<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:
        queue = deque()
        visit = set()
        row,col = len(grid), len(grid[0])
        def dfs(r,c):
            if r< 0 or c< 0 or r >= row or c>= col or grid[r][c] == 0 or (r,c) in visit:
                return
            visit.add((r,c))
            queue.append([r,c,0])
            dfs(r+1,c)
            dfs(r-1,c)
            dfs(r,c+1)
            dfs(r,c-1)
        def bfs():
            directions = [[0,1],[0,-1],[1,0],[-1,0]]
            while queue:
                r,c,length = queue.popleft()

                for dr,dc in directions:
                    nr,nc = r+dr, c+dc
                    if 0<=nr<row and 0<=nc<col and (nr,nc) not in visit:
                        if grid[nr][nc] == 1:
                            return length
                        queue.append([nr,nc,length+1])
                        visit.add((nr,nc))

        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 1:
                    dfs(i,j)
                    return bfs()

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