<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 = []
        
        queue = deque()
        
        def dfs(row,col):
            grid[row][col] = 2
            queue.append((row, col, 0))
            island.append((row,col,0))
            moves = ((row+1,col),(row-1,col),(row,col+1),(row,col-1))
            for x,y in moves:
                if 0<=x<n and 0<=y<n and grid[x][y] == 1:
                    dfs(x,y)
        
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    first_i, first_j = i, j
                    break
        dfs(first_i, first_j)
      
        while queue:
            r,c,dis = queue.popleft()
            if grid[r][c] == 1:return dis
            moves = ((r+1,c),(r-1,c),(r,c-1),(r,c+1))
            for x,y in moves:
                if 0<=x<n and 0<=y<n and grid[x][y]!=2:
                    if grid[x][y] == 1:return dis
                    grid[x][y] = 2
                    queue.append((x,y,dis+1))
        return -1
        
        
19300