# Program to remove all islands which are completely surrounded by water in Python

PythonServer Side ProgrammingProgramming

Suppose we have a binary matrix where 1 represents land and 0 represents water. And an island is a group of 1's that are surrounded by 0s (water) or by the edges. We have to find all of the islands that are completely surrounded by water and modify them into 0s. As we know an island is completed surrounded by water if all of the neighbors (horizontal and vertical not diagonal) are 0s (none of the neighbors are edges).

So, if the input is like

 1 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 0 1

then the output will be

 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

To solve this, we will follow these steps −

• row := row count of A

• col := column count of A

• B := a matrix of size A and fill with 0

• seen := a new set

• for i in range 0 to row, do

• for j in range 0 to col, do

• if i and j are not in range of matrix, then

• go for next iteration

• if (i, j) is seen, then

• go for next iteration

• if A[i, j] is same as 0, then

• go for next iteration

• d := a double ended queue with one element (i, j)

• while d is not empry, do

• (x, y) := left element of d, and delete from d

• B[x, y] := 1

• for each neighbor (x2, y2) of (x, y), do

• if (x2, y2) is not seen, then

• insert (x2, y2) at the end of d

• mark (x2, y2) as seen

• return B

## Example

Let us see the following implementation to get a better understanding −

Live Demo

from collections import deque
class Solution:
def solve(self, A):
row = len(A)
col = len(A[0])
B = [[0 for _ in range(col)] for _ in range(row)]
seen = set()
def nei(i, j):
if i + 1 < row and A[i + 1][j]:
yield (i + 1, j)
if j + 1 < col and A[i][j + 1]:
yield (i, j + 1)
if i - 1 >= 0 and A[i - 1][j]:
yield (i - 1, j)
if j - 1 >= 0 and A[i][j - 1]:
yield (i, j - 1)
for i in range(row):
for j in range(col):
if i not in (0, row - 1) and j not in (0, col - 1):
continue
if (i, j) in seen:
continue
if A[i][j] == 0:
continue
d = deque([(i, j)])
while d:
x, y = d.popleft()
B[x][y] = 1
for x2, y2 in nei(x, y):
if (x2, y2) not in seen:
d.append((x2, y2))
return B
ob = Solution()
matrix = [
[1, 0, 0, 0],
[0, 1, 1, 0],
[0, 1, 1, 0],
[0, 1, 1, 0],
[0, 0, 0, 1],
]
print(ob.solve(matrix))

## Input

[
[1, 0, 0, 0],
[0, 1, 1, 0],
[0, 1, 1, 0],
[0, 1, 1, 0],
[0, 0, 0, 1],
]

## Output

[
[1, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 1]
]
Published on 22-Dec-2020 08:55:35