<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 isvalid(self, i, j, n):
        return (i<n and i>=0 and j>=0 and j<n)
    
    def dfs(self, q, grid, i, j):
        n = len(grid)
        if grid[i][j]!=1:
            return
        q.append((i,j))
        grid[i][j] = -1
        move = [[-1,0],[1,0],[0,-1],[0,1]]
        for c in move:
            if self.isvalid(i+c[0], j+c[1], n): 
                self.dfs(q, grid, i+c[0], j+c[1])       

    def shortestBridge(self, grid: List[List[int]]) -> int:
        n = len(grid)
        move = [[-1,0],[1,0],[0,-1],[0,1]]
        q = []
        flag = 0
        for i in range(n):
            for j in range(n):
                if grid[i][j]==1:
                    self.dfs(q, grid, i, j)
                    flag=1
                    break
            if flag==1:
                break
        level=0
        while len(q)>0:
            level+=1
            for k in range(len(q)):
                x, y = q.pop(0)
                for c in move:
                    i = x+c[0]
                    j = y+c[1]
                    if self.isvalid(i, j, n) and grid[i][j]!=-1:
                        if grid[i][j] == 1:
                            return level-1
                        grid[i][j]=-1
                        q.append((i, j))    
        return level





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