<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:
 
        directions = [(0,1), (0,-1), (1,0), (-1,0)]  # right, left, up, down
        
        n = len(grid)
        m = len(grid[0])
        
        # find one point of a land
        def one_land(grid):
            for in_x, row in enumerate(grid):
                for in_y, val in enumerate(row):
                    if val:
                        return (in_x, in_y)
                    
        # start from one point and get all position of a land
        in_x, in_y = one_land(grid)
        queue = [(in_x, in_y)]
        first_land = [(in_x, in_y)]     # temp queue to find the first land
        
        from collections import defaultdict
        visited = defaultdict(lambda:False, {})
        visited[(in_x, in_y)] = True        
        
        # find positions of the first land
        while first_land:
            x, y = first_land.pop(0)
            for dir_x, dir_y in directions:
                nei_x = x + dir_x
                nei_y = y + dir_y
                if nei_x in range(n) and nei_y in range(m):       # point in range
                    if grid[nei_x][nei_y]:   # it is land
                        if not visited[(nei_x, nei_y)]:
                            first_land.append((nei_x, nei_y))
                            queue.append((nei_x, nei_y))
                            visited[(nei_x, nei_y)] = True
        
       
        
        # BFS
        cost = 0
        layer_size = 0
        while queue:
            layer_size = len(queue)
            for i in range(layer_size):
                x, y = queue.pop(0)
                for dir_x, dir_y in directions:
                    nei_x = x + dir_x
                    nei_y = y + dir_y
                    if nei_x in range(n) and nei_y in range(m):       # point in range
                        if not visited[(nei_x, nei_y)]:
                            if grid[nei_x][nei_y]:   # it is land (second land)
                                return cost
                            else:           # it is water
                                queue.append((nei_x, nei_y))
                                visited[(nei_x, nei_y)] = True
            cost += 1
        return None
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