<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(vector<vector<int>>&grid, set<pair<int,int>>&visited , int i, int j){
         int n = grid.size();
         if(i<0 or j<0 or i>=n or j>=n or grid[i][j]==0)return;
         visited.insert({i,j});
         grid[i][j]=0;
         dfs(grid, visited, i+1, j);
         dfs(grid, visited, i-1, j);
         dfs(grid, visited, i, j-1);
         dfs(grid, visited, i,j+1);
    }

    int bfs(vector<vector<int>>& grid,set<pair<int,int>> &visited){
        int len=0, n = grid.size();
        queue<pair<int,int>>q;
        for(auto v : visited){
            q.push(v);
        }

        while(!q.empty()){
            
            int sz = q.size();
            for(int i=0;i<sz;i++){
             auto [r,c]=q.front();
             q.pop();
             if(grid[r][c]==1)return len-1;
             int drow[] = {1,-1,0,0} , dcol[] = {0,0,-1, 1};
              for(int j=0;j<4;j++){
                 int nR= r+drow[j] ,nC= c+dcol[j];
                 if(nR<0 or nC < 0 or nR >= n or nC >=n or visited.count({nR,nC})> 0)continue;
                 visited.insert({nR,nC});
                 q.push({nR,nC});

              }
              

            }

            len++;
        }
    return -1;
    }

    int shortestBridge(vector<vector<int>>& grid) {
         int n = grid.size();
         set<pair<int,int>>visited;
         for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]){
                    dfs(grid,visited, i,j);
                    return bfs(grid,visited);
                }
            }
         }
      return -1;
    }
};
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