# GRPASSN - EDITORIAL

Setter: Utkarsh Darolia
Testers: Nishank Suresh and Abhinav sharma
Editorialist: Utkarsh Darolia

1092

None

# PROBLEM

There are N people and the group size preference of i^{th} person is A_{i}. The objective is to find out if we can assign every person to a group of their preferred size.

# EXPLANATION

To solve this problem, let’s first try solving a subproblem, which is to find if all the people with a particular value of group size preference can remain happy.

Let’s define happy(r) as a boolean, which is true if all people with group size preference r can remain happy, and false otherwise. And define count(r) as the count of people which have group size preference r.

Let k=count(r) and k\gt 0. Then, every one with group size preference r can remain happy only —

• If they can be distributed into groups of size r where each of the k people is in exactly 1 group, or
• If k=r or k=2 \times r or k=3 \times r…, or
• If k is a multiple of r, or
• If k \% r=0

So, if k > 0 and k \%r =0, then happy(r)=true, otherwise happy(r)=false. And we can say that everyone remains happy, if for all values of r from 2 to n (where count(r) \gt 0) happy(r) is true.

We’re only checking values of r with count(r)>0 as when count(r)=0 there is no sense in saying that people with group size preference r are happy or not as there are no people!

# TIME COMPLEXITY

The time complexity is O(N) per test case.

# SOLUTIONS

Editorialist's Solution
// author: Utkarsh Darolia
#include <bits/stdc++.h>
using namespace std;

int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin >> t;
for (int i = 1; i <= t; ++i)
{
int n;
cin >> n;

bool happy = true;
int count[n+1] = {};

for (int j = 1; j <= n; ++j)
{
int groupSizePreference;
cin >> groupSizePreference;
count[groupSizePreference]++;
}

for (int j = 2; j <= n; ++j)
{
if ((count[j]%j) != 0)
{
happy = false;
}
}

if (happy == true)
{
cout << "YES" << endl;
}
else
{
cout << "NO" << endl;
}
}
return 0;
}


Feel free to share your approach. Suggestions are welcome.

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