# CHFPLN - Editorial

Tester: Aryan Choudhary
Editorialist: Vichitr Gandas

SIMPLE

Map or Hashmap

# PROBLEM:

Given N integers A_1,A_2,\ldots,A_N. For each index i (1≤i≤N), divide A_i into two positive integers x and y such that x+y=A_i, then place this as a point (x,y) in the infinite 2-dimensional coordinate plane. Find the maximum number of distinct points that can be put in the plane, if we optimally split the values A_i. Note that you can only perform one split for each index.

# EXPLANATION

For a given number K, how many positive integer pairs (x,y) exist such that x+y=K, where x,y>0.
The answer is K-1. The following pairs: (1, K-1), (2, K-2), \ldots,(K-1,1).

Now using this fact, if a number X appears C_X times in A then we can get min(C_X, X-1) distinct coordinate pairs from them.
Find the sum of min(C_X, X-1) for every distinct number X in A.

# TIME COMPLEXITY:

O(N) if using hash maps
or O(N \log N) if using maps to find frequency.

# SOLUTIONS:

Setter's Solution
``````#include <bits/stdc++.h>
using namespace std;

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define int long long

const int N = 2e5 + 5;

int n;
int a[N];

int32_t main()
{
IOS;
int t;
cin >> t;
while(t--)
{
cin >> n;
map<int, int> m;
for(int i = 1; i <= n; i++)
{
cin >> a[i];
m[a[i]]++;
}
int ans = 0;
for(auto &it:m)
ans += min(it.second, it.first - 1);
cout << ans << endl;
}
return 0;
}
``````
Tester's Solution
``````/* in the name of Anton */

/*
Compete against Yourself.
Author - Aryan (@aryanc403)
Atcoder library - https://atcoder.github.io/ac-library/production/document_en/
*/

#ifdef ARYANC403
#else
#pragma GCC optimize ("Ofast")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
//#pragma GCC optimize ("-ffloat-store")
#include<bits/stdc++.h>
#define dbg(args...) 42;
#endif

using namespace std;
#define fo(i,n)   for(i=0;i<(n);++i)
#define repA(i,j,n)   for(i=(j);i<=(n);++i)
#define repD(i,j,n)   for(i=(j);i>=(n);--i)
#define all(x) begin(x), end(x)
#define sz(x) ((lli)(x).size())
#define pb push_back
#define mp make_pair
#define X first
#define Y second
#define endl "\n"

typedef long long int lli;
typedef long double mytype;
typedef pair<lli,lli> ii;
typedef vector<ii> vii;
typedef vector<lli> vi;

const auto start_time = std::chrono::high_resolution_clock::now();
void aryanc403()
{
#ifdef ARYANC403
auto end_time = std::chrono::high_resolution_clock::now();
std::chrono::duration<double> diff = end_time-start_time;
cerr<<"Time Taken : "<<diff.count()<<"\n";
#endif
}

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;
}
assert(l<=x&&x<=r);
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) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

assert(getchar()==EOF);
}

vi a(n);
for(int i=0;i<n-1;++i)
return a;
}

const lli INF = 0xFFFFFFFFFFFFFFFL;

lli seed;
inline lli rnd(lli l=0,lli r=INF)
{return uniform_int_distribution<lli>(l,r)(rng);}

class CMP
{public:
bool operator()(ii a , ii b) //For min priority_queue .
{    return ! ( a.X < b.X || ( a.X==b.X && a.Y <= b.Y ));   }};

void add( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt==m.end())         m.insert({x,cnt});
else                    jt->Y+=cnt;
}

void del( map<lli,lli> &m, lli x,lli cnt=1)
{
auto jt=m.find(x);
if(jt->Y<=cnt)            m.erase(jt);
else                      jt->Y-=cnt;
}

bool cmp(const ii &a,const ii &b)
{
return a.X<b.X||(a.X==b.X&&a.Y<b.Y);
}

const lli mod = 1000000007L;
// const lli maxN = 1000000007L;

lli T,n,i,j,k,in,cnt,l,r,u,v,x,y;
lli m;
string s;
vi a;
//priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue .

int main(void) {
ios_base::sync_with_stdio(false);cin.tie(NULL);
// freopen("txt.in", "r", stdin);
// freopen("txt.out", "w", stdout);
// cout<<std::fixed<<std::setprecision(35);
while(T--)
{

map<lli,lli> mm;
for(auto x:a)
mm[x]++;
lli ans=0;
for(auto x:mm){
ans+=min(x.X-1,x.Y);
}
cout<<ans<<endl;
}   aryanc403();
return 0;
}
``````
Editorialist's Solution
``````/*
* @author: vichitr
* @date: 25th July 2021
*/

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define fast ios::sync_with_stdio(0); cin.tie(0);

void solve() {
int n; cin >> n;
int a[n];
map<int, int> C;
for (int i = 0; i < n; i++) {
cin >> a[i];
C[a[i]]++;
}
int ans = 0;
for (auto i : C) {
ans += min(i.second, i.first - 1);
}
cout << ans << '\n';
}

signed main() {
fast;

#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif

int t = 1;
cin >> t;
for (int tt = 1; tt <= t; tt++) {
// cout << "Case #" << tt << ": ";
solve();
}
return 0;
}
``````

If you have other approaches or solutions, let’s discuss in comments.If you have other approaches or solutions, let’s discuss in comments.

1 Like