<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)
        direct = [[0,1],[0,-1],[1,0],[-1,0]]

        visited = set()
        def isInvalid(r,c):
            return r<0 or c<0 or r==N or c==N
        
        def dfs(r,c):
            if isInvalid(r,c) or not grid[r][c] or (r,c) in visited:
                return
            visited.add((r,c))
            for dirr,dirc in direct:
                dfs(dirr + r,dirc + c)

        def bfs():
            res, q = 0, deque(visited)
            while q:
                for i in range(len(q)):
                    r, c = q.popleft()
                    for dr, dc in direct:
                        if isInvalid(r+dr,c+dc) or (r+dr,c+dc) in visited:
                            continue
                        if grid[r+dr][c+dc]:
                            return res
                        q.append([r+dr,c+dc])
                        visited.add((r+dr,c+dc))
                res+=1
            
        for i in range(N):
            for j in range(N):
                if grid[i][j]:
                    dfs(i,j)
                    return bfs()



        
19800