<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):
        isl1 = []

        n = len(grid)

        traversal = ((1,0),(0,1),(-1,0),(0,-1))

        for i in range(n):
            for j in range(n):
                if grid[i][j] == 1 and (i,j) not in isl1:
                    q = deque()
                    q.append((i,j))
                    isl1.append((i,j))

                    while q:
                        node = q.popleft()
                        for y,x in traversal:
                            newy,newx = node[0]+y, node[1]+x
                            if 0<=newx<n and 0<=newy<n and grid[newy][newx] and (newy,newx) not in isl1:
                                q.append((newy,newx))
                                isl1.append((newy,newx))
                    break
            else:
                continue
            break
        best_dist = -1
        q = deque(isl1)
        # for i in isl1:
        #     q.append(i)
        dist = 0
        visited = []
        while q:
            for _ in range(len(q)):
                node = q.popleft()
                for y,x in traversal:
                    newy, newx = node[0]+y, node[1]+x
                    if 0<=newx<n and 0<=newy<n:
                        if grid[newy][newx]==0 and (newy,newx) not in visited:
                            q.append((newy,newx))
                            visited.append((newy,newx))
                        elif grid[newy][newx]==1 and (newy,newx) not in isl1:
                            return dist
            dist+=1
        return best_dist
17400