<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:
        from collections import deque

        # Helper function to perform DFS and mark all land cells of an island
        def dfs(x, y, mark):
            stack = [(x, y)]
            while stack:
                cx, cy = stack.pop()
                if 0 <= cx < n and 0 <= cy < n and grid[cx][cy] == 1:
                    grid[cx][cy] = mark
                    for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                        stack.append((cx + dx, cy + dy))

        # Helper function to find the first island and mark it
        def find_and_mark_first_island():
            for i in range(n):
                for j in range(n):
                    if grid[i][j] == 1:
                        dfs(i, j, -1)
                        return

        # Start of the main function
        n = len(grid)
        find_and_mark_first_island()

        # BFS to expand the first island
        queue = deque()
        for i in range(n):
            for j in range(n):
                if grid[i][j] == -1:
                    queue.append((i, j, 0))

        # Perform BFS to find the shortest bridge
        while queue:
            x, y, dist = queue.popleft()
            for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                nx, ny = x + dx, y + dy
                if 0 <= nx < n and 0 <= ny < n:
                    if grid[nx][ny] == 1:
                        return dist
                    elif grid[nx][ny] == 0:
                        grid[nx][ny] = -1
                        queue.append((nx, ny, dist + 1))

        return -1  # Should never reach here because there are always two islands
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