<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 row,int col,vector<vector<int>>& vis,vector<vector<int>>& grid,queue<pair<int,pair<int,int>>>& q){
        vis[row][col]=1;
        q.push({0,{row,col}});
        int delrow[]={-1,0,1,0};
        int delcol[]={0,1,0,-1};
        for(int i=0;i<4;i++){
            int nrow=row+delrow[i];
            int ncol=col+delcol[i];
            if(nrow>=0 && nrow<grid.size() && ncol>=0 && ncol<grid[0].size() && !vis[nrow][ncol] && grid[nrow][ncol]==1){
                dfs(nrow,ncol,vis,grid,q);
            }
        }
    }
    int shortestBridge(vector<vector<int>>& grid) {
        int n=grid.size();
        int startrow=0,startcol=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    startrow=i;
                    startcol=j;
                    break;
                }
            }
        }
        queue<pair<int,pair<int,int>>> q;
        vector<vector<int>> vis(n,vector<int>(n,0));
        dfs(startrow,startcol,vis,grid,q);
        while(!q.empty()){
            auto it=q.front();
            q.pop();
            int cnt=it.first;
            int row=it.second.first;
            int col=it.second.second;
            int delrow[]={-1,0,1,0};
            int delcol[]={0,1,0,-1};
            for(int i=0;i<4;i++){
                int nrow=row+delrow[i];
                int ncol=col+delcol[i];
                if(nrow>=0 && nrow<n && ncol>=0 && ncol<n && !vis[nrow][ncol]){
                    vis[nrow][ncol]=1;
                    q.push({cnt+1,{nrow,ncol}});
                    if(grid[nrow][ncol]==1) return cnt;
                }
            }
        }
        return -1;
    }
};
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