<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 getNeighbors(row, col):
            directions = [(-1,0),(0,1),(1,0),(0,-1)]
            for d in directions:
                new_row = row + d[0]
                new_col = col + d[1]
                if not(0 <= new_row < n and 0 <= new_col < n):
                    continue
                yield (new_row, new_col)
        
        def dfs(row, col):
            grid[row][col] = 2 # Colorizing 1st island
            bfs_queue.append((row,col))
            for nei_row, nei_col in getNeighbors(row,col):
                if grid[nei_row][nei_col] != 1:
                    continue
                dfs(nei_row, nei_col)
            
        def bfs(queue):
            distance = 0
            while queue:
                next_layer = []
                for row, col in queue:
                    for nei_row, nei_col in getNeighbors(row, col):
                        if grid[nei_row][nei_col] == 1: # We reached the 2nd island
                            return distance
                        if grid[nei_row][nei_col] == 0:
                            next_layer.append((nei_row,nei_col))
                            grid[nei_row][nei_col] = -1
                    
                distance += 1
                queue = next_layer

        # Primary logic
        row = col = -1
        n= len(grid)
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    row, col = i, j
                    break

        bfs_queue = []
        dfs(row, col)
        return bfs(bfs_queue)        
            
            
18900