HIGHFREQ - Editorial

PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4

Setter: Abhinav Gupta
Tester: Jeevan Jyot Singh, Nishank Suresh
Editorialist: Yash Kulkarni

DIFFICULTY:

1443

PREREQUISITES:

Hash Maps

PROBLEM:

Chef has an array A of length N.

Let F(A) denote the maximum frequency of any element present in the array.

For example:

  • If A = [1, 2, 3, 2, 2, 1], then F(A) = 3 since element 2 has the highest frequency = 3.
  • If A = [1, 2, 3, 3, 2, 1], then F(A) = 2 since highest frequency of any element is 2.

Chef can perform the following operation at most once:

  • Choose any subsequence S of the array such that every element of S is the same, say x. Then, choose an integer y and replace every element in this subsequence with y.

For example, let A = [1, 2, 2, 1, 2, 2]. A few examples of the operation that Chef can perform are:

  • [1, \textcolor{red}{2, 2}, 1, 2, 2] \to [1, \textcolor{blue}{5, 5}, 1, 2, 2]
  • [1, \textcolor{red}{2}, 2, 1, \textcolor{red}{2, 2}] \to [1, \textcolor{blue}{19}, 2, 1, \textcolor{blue}{19, 19}]
  • [\textcolor{red}{1}, 2, 2, 1, 2, 2] \to [\textcolor{blue}{2}, 2, 2, 1, 2, 2]

Determine the minimum possible value of F(A) Chef can get by performing the given operation at most once.

EXPLANATION:

Let P be the element with the maximum frequency in A (having frequency freq_P) and Q be the element with the second maximum frequency in A (having frequency freq_Q) . It is possible that Q does not exist. Let us consider these 2 cases:

  • Q exists - Initially F(A) = freq_P. We should choose x = P because we need to minimize the final value of F(A) as in the question and it is always better to have y \neq Q. We should only replace a maximum of \lfloor\frac{freq_P}{2} \rfloor occurrences of P to y because if we replace more, then y becomes the new element with the maximum frequency. Also, replacing less than \lfloor\frac{freq_P}{2} \rfloor occurrences of P to y is not optimal because frequency of P can be further reduced by replacing more occurrences of P and hence F(A) can be reduced further. Hence, it is optimal to replace \lfloor\frac{freq_P}{2} \rfloor occurrences of P to some y \neq Q. Now, freq_P - \lfloor\frac{freq_P}{2} \rfloor is the final frequency of P and freq_Q is the final frequency of Q. The maximum among these two i.e. Max( freq_P - \lfloor\frac{freq_P}{2} \rfloor, freq_Q) has to be the answer because frequency of y = \lfloor\frac{freq_P}{2} \rfloor \leq freq_P - \lfloor\frac{freq_P}{2} \rfloor = frequency of P.
  • Q does not exist - Everything will remain same like the above point just freq_Q is 0. So, the answer is freq_P - \lfloor\frac{freq_P}{2} \rfloor.

This can be implemented very easily using a hash map, which stores the number of occurrences of all elements in A.

TIME COMPLEXITY:

O(N) for each test case.

SOLUTION:

Setter's solution
#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(2,100000,'\n');
    sumN+=n;
    assert(sumN<=200000);
    int A[n+1]={0};
    int freq[n+1]={0};
    for(int i=1;i<=n;i++)
    {
        if(i!=n)
            A[i]=readInt(1,n,' ');
        else
            A[i]=readInt(1,n,'\n');
        freq[A[i]]++;
    }
    vector <int> v;
    for(int i=1;i<=n;i++)
        v.pb(freq[i]);
    sort(all(v));
    reverse(all(v));
    int x=v[0];
    int y=v[1];
    cout<<max((x+1)/2,y)<<'\n';
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL),cout.tie(NULL);
    int T=readInt(1,5000,'\n');
    while(T--)
        solve();
    assert(getchar()==-1);
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Editorialist's Solution
using namespace std;

int main() {
	int T;
	cin >> T;
	while(T--){
	    int n;
	    cin >> n;
	    unordered_map<int,int>ff;
	    for(int i=0;i<n;i++){
	        int t;
	        cin >> t;
	        ff[t]++;
	    }
	    vector<int>y;
	    for(auto i:ff)y.push_back(i.second);
	    sort(y.begin(),y.end());
	    int m=y.size();
	    int a=y[m-1];
	    int b=0;
	    if(m>1)b=y[m-2];
	    cout << max(a/2+(int)(a%2!=0),b) << endl;
	}
	return 0;
}

2 Likes

I could have got this right if only I wrote max((x+1)/2,y) :frowning:

import math
t = int(input())
while t:
    n = int(input())
    arr = list(map(int, input().split()))
    dict = {}
    for i in arr:
        if dict.get(i) == None:
            dict[i] = 1
        else:
            dict[i]+=1
    maxi = 0
    sec_maxi = 0
    for i in dict:
        if dict[i]>=maxi:
            sec_maxi = maxi
            maxi = dict[i]
    print(max(sec_maxi,math.ceil(maxi/2)))
    t-=1 

Can anyone tell what is wrong with my logic ? This code is failing three test cases.

Can anyone find a testcase for which my code fails? Sorry if it is something obvious

Solution: 70076260 | CodeChef (not the same implementation as editorialist)

Input:

1
6
2 2 2 3 3 3

Output:

2

Expected Output:

3
1 Like

Just realized I had misread the questiom

“Choose any subsequence S of the array such that every element of S is the same, say x”

Thanks though

1 Like

@prakhr98 you need to add one more line for the case if the frequency is less than max freq but greater than second highest freq

if dict[i]>sec_maxi:
sec_maxi = dict[i]

setter code is not working on this testcase
1
9
1 2 2 3 2 2 4 2 2
and in question there’s nothing mentioned that I didn’t take this testcase.

@codeforces2103 - Try the following - with an empty line below the list of numbers

1
9
1 2 2 3 2 2 4 2 2

This should run

#include<bits/stdc++.h>
#include
#include<string.h>
#define ll long long
using namespace std;
int main()
{
int t=1;
cin>>t;
while(t–)
{
ll int n;
cin>>n;
map<int,int>mp;
for(int i=0;i<n;i++)
{
int x;
cin>>x;
mp[x]++;
}
int count=mp.rbegin()->second;
//cout<<count<<endl;
if(count%2!=0)
{
count++;
}
if(mp.size()==1)
{
cout<<count/2<<endl;
continue;
}
mp.erase(mp.rbegin()->first);
int count2=(mp.rbegin())->second;
//cout<<count2<<endl;
cout<<max(count/2,count2)<<endl;

} 
return 0;

}

Anyone tell my error please

please tell