<p>You are given an <code>n x n</code> binary matrix <code>grid</code> where <code>1</code> represents land and <code>0</code> represents water.</p>

<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>&#39;s not connected to any other <code>1</code>&#39;s. There are <strong>exactly two islands</strong> in <code>grid</code>.</p>

<p>You may change <code>0</code>&#39;s to <code>1</code>&#39;s to connect the two islands to form <strong>one island</strong>.</p>

<p>Return <em>the smallest number of </em><code>0</code><em>&#39;s you must flip to connect the two islands</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,1],[1,0]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,1,0],[0,0,0],[0,0,1]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> 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]]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There are exactly two islands in <code>grid</code>.</li>
</ul>

class Solution:
    def shortestBridge(self, grid: List[List[int]]) -> int:
        visit1 = set()
        visit2 = set()
        n = len(grid)
        def dfs(r,c,visit):
            if min(r,c)<0 or max(r,c)>=n or (r,c) in visit or grid[r][c]==0:
                return 
            
            visit.add((r,c))
            dfs(r+1 , c ,visit)
            dfs(r-1 , c ,visit)
            dfs(r , c+1 ,visit)
            dfs(r , c-1 ,visit)
            return
        j=0
        for r in range(n):
            for c in range(n):
                if j==0 and (r,c) not in visit1 and grid[r][c]==1:
                    j+=1
                    dfs(r,c,visit1)
                elif j==1 and (r,c) not in visit2 and (r,c) not in visit1 and grid[r][c]==1:
                    dfs(r,c ,visit2) 
        a ,b = list(visit1) , list(visit2)
        res = float('inf')
        print( a, b)
        for r,c in a:
            for r1,c1 in b:
                res = min(res , abs(r-r1) + abs(c-c1))
        return res-1
18300