<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:
        rows = len(grid)
        cols = len(grid[0])

        q = []

        def convert2(y,x):
            if y < 0 or y >= rows or x < 0 or x >= cols or grid[y][x] == 0 or grid[y][x] == 2:
                return
            q.append((y,x))
            grid[y][x] = 2
            convert2(y+1,x)
            convert2(y-1,x)
            convert2(y,x+1)
            convert2(y,x-1)

        def changeOneIsland():
            for row in range(rows):
                for col in range(cols):
                    if grid[row][col] == 1:
                        convert2(row,col)
                        return
        
        changeOneIsland()

        result = 0
        while q:
            for i in range(len(q)):
                curry, currx = q.pop(0)

                if curry >0: 
                    if grid[curry-1][currx] == 1:
                        return result
                    elif grid[curry-1][currx] == 0:
                        grid[curry-1][currx] = 2
                        q.append((curry-1,currx))

                if curry < rows-1: 
                    if grid[curry+1][currx] == 1:
                        return result
                    elif grid[curry+1][currx] == 0:
                        grid[curry+1][currx] = 2
                        q.append((curry+1,currx))

                if currx < cols-1: 
                    if grid[curry][currx+1] == 1:
                        return result
                    elif grid[curry][currx+1] == 0:
                        grid[curry][currx+1] = 2
                        q.append((curry,currx+1))

                if currx > 0: 
                    if grid[curry][currx-1] == 1:
                        return result
                    elif grid[curry][currx-1] == 0:
                        grid[curry][currx-1] = 2
                        q.append((curry,currx-1))

                
                
            result += 1

        return result

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