➤ Problem Link : 574B. Bear and Three Musketeers
✅ C++ Solution :
#include<bits/stdc++.h>
using namespace std;
int ans;
void dfs(int i,int p,vector<int> adj[],int cnt[],bool visited[])
{
for(int j=0;j<adj[i].size();j++)
{
if(p!=-1 && p!=adj[i][j] && find(adj[p].begin(),adj[p].end(),adj[i][j])!=adj[p].end())
{
if(ans!=-1)
ans=min(ans,cnt[p]+cnt[i]+cnt[adj[i][j]]-6);
else
ans=cnt[p]+cnt[i]+cnt[adj[i][j]]-6;
}
if(!visited[adj[i][j]])
{
visited[adj[i][j]]=1;
dfs(adj[i][j],i,adj,cnt,visited);
}
else
{
for(int k=0;k<adj[adj[i][j]].size();k++)
{
if(adj[adj[i][j]][k]!=i && find(adj[i].begin(),adj[i].end(),adj[adj[i][j]][k])!=adj[i].end())
{
if(ans!=-1)
ans=min(ans,cnt[i]+cnt[adj[i][j]]+cnt[adj[adj[i][j]][k]]-6);
else
ans=cnt[i]+cnt[adj[i][j]]+cnt[adj[adj[i][j]][k]]-6;
}
}
}
}
}
int main()
{
ans=-1;
int n,m,i,j;
cin>>n>>m;
vector<int> adj[n+1];
int cnt[n+1];
bool visited[n+1];
memset(cnt,0,sizeof(cnt));
memset(visited,0,sizeof(visited));
while(m--)
{
cin>>i>>j;
adj[i].push_back(j);
adj[j].push_back(i);
cnt[i]++;
cnt[j]++;
}
for(int i=1;i<=n;i++)
{
if(!visited[i])
{
visited[i]=1;
dfs(i,-1,adj,cnt,visited);
}
}
cout<<ans<<"\n";
}
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I am a Member Technical at D.E. Shaw India Pvt. Ltd. CodingWithArt is a blog where you find tricks and solutions to challenging coding and development problems. I aim to build a universal tech platform for every enthusiast out there.