In this HackerEarth Shil and LCP Pairs problem solution Shil came across N strings - s1 , s2 , s3 .. sN . He wants to find out total number of unordered pairs (i, j) such that LCP(si , sj) = k. You have to do thi for every k in [0, L]. Here L = max ( len(si) ) and LCP denotes longest common prefix.


HackerEarth Shil and LCP Pairs problem solution


HackerEarth Shil and LCP Pairs problem solution.

#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define ll  long long int
#define pi pair<ll,ll>
#define pii pair<pi,int>
#define f first
#define s second
#define pb push_back
struct trie{
    int f;
    trie* child[26];
};
ll A[100011];
trie* insert(trie*t,string &s,int i){

    if(i==s.length()){
        t->f++;
        return t;
    }

    if(t->child[s[i]-'a']==NULL){
        t->child[s[i]-'a']=new trie();
        t->child[s[i]-'a']->f=0;
    }

    t->f-=t->child[s[i]-'a']->f;
    t->child[s[i]-'a']=insert(t->child[s[i]-'a'],s,i+1);
    t->f+=t->child[s[i]-'a']->f;

    return t;
}
void calc(trie* t,int d){
    if(t==NULL) return;
    ll cnt=0;
    ll cnt1=0;
    ll r;
    rep(i,26){
        if(t->child[i]){
            r=t->child[i]->f;
            cnt+=r;
            cnt1+=r*r;
            calc(t->child[i],d+1);
        }
    }
    r=cnt*cnt-cnt1;
    r/=2;
    A[d]+=r;
    r = t->f - cnt;
    A[d]+=r*cnt;
    A[d]+=(r*(r-1))/2;
}
int main(){

    freopen("input-15.txt","r",stdin);
    freopen("output-15.txt","w",stdout);
    int N;
    cin >> N;
    string s[N];
    trie*t=new trie();
    int maxlen=0;
    rep(i,N){
        cin >> s[i];
        maxlen=max(maxlen,(int)s[i].size());
        t=insert(t,s[i],0);
    }
    calc(t,0);
    rep(i,maxlen+1){
        cout<<A[i]<<" ";
    }
}