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

        v.insert({i,j});
        dfs(i-1,j,grid,v);
        dfs(i+1,j,grid,v);
        dfs(i,j-1,grid,v);
        dfs(i,j+1,grid,v);

    }
    int shortestBridge(vector<vector<int>>& grid) {
        set<pair<int,int>> v;
        int n = grid.size();
    


        for(int i=0;i<n;i++)
        {
            bool flag=false;
            for(int j=0;j<n;j++)
            {
                if(grid[i][j]==1)
                {
                    flag = true;
                    dfs(i,j,grid,v);
                    break;
                }

            }
            if(flag){break;}
        }

    
        queue<pair<int,int>> q;
        for(auto i : v)
        {
            q.push({i.first,i.second});
        }

        int ans=0;
        vector<vector<int>> dirs{{-1,0},{1,0},{0,-1},{0,1}};

        cout<<q.size();
        while(!q.empty())
        {
            int size = q.size();
            while(size)
            {
               pair<int,int> cur = q.front();
               q.pop();

               int x = cur.first;
               int y = cur.second;

            //    if(grid[x][y]==1)
            //    {
            //     return ans;
            //    }

               // explore x,y
               for(vector<int>& dir : dirs)
               {
                  int n_x = x+dir[0];
                  int n_y = y+dir[1];

                  if(n_x>=0 && n_y>=0 && n_x<n && n_y<n)
                  {
                    if(v.find({n_x,n_y})!=v.end()) 
                    {
                        continue;
                    }
                    else
                    {
                        if(grid[n_x][n_y]==1)
                        {
                            return ans;
                        }
                        v.insert({n_x,n_y});
                        q.push({n_x,n_y});
                    }
                  }
               }
              size--;

            }
            cout<<ans;
            ans++;
        }

        return -1;
        
    }
};
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