<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<vector<int>>d{{1,0},{0,1},{-1,0},{0,-1}};
    int n;
    int m;
    bool safe(int i,int j){
         return(i>=0 && i<n && j>=0 && j<m);
    }
    void dfs(vector<vector<int>>&grid,vector<vector<bool>>&visited,int i ,int j){
         
        if(!safe(i,j)|| visited[i][j] ||grid[i][j]==0)return;
        visited[i][j]=true;
         for(auto dir:d){
            int x=i+dir[0];
            int y=j+dir[1];
            dfs(grid,visited,x,y);
         }
    }
    int bfs(vector<vector<int>>&grid,vector<vector<bool>>&visited){
        queue<pair<int,int>>q;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(visited[i][j]){
                    q.push({i,j});
                }
            }
        }

        int level=0;
        while(!q.empty()){
            int N=q.size();
            while(N--){
                pair<int,int>temp=q.front();
                q.pop();
                for(auto it:d){
                    int X=temp.first+it[0];
                    int Y=temp.second+it[1];
                    if(safe(X,Y) && !visited[X][Y]){
                        if(grid[X][Y]==1)
                    return level;

                    visited[X][Y]=true;
                    q.push({X,Y});
                }
            }
            }
            level++;
        }
        return -1;
    }
    int shortestBridge(vector<vector<int>>& grid) {
         n=grid.size();
         m=grid[0].size();
        
        vector<vector<bool>> visited(n,vector<bool>(m,false));
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==1){
                    dfs(grid,visited,i,j);
                    return bfs(grid,visited);
                }
            }
        }
         return -1;
    }
};
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