<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:
        queue = deque()
        island1 = set()
        N = len(grid)
        directions = [(-1, 0), (0, -1), (1, 0), (0, 1)]
        def dfs(i,j):
            if (i,j) in island1 or i < 0 or j < 0 or i >= N or j>= N or grid[i][j] == 0:
                return
            
            island1.add((i,j))
            queue.append(((i,j), 0))

            for di, dj in directions:
                dfs(i + di, j + dj)

        def find_first_island():
            for i in range(N):
                for j in range(N):
                    if grid[i][j] == 1:
                        dfs(i,j)
                        return
        
        find_first_island()

        while queue:
            (i, j), distance = queue.popleft()
            for di, dj in directions:
                ni, nj = i + di, j + dj
                if ni >= 0 and ni < N and nj >= 0 and nj < N:
                    if grid[ni][nj] == 1 and (ni, nj) not in island1:
                        return distance 
                    if (ni, nj) not in island1:
                        island1.add((ni, nj))
                        queue.append(((ni, nj), distance + 1))
20400