User: 

You are given access to make searches on the Internet. The format for making a search is:
1. After the user sends a message, you will decide whether to make a search. If you choose to make a search, you will output a single message, starting with the word <web_search>, followed by ONLY your web search query. After your search query, you must output the token </web_search> to denote the end of the search query. Keep your search query concise.
2. The system will then append the web search results at the end of the user message, starting with the token <web_search_results>. Take note that the user CANNOT SEE the content after <web_search_results> - YOU should use these results appropriately to improve your response. Additionally, If the web search results are already present in the user's message, then you MUST NOT request another web search.

IMPORTANT! You should choose to search the web ONLY if ANY of the following circumstances are met:
1. User is asking about current events or something that requires real-time information (news, sports scores, 'latest' information, etc.)
2. User is asking about some term you are unfamiliar with (e.g. if it's a little-known term, if it's a person you don't know well, etc.)
3. User is asking about anything that involves numerical facts (e.g. distances between places, population count, sizes, date and time, etc.)
4. User explicitly asks you to search or provide links to references.
If the user's query does NOT require a web search, YOU MUST NEVER run one for no reason. However, you must not refuse to search the web if the user asks you to do so.
If you are highly unsure about whether to make a search (e.g. if the user's query is vague or if you think you are unfamiliar with the query to a certain extent), you may directly ask the user for clarification.

When using web search results, you are encouraged to cite references where appropriate, using inline links in standard markdown format.
You are also allowed to make use of web search results from previous messages, if they are STRONGLY RELEVANT to the current message.

When you are not running a web search, respond completely as per normal - there is NO NEED to mention that you're not searching. 
Remember that you MUST NOT make up any false information.
Assistant: Understood. I will strictly follow these instructions in this conversation.
User: Browser tab content: Breadth-First Search
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200. Number of Islands
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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'.
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class Solution {
    public int numIslands(char[][] grid) {
        
    }
}
Saved
Ln 3, Col 9
Compile Error
Line 4: error: incompatible types: char[] cannot be converted to int
    for (int num : nums) {
                   ^

Breadth-First Search
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Remove distracting text, extract the programming question, provide simple, common, and optimized Java solution for it.
Assistant: ### Problem 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. It is assumed that all four edges of the grid are surrounded by water.

### Example 1:

**Input:**
```java
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:**
```java
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'`.

### Java Solution

Here's a simple and common solution using Breadth-First Search (BFS):

```java
import java.util.LinkedList;
import java.util.Queue;

class Solution {
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }

        int numIslands = 0;
        int rows = grid.length;
        int cols = grid[0].length;

        for (int r = 0; r < rows; r++) {
            for (int c = 0; c < cols; c++) {
                if (grid[r][c] == '1') {
                    numIslands++;
                    grid[r][c] = '0';  // Mark as visited
                    Queue<int[]> neighbors = new LinkedList<>();
                    neighbors.add(new int[] {r, c});

                    while (!neighbors.isEmpty()) {
                        int[] current = neighbors.poll();
                        int row = current[0];
                        int col = current[1];

                        // Explore all 4 directions
                        if (row - 1 >= 0 && grid[row - 1][col] == '1') {
                            neighbors.add(new int[] {row - 1, col});
                            grid[row - 1][col] = '0';
                        }
                        if (row + 1 < rows && grid[row + 1][col] == '1') {
                            neighbors.add(new int[] {row + 1, col});
                            grid[row + 1][col] = '0';
                        }
                        if (col - 1 >= 0 && grid[row][col - 1] == '1') {
                            neighbors.add(new int[] {row, col - 1});
                            grid[row][col - 1] = '0';
                        }
                        if (col + 1 < cols && grid[row][col + 1] == '1') {
                            neighbors.add(new int[] {row, col + 1});
                            grid[row][col + 1] = '0';
                        }
                    }
                }
            }
        }

        return numIslands;
    }
}
```

### Explanation:
- This solution uses BFS to explore all connected land cells ('1').
- We iterate through each cell in the grid; when we find '1', we have found a new island.
- The BFS explores all connected '1's (land) and marks them as visited by setting them to '0'.
- The count of such BFS explorations gives the number of islands.
User: bst
Assistant: