<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)

        def neighbors(cell):
            i, j = cell
            if i > 0 and grid[i - 1][j] != -1:
                yield i - 1, j
            if i + 1 < n and grid[i + 1][j] != -1:
                yield i + 1, j
            if j > 0 and grid[i][j - 1] != -1:
                yield i, j - 1
            if j + 1 < n and grid[i][j + 1] != -1:
                yield i, j + 1

        def island():
            for i in range(n):
                for j in range(n):
                    if grid[i][j] == 1:
                        grid[i][j] = -1
                        return [(i, j)]

        queue = island()
        queue2 = []
        ans = 1

        while queue:
            queueNext = []
            for cell in queue:
                for i, j in neighbors(cell):
                    (queueNext if grid[i][j] == 1 else queue2).append((i, j))
                    grid[i][j] = -1
            queue = queueNext

        while queue2:
            queueNext = []
            for cell in queue2:
                for i, j in neighbors(cell):
                    if grid[i][j] == 1:
                        return ans

                    queueNext.append((i, j))
                    grid[i][j] = -1
            ans += 1
            queue2 = queueNext

        return ans
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