<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:
        m, n = len(grid), len(grid[0])
        queue = deque()
        visited = set()
        dirs = [[1, 0], [-1, 0], [0, 1], [0, -1]]

        def dfs(i, j):
            queue.append((i, j))
            visited.add((i, j))
            grid[i][j] = 2  # Mark first island with 2
            for di, dj in dirs:
                ni, nj = i + di, j + dj
                if 0 <= ni < m and 0 <= nj < n and grid[ni][nj] == 1:
                    dfs(ni, nj)

        # Find and mark the first island
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    dfs(i, j)
                    break
            else:
                continue
            break

        # BFS to find the shortest bridge
        steps = 0
        while queue:
            size = len(queue)
            for _ in range(size):
                i, j = queue.popleft()
                for di, dj in dirs:
                    ni, nj = i + di, j + dj
                    if 0 <= ni < m and 0 <= nj < n and (ni, nj) not in visited:
                        if grid[ni][nj] == 1:
                            return steps
                        elif grid[ni][nj] == 0:
                            queue.append((ni, nj))
                            visited.add((ni, nj))
            steps += 1

        return -1

20700