<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)

        island = None
        # 1. Find island
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    break
            else:
                continue
            break

        # 2. Find all shore tiles
        shore = set()
        queue = set()
        queue.add((i, j))
        while queue:
            i, j = queue.pop()
            grid[i][j] = -1
            
            if i < n - 1:
                if grid[i + 1][j] == 1:
                    queue.add((i + 1, j))
                elif grid[i + 1][j] == 0:
                    shore.add((i, j))

            if i > 0:
                if grid[i - 1][j] == 1:
                    queue.add((i - 1, j))
                elif grid[i - 1][j] == 0:
                    shore.add((i, j))

            if j < n - 1:
                if grid[i][j + 1] == 1:
                    queue.add((i, j + 1))
                elif grid[i][j + 1] == 0:
                    shore.add((i, j))
            
            if j > 0:
                if grid[i][j - 1] == 1:
                    queue.add((i, j - 1))
                elif grid[i][j - 1] == 0:
                    shore.add((i, j))

        # 3. Expand shore with BFS
        queue = defaultdict(set)
        queue[0] = shore
        dist = 0
        while queue:
            while queue[dist]:
                i, j = queue[dist].pop()
                grid[i][j] = -1
                
                if i < n - 1:
                    if grid[i + 1][j] == 0:
                        queue[dist + 1].add((i + 1, j))
                    elif grid[i + 1][j] == 1:
                        return dist
                
                if i > 0:
                    if grid[i - 1][j] == 0:
                        queue[dist + 1].add((i - 1, j))
                    elif grid[i - 1][j] == 1:
                        return dist

                if j < n - 1:   
                    if grid[i][j + 1] == 0:
                        queue[dist + 1].add((i, j + 1))
                    elif grid[i][j + 1] == 1:
                        return dist
                
                if j > 0:
                    if grid[i][j - 1] == 0:
                        queue[dist + 1].add((i, j - 1))
                    elif grid[i][j - 1] == 1:
                        return dist
            
            dist += 1

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