<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:
        N = len(grid)
        visit = set()
        dirs = [[1, 0], [-1, 0], [0, 1], [0, -1]]

        def in_bounds(r, c):
            return min(r, c) >= 0 and max(r, c) < N
        
        def dfs(r, c):
            if not in_bounds(r, c) or (r, c) in visit or grid[r][c] == 0:
                return
            visit.add((r, c))
            for dr, dc in dirs:
                dfs(r + dr, c + dc)
        
        def bfs():
            res, q = 0, deque(visit)
            while q:
                for i in range(len(q)):
                    r, c = q.popleft()
                    for dr, dc in dirs:
                        r2, c2 = r + dr, c + dc
                        if not in_bounds(r2, c2) or (r2, c2) in visit:
                            continue
                        if grid[r2][c2]:
                            return res
                        visit.add((r2, c2))
                        q.append((r2, c2))
                res += 1
            return res
        
        for r in range(N):
            for c in range(N):
                if grid[r][c]:
                    dfs(r, c)
                    return bfs()
20300