<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:
    int dx[4] = {1, 0, -1, 0};
    int dy[4] = {0, 1, 0, -1};

    void dfs(int i, int j, set<pair<int, int>>& vis, vector<vector<int>>& grid, queue<pair<int, int>>& q) {
  
        if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() || grid[i][j] == 0 || vis.find({i, j}) != vis.end()) {
            return;
        }

   
        vis.insert({i, j});
        q.push({i, j});  
       
        for (int k = 0; k < 4; k++) {
            int nr = i + dx[k];
            int nc = j + dy[k];
            dfs(nr, nc, vis, grid, q);
        }
    }

  
    int bfs(queue<pair<int, int>>& q, set<pair<int, int>>& vis, vector<vector<int>>& grid) {
        int cnt = 0;
        
        while (!q.empty()) {
            int sz = q.size();
            while (sz--) {
                auto [x, y] = q.front();
                q.pop();

         
                for (int k = 0; k < 4; k++) {
                    int nr = x + dx[k];
                    int nc = y + dy[k];


                    if (nr < 0 || nr >= grid.size() || nc < 0 || nc >= grid[0].size() || vis.find({nr, nc}) != vis.end()) {
                        continue;
                    }

                    if (grid[nr][nc] == 1) {
                        return cnt;
                    }

               
                    q.push({nr, nc});
                    vis.insert({nr, nc});
                }
            }
            cnt++; 
        }
        
        return 0;
    }

    int shortestBridge(vector<vector<int>>& grid) {
        set<pair<int, int>> vis;
        queue<pair<int, int>> q;

        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (grid[i][j] == 1) {
                    dfs(i, j, vis, grid, q); 
                    return bfs(q, vis, grid);  
                }
            }
        }
        
        return 0; \
    }
};

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