934. Shortest Bridge

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In a given 2D binary array A, there are two islands.  (An island is a 4-directionally connected group of 1s not connected to any other 1s.)

Now, we may change 0s to 1s so as to connect the two islands together to form 1 island.

Return the smallest number of 0s that must be flipped.  (It is guaranteed that the answer is at least 1.)

Example 1:

Input: [[0,1],[1,0]]
Output: 1

Example 2:

Input: [[0,1,0],[0,0,0],[0,0,1]]
Output: 2

Example 3:

Input: [[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

Note:

1. 1 <= A.length = A[0].length <= 100
2. A[i][j] == 0 or A[i][j] == 1

思路：2遍BFS，第一次先求出2个group，第二次遍历某个group到另外一个group的最短距离

class Solution:
    def shortestBridge(self, A):
        """
        :type A: List[List[int]]
        :rtype: int
        """
        n,m=len(A),len(A[0])
        for i in range(n):
            for j in range(m):
                if A[i][j]==1: break
            if A[i][j]==1: break
        ds=[[0,1],[0,-1],[1,0],[-1,0]]
        A[i][j]=2
        q=[(i,j)]
        while q:
            si,sj=q.pop()
            for di,dj in ds:
                ii,jj=si+di,sj+dj
                if 0<=ii<n and 0<=jj<m and A[ii][jj]==1:
                    A[ii][jj]=2
                    q.append((ii,jj))
        
        q=[(i,j) for i in range(n) for j in range(m) if A[i][j]==2]
        qq=[]
        vis=[[False for _ in range(m)] for _ in range(n)]
        for i,j in q: vis[i][j]=True
        step=0
        while q:
            while q:
                si,sj=q.pop()
                for di,dj in ds:
                    ii,jj=si+di,sj+dj
                    if 0<=ii<n and 0<=jj<m:
                        if A[ii][jj]==1: return step
                        if A[ii][jj]==0 and not vis[ii][jj]:
                            vis[ii][jj]=True
                            qq.append((ii,jj))
            q,qq=qq,q
            step+=1