<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:

        island_points = deque()
        n = len(grid)
        d = [(0, -1), (1, 0), (0, 1), (-1, 0)]
        found = False

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

                    queue = deque([(i, j)])
                    grid[i][j] = -1
                    island_points.append((i, j, 0))

                    while queue:
                        x, y = queue.popleft()

                        for a, b in d:
                            nx, ny = x+a, y+b

                            if 0 <= nx <= n-1 and 0 <= ny <= n-1 and grid[nx][ny] != -1 and grid[nx][ny] != 0:
                                queue.append((nx, ny))
                                grid[nx][ny] = -1
                                island_points.append((nx, ny, 0))

                    found = True
                    break

            if found:
                break
        
        while island_points:
            x, y, steps = island_points.popleft()

            for a, b in d:
                nx, ny = x+a, y+b

                if 0 <= nx <= n-1 and 0 <= ny <= n-1 and grid[nx][ny] != -1:
                    if grid[nx][ny] == 1:
                        return steps
                    island_points.append((nx, ny, steps+1))
                    grid[nx][ny] = -1
17000