<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 {
    #define F first
    #define S second
    using state = pair<int,int>;
    const int dx[4] = {0,1,0,-1};
    const int dy[4] = {1,0,-1,0};
private:
    set<state> bfs(state node,vector<vector<int>>&g,vector<vector<int>> &vis,const int n,const int m)
    {
        queue<state> q;
        q.push(node);
        set<state> ans;
        while(!q.empty())
        {
            state node = q.front();
            q.pop();
            vector<state> temp;
            for(int i=0;i<4;i++)
            {
                int x = node.F + dx[i];
                int y = node.S + dy[i];
                if(x>=0 and x<n and y>=0 and y<m and g[x][y]==1)
                {
                    temp.push_back({x,y});
                }
            }
            if(temp.size()<4)
            {
                for(auto i:temp)ans.insert(i);
                ans.insert(node);
            }
            for(auto &v:temp)
            {
                if(!vis[v.F][v.S])
                {
                    vis[v.F][v.S] = 1;
                    q.push(v);
                }
            }
        }
        return ans;
    }
public:
    int shortestBridge(vector<vector<int>>& grid) 
    {
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> vis(n,vector<int>(m,0));
        set<state> ans;
        for(int i=0;i<n;i++)
        {
            bool f = false;
            for(int j=0;j<m;j++)
            {
                if(grid[i][j]==1)
                {
                    state node = make_pair(i,j);
                    ans = bfs(node,grid,vis,n,m);
                    f = true;
                    break;
                }
            }
            if(f)break;
        }
        // for(auto i:ans)
        // {
        //     cout<<i.first<<" "<<i.second<<endl;
        // }
        vector<vector<int>> dis(n,vector<int>(m,-1));
        queue<state> q;
        for(auto i:ans)
        {
            q.push(i);
            dis[i.F][i.S] = 0;
        }
        while(!q.empty())
        {
            state node = q.front();
            q.pop();
            vector<state> temp;
            for(int i=0;i<4;i++)
            {
                int x = node.F + dx[i];
                int y = node.S + dy[i];
                if(x>=0 and x<n and y>=0 and y<m and !vis[x][y])
                {
                    temp.push_back({x,y});
                }
            }
            for(auto &v:temp)
            {
                // cout<<v.F<<" "<<v.S<<endl;
                if(dis[v.F][v.S]==-1)
                {
                    dis[v.F][v.S] = dis[node.F][node.S] + 1;
                    q.push(v);
                    if(grid[v.F][v.S]==1)
                    {
                        return dis[v.F][v.S]-1;
                    }
                }
            }
        }
        return 0;
    }
};
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