<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)

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

        def dfs(x, y, islandA):
            if x<0 or y<0 or x==n or y==n:
                return
            if grid[x][y]==0:
                return
            if (x,y) in visited:
                return
            visited.add((x,y))
            islandA.append((x,y))
            for dx,dy in directions:
                dfs(x+dx, y+dy, islandA)
        
        islandA = deque()
        found = False
        for i in range(n):
            for j in range(n):
                if grid[i][j]:
                    dfs(i,j, islandA)
                    found = True
                    break
            if found:
                break

        result = 0
        while(islandA):
            for _ in range(len(islandA)):
                x,y = islandA.popleft()
                for d in directions:
                    nextx,nexty = x+d[0], y+d[1]
                    if nextx<0 or nexty<0 or nextx==n or nexty==n:
                        continue
                    if (nextx, nexty) in visited:
                        continue
                    if grid[nextx][nexty]==1:
                        return result
                    visited.add((nextx, nexty))
                    islandA.append((nextx, nexty))
            result+=1
        return result


            


19700