m || j >= n) {
return;
}
if (grid[i][j] == 1) {
// 已经是海水了
return;
}
// 将 (i, j) 变成海水
grid[i][j] = 1;
// 淹没上下左右的陆地
dfs(grid, i + 1, j);
dfs(grid, i, j + 1);
dfs(grid, i - 1, j);
dfs(grid, i, j - 1);
}

}``````
1. Number of Enclaves

You are given an m x n binary matrix grid, where 0 represents a sea cell and 1 represents a land cell.
A move consists of walking from one land cell to another adjacent (4-directionally) land cell or walking off the boundary of the grid.
Return the number of land cells in grid for which we cannot walk off the boundary of the grid in any number of moves.

Example 1:

Input: grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
Output: 3
Explanation: There are three 1s that are enclosed by 0s, and one 1 that is not enclosed because its on the boundary.
Example 2:

Input: grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
Output: 0
Explanation: All 1s are either on the boundary or can reach the boundary.

My Solution

``````class Solution {
public int numEnclaves(int[][] grid) {
// remove the land in boundary
for(int i = 0; i < grid[0].length; i++){
dfs(grid, 0, i);
dfs(grid, grid.length - 1, i);
}
for(int i = 0; i < grid.length; i++){
dfs(grid, i, 0);
dfs(grid, i, grid[0].length - 1);
}

int count = 0;
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[i].length; j++){
if(grid[i][j] == 1){
count++;
}
}
}

return count;
}

public void dfs(int[][] grid, int r, int c){
if(r < 0 || r > grid.length - 1 || c < 0 || c > grid[0].length - 1 || grid[r][c] == 0){
return;
}

grid[r][c] = 0;

int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

for(int[] dir : dirs){
int n_r = r + dir[0];
int n_c = c + dir[1];
dfs(grid, n_r, n_c);
}
}
}``````
1. Max Area of Island

You are given an m x n binary matrix grid. An island is a group of 1’s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.
The area of an island is the number of cells with a value 1 in the island.
Return the maximum area of an island in grid. If there is no island, return 0.

Example 1:

Input: grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
Output: 6
Explanation: The answer is not 11, because the island must be connected 4-directionally.
Example 2:
Input: grid = [[0,0,0,0,0,0,0,0]]
Output: 0

My Solution

``````class Solution {
public int maxAreaOfIsland(int[][] grid) {
int[][] visited = new int[grid.length][grid[0].length];
int ret = 0;

// go over the grid and explore if cell is valid
for(int r = 0; r < grid.length; r++){
for(int c = 0; c < grid[0].length; c++){
// if find valid unvisited cell
if(visited[r][c] == 0 && grid[r][c] == 1){
queue.offer(new int[]{r, c});
int area = 0;
visited[r][c] = 1;
// expand the cell
while(!queue.isEmpty()){
int[] cell = queue.poll();
area++;
int[][] dir = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
for(int[] d : dir){
int n_r = cell[0] + d[0];
int n_c = cell[1] + d[1];
if(n_r >= 0 && n_r < grid.length && n_c >= 0 && n_c < grid[0].length && visited[n_r][n_c] == 0 && grid[n_r][n_c] == 1){
visited[n_r][n_c] = 1;
queue.offer(new int[]{n_r, n_c});
}
}
}

ret = Math.max(ret, area);
}
}
}

return ret;
}
}``````

DFS Solution

``````int maxAreaOfIsland(int[][] grid) {
// 记录岛屿的最大面积
int res = 0;
int m = grid.length, n = grid[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
// 淹没岛屿，并更新最大岛屿面积
res = Math.max(res, dfs(grid, i, j));
}
}
}
return res;
}

// 淹没与 (i, j) 相邻的陆地，并返回淹没的陆地面积
int dfs(int[][] grid, int i, int j) {
int m = grid.length, n = grid[0].length;
if (i < 0 || j < 0 || i >= m || j >= n) {
// 超出索引边界
return 0;
}
if (grid[i][j] == 0) {
// 已经是海水了
return 0;
}
// 将 (i, j) 变成海水
grid[i][j] = 0;

return dfs(grid, i + 1, j)
+ dfs(grid, i, j + 1)
+ dfs(grid, i - 1, j)
+ dfs(grid, i, j - 1) + 1;
}``````
1. Count Sub Islands

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.

Solution

``````int countSubIslands(int[][] grid1, int[][] grid2) {
int m = grid1.length, n = grid1[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid1[i][j] == 0 && grid2[i][j] == 1) {
// 这个岛屿肯定不是子岛，淹掉
dfs(grid2, i, j);
}
}
}
// 现在 grid2 中剩下的岛屿都是子岛，计算岛屿数量
int res = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid2[i][j] == 1) {
res++;
dfs(grid2, i, j);
}
}
}
return res;
}

// 从 (i, j) 开始，将与之相邻的陆地都变成海水
void dfs(int[][] grid, int i, int j) {
int m = grid.length, n = grid[0].length;
if (i < 0 || j < 0 || i >= m || j >= n) {
return;
}
if (grid[i][j] == 0) {
return;
}

grid[i][j] = 0;
dfs(grid, i + 1, j);
dfs(grid, i, j + 1);
dfs(grid, i - 1, j);
dfs(grid, i, j - 1);
}``````
1. Number of Distinct Islands
You are given an m x n binary matrix grid. An island is a group of 1’s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.
An island is considered to be the same as another if and only if one island can be translated (and not rotated or reflected) to equal the other.
Return the number of distinct islands.

Example 1:

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

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

Solution

``````class Solution {
int numDistinctIslands(int[][] grid) {
int m = grid.length, n = grid[0].length;
// 记录所有岛屿的序列化结果
HashSet<String> islands = new HashSet<>();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
// 淹掉这个岛屿，同时存储岛屿的序列化结果
StringBuilder sb = new StringBuilder();
// 初始的方向可以随便写，不影响正确性
dfs(grid, i, j, sb, 1);
}
}
}
// 不相同的岛屿数量
return islands.size();
}

void dfs(int[][] grid, int i, int j, StringBuilder sb, int dir) {
int m = grid.length, n = grid[0].length;
if (i < 0 || j < 0 || i >= m || j >= n
|| grid[i][j] == 0) {
return;
}
// 前序遍历位置：进入 (i, j)
grid[i][j] = 0;
sb.append(dir).append(',');

dfs(grid, i - 1, j, sb, 1); // 上
dfs(grid, i + 1, j, sb, 2); // 下
dfs(grid, i, j - 1, sb, 3); // 左
dfs(grid, i, j + 1, sb, 4); // 右

// 后序遍历位置：离开 (i, j)
sb.append(-dir).append(',');
}
}
``````