<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):
            visited.add((i, j))
            for di, dj in directions:
                ni, nj = i + di, j + dj
                if 0 <= ni < n and 0 <= nj < m and grid[ni][nj] == 1 and (ni, nj) not in visited:
                    dfs(ni, nj)

        def bfs():
            step = 0
            queue = deque(visited)
            while queue:
                level_len = len(queue)
                step += 1
                for _ in range(level_len):
                    i, j = queue.popleft()
                    for di, dj in directions:
                        ni, nj = i + di, j + dj
                        if 0 <= ni < n and 0 <= nj < m:
                            if (ni, nj) in visited:
                                continue
                            elif grid[ni][nj] == 0:
                                grid[ni][nj] = 1
                                visited.add((ni, nj))
                                queue.append((ni, nj))
                            elif grid[ni][nj] == 1:
                                return step

        visited = set()
        n, m = len(grid), len(grid[0])
        directions = ((0, 1), (0, -1), (1, 0), (-1, 0))
        for i in range(n):
            for j in range(m):
                if grid[i][j] == 1:
                    dfs(i, j)
                    return bfs() - 1
                    

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