<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<pair<int,int>>dir={{1,0},{0,1},{-1,0},{0,-1}};
void dfs(vector<vector<int>>& grid,vector<vector<int>>&visited,int r,int c,queue<pair<int,int>>&q){
    for(int i=0;i<4;i++){
        int nr=r+dir[i].first;
        int nc=c+dir[i].second;
        if(nr<0 || nc<0 ||nr>=grid.size()||nc>=grid[0].size()||visited[nr][nc]==1||grid[nr][nc]==0) continue;
        visited[nr][nc]=1;
        q.push({nr,nc});
        dfs(grid,visited,nr,nc,q);
    }
}
    int shortestBridge(vector<vector<int>>& grid) {
        int ans=INT_MAX;
        int m=grid.size();
        int n=grid[0].size();
        queue<pair<int,int>>q;
        vector<vector<int>>visited(m,vector<int>(n,0));
        bool flag=true;
        for(int i=0;i<m && flag;i++){
            for(int j=0;j<n && flag;j++){
                if(grid[i][j]==1){
                    visited[i][j]=1;
                    q.push({i,j});
                    dfs(grid,visited,i,j,q);
                   flag=false;
                }
            }
        }
        int level=0;
        q.push({-1,-1});
        while(q.size()>0){
            int r=q.front().first;
            int c=q.front().second;
            q.pop();
            if(r==-1 && c==-1){
                if(q.size()!=0){
                    q.push({-1,-1});
                   level++;
                }
                else{
                    break;
                }
            }
            for(int i=0;i<4;i++){
          int nr=r+dir[i].first;
          int nc=c+dir[i].second;
        if(nr<0 || nc<0 ||nr>=grid.size()||nc>=grid[0].size()||visited[nr][nc]==1) continue;
        if(grid[nr][nc]==1){
            ans=min(ans,level);
        }
        visited[nr][nc]=1;
        q.push({nr,nc});
        }
        }
        return ans;
    }
};
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