<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>

from collections import deque
class Solution:
    def shortestBridge(self, grid: List[List[int]]) -> int:
        n = len(grid)

        directions = [(1,0),(0,1),(-1,0),(0,-1)]
        island1 = []
        findisland1 = False

        def dfs(x,y):
            grid[x][y] = 2
            island1.append((x,y))
            for dx, dy in directions:
                nx = x+dx
                ny = y+dy
                if nx >=0 and nx < n and ny >= 0 and ny < n and grid[nx][ny] == 1:
                    dfs(nx,ny)
        
        for i in range(n):
            if findisland1:
                break
            for j in range(n):
                if grid[i][j] == 1:
                    dfs(i,j)
                    findisland1 = True
                    break
        
        queue = deque(island1)
        res = 0

        while queue:
            for _ in range(len(queue)):
                x, y = queue.popleft()
                for dx, dy in directions:
                    nx = x + dx 
                    ny = y + dy
                    if nx >=0 and nx < n and ny >= 0 and ny < n:
                        if grid[nx][ny] == 1:
                            return res
                        elif grid[nx][ny] == 0:
                            grid[nx][ny] = -1
                            queue.append((nx,ny))
            
            res += 1
        
        return -1
                        

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