<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 {
private:
    int n;
    int dr[4] = {-1, 0, 1, 0};
    int dc[4] = {0, 1, 0, -1};
    vector<vector<int>> g;
    set<pair<int, int>> sp; 
public:
    void dfs(int r, int c) {
        if (r < 0 || c < 0 || r >= n || c >= n || g[r][c] != 1 || sp.find({r, c}) != sp.end()) {
            return;
        }
        sp.insert({r, c});
        for (int k = 0; k < 4; k++) {
            dfs(r + dr[k], c + dc[k]);
        }
    }
    int bfs() {
        int ans = 0;
        queue<pair<int, int>> q; 
        for (auto &p : sp) {
            q.push(p); 
        }
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                auto [r, c] = q.front();
                q.pop();
                for (int k = 0; k < 4; k++) {
                    int nr = r + dr[k];
                    int nc = c + dc[k];
                    if (nr < 0 || nc < 0 || nr >= n || nc >= n || sp.find({nr, nc}) != sp.end()) {
                        continue;
                    }
                    if (g[nr][nc] == 1) {
                        return ans; 
                    }
                    sp.insert({nr, nc}); 
                    q.push({nr, nc}); 
                }
            }
            ans++;
        }
        return -1; 
    }

    int shortestBridge(vector<vector<int>>& grid) {
        n = grid.size();
        this->g = grid;
        
        // Find the first island using DFS
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) { 
                    dfs(i, j); // Mark the first island
                    return bfs(); // Start BFS to find the shortest bridge to the second island
                }
            }
        }
        
        return -1; // Should not reach here
    }
};

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