The problem is asking to find the number of closed islands in a 2D grid. An island is a group of 0s (land) that are connected in the four directions (up, down, left, right). A closed island is an island that is completely surrounded by 1s (water).

Here are the steps to solve this problem:

1. Initialize a counter to keep track of the number of closed islands.

2. Iterate over the 2D grid. For each cell that contains a 0, perform a depth-first search (DFS) to visit all the cells in the same island. If the DFS visits a cell on the border of the grid, it means the island is not closed, so we should ignore it. If the DFS finishes without visiting a border cell, increment the counter.

3. To implement the DFS, use a stack or recursion. Start from the current cell and visit all its unvisited neighbors that contain a 0. To avoid visiting the same cell multiple times, mark it as visited after pushing it to the stack or calling the recursive function. You can mark a cell by changing its value to 1.

4. After visiting all the cells in the grid, the counter will be the number of closed islands.

Here is a Python solution for the problem:

```python
def closedIsland(grid):
def dfs(i, j):
if not (0

Question

The problem is asking to find the number of closed islands in a 2D grid. An island is a group of 0s (land) that are connected in the four directions (up, down, left, right). A closed island is an island that is completely surrounded by 1s (water).

Here are the steps to solve this problem:

1. Initialize a counter to keep track of the number of closed islands.

2. Iterate over the 2D grid. For each cell that contains a 0, perform a depth-first search (DFS) to visit all the cells in the same island. If the DFS visits a cell on the border of the grid, it means the island is not closed, so we should ignore it. If the DFS finishes without visiting a border cell, increment the counter.

3. To implement the DFS, use a stack or recursion. Start from the current cell and visit all its unvisited neighbors that contain a 0. To avoid visiting the same cell multiple times, mark it as visited after pushing it to the stack or calling the recursive function. You can mark a cell by changing its value to 1.

4. After visiting all the cells in the grid, the counter will be the number of closed islands.

Here is a Python solution for the problem:

```python
def closedIsland(grid):
    def dfs(i, j):
        if not (0 <= i < len(grid)) or not (0 <= j < len(grid[0])) or grid[i][j] != 0:
            return False
        grid[i][j] = 1
        return dfs(i-1, j) and dfs(i+1, j) and dfs(i, j-1) and dfs(i, j+1)

return sum(dfs(i, j) for i in range(len(grid)) for j in range(len(grid[0])) if grid[i][j] == 0)
```

This function first defines a helper function `dfs` to perform the depth-first search. Then it iterates over the grid and calls `dfs` for each cell that contains a 0. The `dfs` function returns `True` if the island is closed and `False` otherwise. The `sum` function counts the number of `True` values returned by `dfs`, which is the number of closed islands.

Knowee AI · Accepted Answer