<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>>directions={{-1,0},{1,0},{0,1},{0,-1}};
    bool isSafe(vector<vector<int>>&grid,int x,int y){
        int n=grid.size();
        int m=grid[0].size();
        return x>=0 && x<n && y>=0 && y<m;
    }
    void dfs(int i,int j,vector<vector<int>>&grid){
        grid[i][j]=2;
        for(auto it:directions){
            int ni=i+it[0];
            int nj=j+it[1];
            if(isSafe(grid,ni,nj) && grid[ni][nj]==1){
                dfs(ni,nj,grid);
            }
        }
    }
    int shortestBridge(vector<vector<int>>&grid) {
        int n=grid.size();
        int m=grid[0].size();
        queue<pair<int,int>>q;
        for(int i=0;i<n;i++){
            bool flag=false;
            for(int j=0;j<m;j++){
                if(grid[i][j]==1){
                    dfs(i,j,grid);
                    flag=true;
                    break;
                }
            }if(flag)break;
        }
        for(auto it:grid){
            for(auto j:it){
                cout<<j<<" ";
            }cout<<endl;
        }

        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]==2){
                    q.push({i,j});
                    visited[i][j]=true;
                }
            }
        }int step=0;
        while(!q.empty()){
            int size=q.size();
            while(size--){
                auto p=q.front();
                int x=p.first;
                int y=p.second;
                if(grid[x][y]==1)return step-1;
                q.pop();
                for(auto it:directions){
                    int nx=x+it[0];
                    int ny=y+it[1];
                    if(isSafe(grid,nx,ny) && !visited[nx][ny]){
                        q.push({nx,ny});
                        visited[nx][ny]=true;
                    }
                }
            }step++;
        }return -1;
    }
};
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