<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 bfs(vector<vector<int>>& grid,vector<vector<int>>&vis,int srow,int scol, vector<pair<int,int>> &island){
        int n=grid.size();
        queue<pair<int,int>>q;
        int dx[]={1,-1,0,0};
        int dy[]={0,0,1,-1};
        island.push_back({srow,scol});
        vis[srow][scol]=1;
        q.push({srow,scol});
        while(!q.empty()){
            auto it =q.front();
            int r=it.first;
            int c=it.second;
            
            q.pop();
            
            for(int i=0;i<4;i++){
                int nr=r+dx[i];
                int nc=c+dy[i];
                if(nr>=0 && nr<n && nc>=0 && nc<n){
                    if(grid[nr][nc]==1 && vis[nr][nc]==-1){
                        island.push_back({nr,nc});
                        q.push({nr,nc});
                        vis[nr][nc]=1;
                    }
                    
                }
            }

        }
    }
    int shortestBridge(vector<vector<int>>& grid) {
        queue<pair<pair<int,int>,int>>q; // row,col,moves
        int n=grid.size();
        int dx[]={1,-1,0,0};
        int dy[]={0,0,1,-1};
        vector<pair<int,int>> island1,island2;
        vector<vector<int>> vis(grid.size(),vector<int> (grid.size(),-1));
        vector<vector<int>> vis2(grid.size(),vector<int> (grid.size(),-1));
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1 && vis[i][j]==-1){
                    if(island1.size()==0){

                    bfs(grid,vis,i,j,island1);
                    }
                    else{
                        bfs(grid,vis,i,j,island2);
                    }
                }
            }
        }
        
        for (auto p : island1) {
            q.push({p, 0});
            vis2[p.first][p.second] = 1; 
        }
        while(!q.empty()){
            auto it=q.front();
            int r=it.first.first;
            int c=it.first.second;
            int m=it.second;
            q.pop();
            for(auto i:island2){
                if(i.first==r && i.second==c){
                    return m;
                }
            }
            
            for(int i=0;i<4;i++){
                int nr=r+dx[i];
                int nc=c+dy[i];
                if(nr>=0 && nr<n && nc>=0 && nc<n && vis2[nr][nc]==-1){
                    if(grid[nr][nc]==1){
                        q.push({{nr,nc},m});
                    }else{
                        q.push({{nr,nc},m+1});
                        vis2[nr][nc]=1;
                    }
                    
                }
            }
        }
        
        return -1;
    }
};
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