<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:


        m = len(grid)
        n = len(grid)

        directions = [[0,1],[1,0],[-1,0],[0,-1]]

        first_x = -1
        first_y = -1       

        bfs_queue = []
        for i in range(m):
            for j in range(n):
                if(grid[i][j]==1):
                    first_x = i
                    first_y = j
                    break
        self.dfs(grid,first_x,first_y,m,n,bfs_queue)

        distance = 0
        
        while(bfs_queue):
            new_queue = []
            for curr_x,curr_y in bfs_queue:
                for direction in directions:
                    x  = curr_x+direction[0]
                    y = curr_y+direction[1]
                    if(x>=m or x<0 or y>=n or y<0 or grid[x][y]==-1):
                        continue                
                    if(grid[x][y]==1):
                        return distance
                    elif(grid[x][y]==0):
                        new_queue.append([x,y])
                        grid[x][y] = -1

            bfs_queue = new_queue
            distance+=1

        return distance

    def dfs(self,grid,i,j,m,n,bfs_queue):
        
        if(i>=m or i<0 or j>=n or j<0 or grid[i][j]==0 or grid[i][j]==2):
            return
        grid[i][j] = 2
        bfs_queue.append([i,j])
        self.dfs(grid,i+1,j,m,n,bfs_queue)  
        self.dfs(grid,i,j+1,m,n,bfs_queue)
        self.dfs(grid,i-1,j,m,n,bfs_queue)
        self.dfs(grid,i,j-1,m,n,bfs_queue)


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