<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:
        def dfs(i, j):
            if i < 0 or j < 0 or i >= m or j >= n or (i, j) in seen or grid[i][j] != 1: return
            seen.add((i, j))
            q.append((i, j, 0))
            grid[i][j] = -1
            for x, y in (i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1):
                dfs(x, y)
        
        seen, q, m, n = set(), deque(), len(grid), len(grid[0])
        for i, j in product(range(m), range(n)):
            if grid[i][j]:
                dfs(i, j)
                break
        
        while q:
            i, j, h = q.popleft()
            if grid[i][j] == 1:
                return h - 1
            for x, y in (i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1):
                if 0 <= x < m and 0 <= y < n and (x, y) not in seen:
                    seen.add((x, y))
                    q.append((x, y, h + 1))
        
        return -1
20600