Formatted question description: https://leetcode.ca/all/1905.html

# 1905. Count Sub Islands

## Level

Medium

## Description

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.length`

`n == grid1[i].length == grid2[i].length`

`1 <= m, n <= 500`

`grid1[i][j]`

and`grid2[i][j]`

are either`0`

or`1`

.

## Solution

Do breadth first search on the islands in `grid2`

. For each island in `grid2`

, it is a sub-island if and only if all the cells in the island are lands in `grid1`

. After each island in `grid2`

is visited, it can be determined whether the island is a sub-island. The number of sub-islands can be calculated as well.

```
class Solution {
static int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
public int countSubIslands(int[][] grid1, int[][] grid2) {
int subIslands = 0;
int m = grid1.length, n = grid1[0].length;
boolean[][] visited = new boolean[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid2[i][j] == 1 && !visited[i][j]) {
boolean isSubisland = breadthFirstSearch(grid1, grid2, visited, m, n, i, j);
if (isSubisland)
subIslands++;
}
}
}
return subIslands;
}
public boolean breadthFirstSearch(int[][] grid1, int[][] grid2, boolean[][] visited, int m, int n, int startRow, int startColumn) {
boolean isSubisland = true;
Queue<int[]> queue = new LinkedList<int[]>();
queue.offer(new int[]{startRow, startColumn});
visited[startRow][startColumn] = true;
while (!queue.isEmpty()) {
int[] cell = queue.poll();
int row = cell[0], column = cell[1];
if (grid1[row][column] == 0)
isSubisland = false;
for (int[] direction : directions) {
int newRow = row + direction[0], newColumn = column + direction[1];
if (newRow >= 0 && newRow < m && newColumn >= 0 && newColumn < n && grid2[newRow][newColumn] == 1 && !visited[newRow][newColumn]) {
visited[newRow][newColumn] = true;
queue.offer(new int[]{newRow, newColumn});
}
}
}
return isSubisland;
}
}
```