<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 {
public:
    void dfs(vector<vector<int>>& grid, int x, int y, queue<pair<int,int>>& q) {
        int n = grid.size();
        if (x < 0 || y < 0 || x >= n || y >= n || grid[x][y] != 1) return;
        grid[x][y] = 2;
        q.push({x, y});
        dfs(grid, x + 1, y, q);
        dfs(grid, x - 1, y, q);
        dfs(grid, x, y + 1, q);
        dfs(grid, x, y - 1, q);
    }
    
    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        queue<pair<int, int>> q;
        bool found = false;
        
        // Step 1: Find one island and mark it as 2
        for (int i = 0; i < n && !found; ++i) {
            for (int j = 0; j < n && !found; ++j) {
                if (grid[i][j] == 1) {
                    dfs(grid, i, j, q);
                    found = true;
                }
            }
        }
        
        // Step 2: Expand the first island using BFS until we reach the second island
        int steps = 0;
        vector<pair<int, int>> directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        while (!q.empty()) {
            int size = q.size();
            while (size-- > 0) {
                auto [x, y] = q.front(); q.pop();
                for (auto [dx, dy] : directions) {
                    int nx = x + dx, ny = y + dy;
                    if (nx >= 0 && ny >= 0 && nx < n && ny < n) {
                        if (grid[nx][ny] == 1) return steps;
                        if (grid[nx][ny] == 0) {
                            grid[nx][ny] = 2;
                            q.push({nx, ny});
                        }
                    }
                }
            }
            ++steps;
        }
        return -1;
    }
};
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