<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:
        n = len(grid)
        directions = [[1,0], [-1,0], [0,1], [0,-1]]
        
        # get the first cell that is land
        def get_first_land():
            for i in range(n):
                for j in range(n):
                    if grid[i][j] == 1:
                        grid[i][j] = 2
                        return [i,j]

        first_x, first_y = get_first_land()
        
        # iterate the first land and mark as visited (2) - using bfs
        queue = deque([[first_x, first_y]])  # add all first land nodes here
        first_land_queue = deque([[first_x, first_y]]) # use this queue to iterate all land of first land

        while first_land_queue:
            x, y = first_land_queue.popleft()

            # if is in the first land, and mark as visited
            for di, dj in directions:
                i, j = x+di, y+dj
                if not (0<=i<n) or not (0<=j<n): continue
                if grid[i][j] == 1:
                    grid[i][j] = 2
                    queue.append([i,j])
                    first_land_queue.append([i,j])

        # after marked the first land, perform BFS to find the second land
        # also record the distance

        print(grid)
        print(queue)
        distance = 0

        while queue:

            for _ in range(len(queue)):
                x, y = queue.popleft()

                for di, dj in directions:
                    i, j = x + di, y + dj
                    if not (0<=i<n) or not (0<=j<n): continue
                    if grid[i][j] == 0:  # if the location is sea, change to -1 as visited
                        grid[i][j] = -1
                        queue.append([i, j])
                    elif grid[i][j] == 1: return distance
            
            distance += 1
            
        return -1
17100