<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)
        directions = [[0, 1], [1,0], [0, -1], [-1, 0]]
        visited = set()

        def dfs(i, j):
            if invalid(i, j) or not grid[i][j] or (i, j) in visited:
                return
            
            visited.add((i, j))
            for dirs in directions:
                dfs(i+dirs[0], j+dirs[1])
            
        def invalid(i, j):
            return (i < 0 or i >= n or j < 0 or j >= n)
        

        def bfs():
            result, q = 0, deque(visited)

            while q:
                for i in range(len(q)):
                    r, c = q.popleft()
                    for dr, dc in directions:
                        nr = r + dr
                        nc = c + dc

                        if invalid(nr, nc) or (nr, nc) in visited:
                            continue
                        
                        if grid[nr][nc] == 1:
                            return result
                        
                        q.append((nr, nc))
                        visited.add((nr, nc))
                result += 1
                
            
            return result

        for i in range(n):
            for j in range(n):
                if grid[i][j]:
                    dfs(i, j)
                    return bfs()
20000