<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):
        m, n = len(grid), len(grid[0])

        #dfs the first island
        #having all visited point as condidates
        #bfs until reach another island
        start = None
        for i in range(m):
            for j, p in enumerate(grid[i]):
                if p == 1:
                    start = (i, j)
        
        stack = [start]
        visited = set(stack)
        while stack:
            i, j = stack.pop()
            visited.add((i, j))  
            for ii, jj in (i-1, j), (i, j-1), (i, j+1), (i+1, j):
                if 0 <= ii < m and 0 <= jj < n and grid[ii][jj] and (ii, jj) not in visited:
                    stack.append((ii, jj))
                    visited.add((ii, jj))
        
        
        ans = 0
        queue = list(visited)
        while queue:
            new_queue = []
            for i, j in queue:
                for ii, jj in (i-1, j), (i, j-1), (i, j+1), (i+1, j):
                    if 0 <= ii < m and 0 <= jj < n and (ii, jj) not in visited:
                        if grid[ii][jj] == 1:
                            return ans
                        new_queue.append((ii, jj))
                        visited.add((ii, jj))
            queue = new_queue
            ans += 1
17800