# Program to find distance of shortest bridge between islands in Python

PythonServer Side ProgrammingProgramming

Suppose we have a binary matrix, where 0 represents water and 1 represents the land. An island is a group of connecting 1s in 4 directions. Islands are either surrounded by 0s (water) or by the edges. We have to find the length of shortest bridge that connects two islands.

So, if the input is like

 0 0 1 1 0 1 1 0 0

then the output will be 1. This will connect (1,0) to (1,2) points.

To solve this, we will follow these steps −

• row := row count of matrix

• col := column count of matrix

• Define a function dfs() . This will take i, j, s

• if (i, j) is in s, then

• return

• if mat[i, j] is same as 0, then

• return

• insert (i, j) into s

• if i - 1 >= 0, then

• dfs(i - 1, j, s)

• if i + 1 < row, then

• dfs(i + 1, j, s)

• if j - 1 >= 0, then

• dfs(i, j - 1, s)

• if j + 1 < col, then

• dfs(i, j + 1, s)

• From the main method do the following −

• seen := a new set

• for i in range 0 to row, do

• if size of seen > 0, then

• come out from the loop

• for j in range 0 to col, do

• if mat[i, j] is same as 1, then

• dfs(i, j, seen)

• come out from the loop

• q := a double ended queue

• for each land in seen, do

• (i, j) := land

• if i - 1 >= 0 and mat[i - 1, j] is same as 0, then

• insert (i - 1, j, 1) at the end of q

• if i + 1 < row and mat[i + 1, j] is same as 0, then

• insert (i + 1, j, 1) at the end of q

• if j - 1 >= 0 and mat[i, j - 1] is same as 0, then

• insert (i, j - 1, 1) at the end of q

• if j + 1 < col and mat[i, j + 1] is same as 0, then

• insert (i, j + 1, 1) at the end of q

• while size of q > 0, do

• (i, j, dist) := left item of q, and delete item from left of q

• if (i, j) is seen, then

• go for next iteration

• mark (i, j) as seen

• if mat[i, j] is same as 1, then

• return dist - 1

• if i - 1 >= 0, then

• insert (i - 1, j, dist + 1) at the end of q

• if i + 1 < row is non-zero, then

• insert (i + 1, j, dist + 1) at the end of q

• if j - 1 >= 0, then

• insert (i, j - 1, dist + 1) at the end of q

• if j + 1 < col is non-zero, then

• insert (i, j + 1, dist + 1) at the end of q

## Example

Let us see the following implementation to get better understanding −

Live Demo

import collections
class Solution:
def solve(self, mat):
row = len(mat)
col = len(mat[0])
def dfs(i, j, s):
if (i, j) in s:
return
if mat[i][j] == 0:
return
if i - 1 >= 0:
dfs(i - 1, j, s)
if i + 1 < row:
dfs(i + 1, j, s)
if j - 1 >= 0:
dfs(i, j - 1, s)
if j + 1 < col:
dfs(i, j + 1, s)
seen = set()
for i in range(row):
if len(seen) > 0:
break
for j in range(col):
if mat[i][j] == 1:
dfs(i, j, seen)
break
q = collections.deque()
for land in seen:
i, j = land
if i - 1 >= 0 and mat[i - 1][j] == 0:
q.append((i - 1, j, 1))
if i + 1 < row and mat[i + 1][j] == 0:
q.append((i + 1, j, 1))
if j - 1 >= 0 and mat[i][j - 1] == 0:
q.append((i, j - 1, 1))
if j + 1 < col and mat[i][j + 1] == 0:
q.append((i, j + 1, 1))
while len(q) > 0:
i, j, dist = q.popleft()
if (i, j) in seen:
continue
if mat[i][j] == 1:
return dist - 1
if i - 1 >= 0:
q.append((i - 1, j, dist + 1))
if i + 1 < row:
q.append((i + 1, j, dist + 1))
if j - 1 >= 0:
q.append((i, j - 1, dist + 1))
if j + 1 < col:
q.append((i, j + 1, dist + 1))
ob = Solution()
matrix = [
[0, 0, 1],
[1, 0, 1],
[1, 0, 0],
]
print(ob.solve(matrix))

## Input

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

## Output

1
Published on 22-Dec-2020 08:50:22