<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 init(self, grid):
        x = 0
        y = 0
        n = len(grid)
        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1:
                    x = i
                    y = j
                    return (x, y)
         
    def shortestBridge(self, grid: List[List[int]]) -> int:
        n = len(grid)
        visited = [[False]*n for _ in range(n)]
        direction = [(1,0), (-1, 0), (0, 1), (0, -1)]
        x, y = self.init(grid)
        dq = collections.deque()

        def dfs( i, j):
            if i < 0 or i >= len(grid) or j < 0 or j >=len(grid) or visited[i][j] == True or grid[i][j] == 0:
                return
            dq.append((i, j))
            visited[i][j] = True
            for di , dj in direction:
                dfs(i + di, j + dj)
            #return dq


        dfs(x, y)
        length = 0

        while dq:
            dq2 = collections.deque()
            for i in range(len(dq)):
                print(len(dq))
                x, y = dq.popleft()
                for dx, dy in direction:
                    nx = x + dx
                    ny = y + dy
                    if nx >= 0 and nx < n and ny >= 0 and ny < n and visited[nx][ny] == False:
                        if grid[nx][ny] != 0:
                            return length
                        else:
                            dq2.append((nx, ny))
                            visited[nx][ny] = True
            dq = dq2
            length = length + 1
        return -1




       
        

        
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