Grokking the Coding Interview: Patterns for Coding Questions
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Solution: Number of Islands

Problem Statement

Solution

Code  (DFS)

Code  (BFS)

Code  (BFS with visited matrix)

Problem Statement

Given a 2D array (i.e., a matrix) containing only 1s (land) and 0s (water), count the number of islands in it.

An island is a connected set of 1s (land) and is surrounded by either an edge or 0s (water). Each cell is considered connected to other cells horizontally or vertically (not diagonally).

Example 1

Input: matrix =

Image

Output: 3
Explanation: The matrix has three islands. See the highlighted cells below.

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Example 2

Input: matrix =

Image

Output: 1
Explanation: The matrix has only one island. See the highlighted cells below.

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Constraints:

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 300
  • matrix[i][j] is '0' or '1'.

Solution

We can traverse the matrix linearly to find islands.

Whenever we find a cell with the value '1' (i.e., land), we have found an island. Using that cell as the root node, we will perform a Depth First Search (DFS) or Breadth First Search (BFS) to find all of its connected land cells. During our DFS or BFS traversal, we will find and mark all the horizontally and vertically connected land cells. 

We need to have a mechanism to mark each land cell to ensure that each land cell is visited only once. To mark a cell visited, we have two options:

  1. We can update the given input matrix. Whenever we see a '1', we will make it '0'.
  2. A separate boolean matrix can be used to record whether or not each cell has been visited. 

Following is the DFS or BFS traversal of the example-2 mentioned above:

Image

By following the above algorithm, every time DFS or BFS is triggered, we are sure that we have found an island. We will keep a running count to calculate the total number of islands.

Below, we will see three solutions based on:

  1. DFS
  2. BFS
  3. BFS with visited matrix

Code  (DFS)

Here is what our DFS algorithm will look like. We will update the input matrix to mark cells visited.

Python3
Python3
. . . .

Time Complexity
Time complexity of the above algorithm will be O(M*N), where ‘M’ is the number of rows and 'N' is the number of columns of the input matrix. This is due to the fact that we have to traverse the whole matrix to find the islands.

Space Complexity
DFS recursion stack can go M*N deep when the whole matrix is filled with '1's. Hence, the space complexity will be O(M*N), where ‘M’ is the number of rows and 'N' is the number of columns of the input matrix.

Code  (BFS)

Here is what our BFS algorithm will look like. We will update the input matrix to mark cells visited.

Python3
Python3
. . . .

Time Complexity
Time complexity of the above algorithm will be O(M*N), where ‘M’ is the number of rows and 'N' is the number of columns.

Space Complexity
Space complexity of the above algorithm will be O(min(M,N). In the worst case, when the matrix is completely filled with land cells, the size of the queue can grow up to min(M,N).

Code  (BFS with visited matrix)

Here is what our BFS algorithm will look like. We will keep a separate boolean matrix to record whether or not each cell has been visited.

Python3
Python3
. . . .

Time Complexity
Time complexity of the above algorithm will be O(M*N), where ‘M’ is the number of rows and 'N' is the number of columns.

Space Complexity
Because of the visited array and max size of the queue, the space complexity will be O(M*N), where ‘M’ is the number of rows and 'N' is the number of columns of the input matrix.

S

Smoke

· 4 years ago

Introduction/Solution lacks additional explanation. Why use DFS/BFS? What is the connection to those technique is not explained at all.

M

Mikhail Putilov

· 4 years ago

Why space complexity for BFS is min (M*N) ? what does it even mean?

Show 3 replies
R

Runyao Fan

· 4 years ago

Why is it that the BFS solution has a space complexity of O(min(M, N))? Why doesn't the queue size grow beyond min(M, N)?

Show 7 replies
M

Mike Xu

· 3 years ago

To clarify the space complexity of BFS traversal of a matrix:

When you start traversing a matrix from the corner, the maximum number of cells/nodes you can have in the queue is k where k is the number of cells on a diagonal line in the matrix, which means k = min(M, N).

When you start traversing a matrix from the centre, the maximum number of cells/nodes you can have in the queue is {1, 4, 8, 12, 16, ..., 4i} where i is the i-th layer. And such cells fit in a matrix of min size {1, 4, 9, 16, 25, ..., i*i} respectively. We know that i is min(M, N), so yet again we have space complexity of O(4 * min(M, N)) which is O(min(M,N)).

https://stackoverflow.com/a/75422161/9406517

S

sweetykumari

· 4 years ago

Hi , pls anyone share BFS code in C#.

S

Sheen Goh

· 4 years ago

Hello

E

evmorov

· 4 years ago

When can BFS be useful? For me, it looks like a more complicated version of DFS.

Show 2 replies
M

Manthan

· 4 years ago

can someone please explain what does this do and why are we doing this? Queue neighbors = new LinkedList(); neighbors.add(new int[] { x, y }); while (!neighbors.isEmpty()) { int row = neighbors.peek()[0]; int col = neighbors.peek()[1]; neighbors.remove();

Show 1 reply
R

Richard Yuan

· 4 years ago

Is there a specific reason why you used neighbors.extend() for a deque rather than neighbors.append()? The iterable only contains a single tuple in this case.

Show 3 replies
A

Anand Mohan

· 3 years ago

Given :

DFS recursion stack can go  deep when the whole matrix is filled with '1's. Hence, the space complexity will be , where ‘M’ is the number of rows and 'N' is the number of columns of the input matrix.

All the matrix element will not be in one recursion stack. It will part of call back to left and right sub-path. So, the space complexity should be less then (M*N)

I think for DFS as well the Space Complexity would be O(min(M, N))

Please correct me If I am wrong.

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On This Page

Problem Statement

Solution

Code  (DFS)

Code  (BFS)

Code  (BFS with visited matrix)