<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:
    vector<pair<int,int>> tee;
    int ans = INT_MAX;
    
    void sol(vector<vector<int>>& grid, vector<vector<int>>& visi, int h) {
        queue<vector<int>> qu;

        // Initialize the BFS with all points from the first island
        for (auto& p : tee) {
            qu.push({p.first, p.second, 0});
        }
        
        vector<vector<int>> directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        
        while (!qu.empty()) {
            int i = qu.front()[0];
            int j = qu.front()[1];
            h = qu.front()[2];
            qu.pop();
            
            visi[i][j] = 1;
            
            for (auto& d : directions) {
                int ni = i + d[0], nj = j + d[1];
                
                if (ni >= 0 && ni < grid.size() && nj >= 0 && nj < grid[0].size() && visi[ni][nj] == 0) {
                    if (grid[ni][nj] == 1 && (find(tee.begin(), tee.end(), make_pair(ni, nj)) == tee.end())) {
                        ans = min(ans, h);
                        return;
                    }
                    qu.push({ni, nj, h + 1});
                    visi[ni][nj] = 1;
                }
            }
        }
    }

    void ddfs(int i, int j, vector<vector<int>>& grid, vector<vector<int>>& visi) {
        if (visi[i][j] == 1) return;
        
        visi[i][j] = 1;
        tee.push_back({i, j});
        
        for (int di = -1; di <= 1; di++) {
            for (int dj = -1; dj <= 1; dj++) {
                if (abs(di) != abs(dj)) {
                    int ni = i + di, nj = j + dj;
                    if (ni >= 0 && ni < grid.size() && nj >= 0 && nj < grid[0].size() && grid[ni][nj] == 1) {
                        ddfs(ni, nj, grid, visi);
                    }
                }
            }
        }
    }

    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        
        // Find the first island and mark it
        vector<vector<int>> visi(n, vector<int>(m, 0));
        bool found = false;
        
        for (int i = 0; i < n && !found; i++) {
            for (int j = 0; j < m && !found; j++) {
                if (grid[i][j] == 1) {
                    ddfs(i, j, grid, visi);
                    found = true;
                }
            }
        }
        
        // Perform BFS to find the shortest bridge
        fill(visi.begin(), visi.end(), vector<int>(m, 0));
        sol(grid, visi, 0);
        
        return ans;
    }
};

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