<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:
        
        def findFirstIsland() -> (int, int):
            for i in range(len(grid)):
                for j in range(len(grid[0])):
                    if grid[i][j] == 1:
                        grid[i][j] = 2
                        return (i, j)
            return [-1, -1]
        
        first_x, first_y = findFirstIsland()

        first_queue = collections.deque([(first_x, first_y)])
        second_queue = collections.deque([(first_x, first_y, 0)])

        # find all points in the first island and treat them as the initial starting pos of second queue search
        while first_queue:
            x, y = first_queue.popleft()
            for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
                nx, ny = x + dx, y + dy
                if 0 <= nx < len(grid) and 0 <= ny < len(grid[0]) and grid[nx][ny] == 1:
                    grid[nx][ny] = 2
                    first_queue.append((nx, ny))
                    second_queue.append((nx, ny, 0))
        
        while second_queue:
            x, y, steps = second_queue.popleft()
            
            for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
                nx, ny = x + dx, y + dy
                if 0 <= nx < len(grid) and 0 <= ny < len(grid[0]):
                    if grid[nx][ny] == 1:
                        return steps
                    if grid[nx][ny] == 0:
                        grid[nx][ny] = -1
                        second_queue.append((nx, ny, steps + 1))
        return -1
        
        
17100