PROBLEM LINK:
Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4
Setter: Utkarsh Gupta
Tester: Manan Grover, Abhinav sharma
Editorialist: Devendra Singh
DIFFICULTY:
1469
PREREQUISITES:
PROBLEM:
You are given 2 arrays A and B of length N each. Determine the number of good pairs.
A pair (i,j) is said to be good if:
- i \lt j
- A_i = B_j
- A_j = B_i
EXPLANATION:
Iterate over the arrays A and B from j=1 to N and keep count of the pair formed by (A_j,B_j). Let us now count the number of indices i for each index j (i<j) such that (i,j) form a good pair.
If (i,j) form a good pair then :
A_j=B_i and B_j=A_i;
This means the pair (A_i,B_i) is same as the pair (B_j,A_j);
Thus for each j add count of the pair (B_j,A_j) to the answer and increase the count of (A_j,B_j) by 1
TIME COMPLEXITY:
O(Nlog(N)) for each test case.
SOLUTION:
Setter's solution
//Utkarsh.25dec
#include <bits/stdc++.h>
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
using namespace std;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
vector <int> adj[N];
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
return x;
} else {
assert(false);
}
}
}
string readString(int l,int r,char endd){
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
int sumN=0;
void solve()
{
int n=readInt(1,100000,'\n');
sumN+=n;
assert(sumN<=200000);
int A[n+1]={0}, B[n+1]={0};
for(int i=1;i<=n;i++)
{
int c;
if(i<n)
c=readInt(1,1000000000,' ');
else
c=readInt(1,1000000000,'\n');
A[i]=c;
}
for(int i=1;i<=n;i++)
{
int c;
if(i<n)
c=readInt(1,1000000000,' ');
else
c=readInt(1,1000000000,'\n');
B[i]=c;
}
map <pair<ll,ll>,ll> cnt;
ll ans=0;
for(int i=1;i<=n;i++)
cnt[mp(A[i],B[i])]++;
for(int i=1;i<=n;i++)
{
cnt[mp(A[i],B[i])]--;
ans+=cnt[mp(B[i],A[i])];
}
cout<<ans<<'\n';
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
int T=readInt(1,1000,'\n');
while(T--)
solve();
assert(getchar()==-1);
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Editorialist's Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
const int N = 1e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void sol(void)
{
int n;
cin >> n;
int a[n], b[n];
for (int i = 0; i < n; i++)
cin >> a[i];
for (int i = 0; i < n; i++)
cin >> b[i];
map<pll, ll> cnt;
ll ans = 0;
for (int i = 0; i < n; i++)
{
pll p = {b[i], a[i]};
ans += cnt[p];
p = {p.S, p.F};
cnt[p]++;
}
cout << ans << '\n';
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}
