# PROBLEM LINK:

Contest Division 1

Contest Division 2

Contest Division 3

Contest Division 4

Setter:Tejas Pandey

Tester: Manan Grover, Abhinav Sharma

Editorialist: Devendra Singh

# DIFFICULTY:

1390

# PREREQUISITES:

None

# PROBLEM:

Chef is teaching his class of N students at Hogwarts. He groups students with the same height together for an activity. Some of the students end up in a groups with only themselves and are saddened by this.

With the help of his magic wand, Chef can **increase** the height of any student to any value he likes. Now Chef wonders, what is the minimum number of students whose height needs to be increased so that there are no sad students?

# EXPLANATION:

For all students we first need to find how many students are going to end up in a group with only themselves which is same as counting students with unique heights. Let K be the number of students which have a height H such that count of H in the array is 1.

Now there are four cases :

- K=1 and H is not the maximum height of all the students: The answer is 1 in this case , just increase the height of this particular student to the maximum value of heights of all students.
- K=1 and H is the maximum height of all students but there exists a group of at least 3 students grouped together: The answer is 1 in this case just pick any student from any group with at least 3 students and increase his/her height to the maximum height of all students.
- K=1 and H is the maximum height of all students and all groups consist of 2 students only. In this case the answer cannot be less than 2 as increasing any students height to the maximum height still results in a student who has a unique height. Thus we need to increase the height of at least 2 students to the maximum height of all students in this case.
- K>1 : The answer is ceil((K+1)/2) in this case. If K is even we can form K/2 pairs by increasing the height of students.(Sort the students according to height, increase the height of first student to make it equal to second students height and so on from the third student). If K is odd form (K-3)/2 pairs and perform 2 operations to a make of group 3 students (Sort the students according to height, increase the height of first and second student to make it equal to third student’s height and then follow the same method for even case).

# TIME COMPLEXITY:

O(N) for each test case.

# SOLUTION:

## Setter's solution

```
#include <bits/stdc++.h>
using namespace std;
int main() {
//freopen("inp7.in", "r", stdin);
//freopen("out7.out", "w", stdout);
int t;
cin >> t;
while(t--) {
int n;
cin >> n;
map<int, int> cnt;
int a[n], mx = 0;
for(int i = 0; i < n; i++) cin >> a[i], mx = max(mx, a[i]), cnt[a[i]]++;
int bad = 0, g2 = 0, largest = 0;
for(int i = 0; i < n; i++) {
if(cnt[a[i]] == 1) {
bad++;
if(mx == a[i]) largest = 1;
}
else if(cnt[a[i]] > 2) g2++;
}
if(bad == 1) {
if(g2 || !largest) cout << 1 << "\n";
else cout << 2 << "\n";
} else cout << (bad + 1)/2 << "\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>
ll INF = 1e18;
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, cnt = 0, gp3 = 0;
cin >> n;
map<int, int> mp;
vll v(n);
for (int i = 0; i < n; i++)
{
cin >> v[i];
mp[v[i]]++;
}
sort(all(v));
for (auto x : mp)
{
if (x.S == 1)
cnt++;
if (x.S > 2)
gp3++;
}
if (cnt ==1 && !gp3 && mp[v.back()]==1)
{
cout<<2<<'\n';
return ;
}
cout << (cnt + 1) / 2 << '\n';
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}
```