Program to find distance of shortest bridge between islands in Python

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

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
         s.add((i, j))
         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
         seen.add((i, j))
         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