Shortest Bridge
Problem Statement
You are given an n x n binary matrix grid where 1 represents land and 0 represents water.
An island is a 4-directionally connected group of 1's not connected to any other 1's. There are exactly two islands in grid.
You may change 0's to 1's to connect the two islands to form one island.
Return the smallest number of 0's you must flip to connect the two islands.
Example 1:
Input: grid = [[0,1],[1,0]]
Output: 1
Example 2:
Input: grid = [[0,1,0],[0,0,0],[0,0,1]]
Output: 2
Example 3:
Input: grid = [[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]]
Output: 1
Constraints:
- n == grid.length == grid[i].length
- 2 <= n <= 100
- grid[i][j] is either 0 or 1.
- There are exactly two islands in grid.
Code
Python
class Solution:
def dfs(self,x,y,m,n,grid,visited,lst):
visited[x][y]=1
lst.append([x,y,0])
row=[-1,1,0,0]
col=[0,0,-1,1]
for i in range(4):
if x+row[i]>=0 and x+row[i]<m and y+col[i]>=0 and y+col[i]<n and visited[x+row[i]][y+col[i]]==-1 and grid[x+row[i]][y+col[i]]==1:
self.dfs(x+row[i],y+col[i],m,n,grid,visited,lst)
def shortestBridge(self, grid: List[List[int]]) -> int:
m=len(grid)
n=len(grid[0])
visited=[[-1]*n for i in range(m)]
lst=[]
for i in range(m):
for j in range(n):
if visited[i][j]==-1 and grid[i][j]==1:
self.dfs(i,j,m,n,grid,visited,lst)
break
else:
continue
break
mn=float("infinity")
while lst:
x,y,d=lst.pop(0)
row=[-1,1,0,0]
col=[0,0,-1,1]
for i in range(4):
if x+row[i]>=0 and x+row[i]<m and y+col[i]>=0 and y+col[i]<n and visited[x+row[i]][y+col[i]]==-1:
if grid[x+row[i]][y+col[i]]==1:
mn=min(mn,d)
else:
lst.append([x+row[i],y+col[i],d+1])
visited[x+row[i]][y+col[i]]=1
return mn