<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:
        # essentially find the shortest path from island1 to island2

        # should be able to run a single BFS
        
        # don't increment distance on a path until we leave the island (can start searching from
        # arbitrary 1 cell)
        def getIslandN(cell):
            r, c = cell
            n = []
            if r > 0 and grid[r-1][c] == 1: n.append((r-1, c))
            if r < len(grid)-1 and grid[r+1][c] == 1: n.append((r+1, c))
            if c > 0 and grid[r][c-1] == 1: n.append((r, c-1))
            if c < len(grid)-1 and grid[r][c+1] == 1: n.append((r, c+1))
            return n
        
        def getIsland(cell):
            islandCells = set()
            q = [cell]
            while len(q) > 0:
                cur = q.pop()
                islandCells.add(cur)
                for n in getIslandN(cur):
                    if n not in islandCells: q.append(n)
            return islandCells
    
        def getN(cell):
            r, c = cell
            n = []
            if r > 0: n.append((r-1, c))
            if r < len(grid)-1: n.append((r+1, c))
            if c > 0: n.append((r, c-1))
            if c < len(grid[0])-1: n.append((r, c+1))
            return n

        def bfs(start, island):
            q = [(c, 0) for c in island]
            vis = set()
            shortest = float('inf')
            while len(q) > 0:
                cur = q.pop(0)
                cell, dist = cur
                if grid[cell[0]][cell[1]] == 1 and cell not in island: shortest = min(shortest, dist)
                vis.add(cell)
                for n in getN(cell):
                    if n not in vis:
                        r, c = n
                        if grid[r][c] == 0: q.append((n, dist + 1))
                        else: q.append((n, dist))
                        vis.add(n)
            return shortest
        

        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if grid[i][j] == 1:
                    island = getIsland((i, j))
                    return bfs((i, j), island)

        return -1

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