<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)
        q = deque()
        directions = [(0,1),(1,0),(-1,0),(0,-1)]
        def dfs(i,j):

            if i < 0 or i == n or j < 0 or j == n or grid[i][j] != 1:

                return

            q.append((i,j,0))

            grid[i][j] = 'T'

            dfs(i+1,j)
            dfs(i-1,j)
            dfs(i,j-1)
            dfs(i,j+1)


        for i in range(n):
            c = 0
            for j in range(n):
                
                if grid[i][j] == 1:

                    dfs(i,j)
                    c = 1
                    break
            if c == 1:
                break
        

        visited = set()

        while q :

            i,j,d = q.popleft()

            if grid[i][j] == 1:

                return d -1

            if (i,j) in visited:

                continue

            visited.add((i,j))

            for di,dj in directions:

                ni = i + di
                nj = j + dj

                if 0 <= ni < n and 0 <= nj < n :
                    q.append((ni,nj,d+1))


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