<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 m,n;
    vector<vector<int>>directions={
                {-1,0},
        {0,-1},         {0,1},
                 {1,0}
    };
    bool isSafe(int i,int j){
       return i >= 0 && i < m && j >= 0 && j < n;
    }
    void dfs(int i,int j,vector<vector<int>>&grid,set<pair<int,int>>&visited){
         if(!isSafe(i,j)||grid[i][j]==0||visited.find({i,j})!=visited.end()) return ;
         visited.insert({i,j});
         for(auto dir:directions){
            int i_=i+dir[0];
            int j_=j+dir[1];
                dfs(i_,j_,grid,visited);
            }
    }
    int bfs(vector<vector<int>>& grid, set<pair<int,int>>&visited){
        queue<pair<int,int>>q;
        for(auto &it:visited){
            q.push(it);
        }
        int level=0;
        while(!q.empty()){
           int L=q.size();
           while(L--){
              pair<int,int>P=q.front();
              q.pop();
              int x=P.first;
              int y=P.second;
              for(auto dir:directions){
                int x_=x+dir[0];
                int y_=y+dir[1];
                if(isSafe(x_,y_)&& visited.find({x_,y_})==visited.end()){
                    if(grid[x_][y_]==1) return level;
                    visited.insert({x_,y_});
                    q.push({x_,y_});
                }
              }
           }
           level++;
        }
        return -1;
    }
    int shortestBridge(vector<vector<int>>& grid) {
        m=grid.size();
        n=grid[0].size();
        set<pair<int,int>>visited;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    dfs(i,j,grid,visited);
                    return bfs(grid,visited);
                }
            }
        }
        return -1;
    }
};
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