Problem:
Score : 700 points

Problem Statement

There are N cats.
We number them from 1 through N.

Each of the cats wears a hat.
Cat i says: "there are exactly a_i different colors among the N - 1 hats worn by the cats except me."

Determine whether there exists a sequence of colors of the hats that is consistent with the remarks of the cats.

Constraints

2 ≤ N ≤ 10^5

1 ≤ a_i ≤ N-1

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print Yes if there exists a sequence of colors of the hats that is consistent with the remarks of the cats; print No otherwise.

Sample Input 1

3
1 2 2

Sample Output 1

Yes

For example, if cat 1, 2 and 3 wears red, blue and blue hats, respectively, it is consistent with the remarks of the cats.

Sample Input 2

3
1 1 2

Sample Output 2

No

From the remark of cat 1, we can see that cat 2 and 3 wear hats of the same color.
Also, from the remark of cat 2, we can see that cat 1 and 3 wear hats of the same color.
Therefore, cat 1 and 2 wear hats of the same color, which contradicts the remark of cat 3.

Sample Input 3

5
4 3 4 3 4

Sample Output 3

No

Sample Input 4

3
2 2 2

Sample Output 4

Yes

Sample Input 5

4
2 2 2 2

Sample Output 5

Yes

Sample Input 6

5
3 3 3 3 3

Sample Output 6

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <deque>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
using namespace std;
 
typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, ll> Pl;
const double PI = 3.1415926535897932;   // acos(-1)
const double EPS = 1e-15;
const int mod = 1e+9 + 7;
const int INF = 1e+9;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    vector<int> a(n);
    for(int i = 0; i < n; ++i){
        cin >> a[i];
    }
    int max_e = *max_element(a.begin(), a.end());
    int min_e = *min_element(a.begin(), a.end());
    int dif = max_e - min_e;
    if(dif > 1){
        cout << "NO\n";
        return 0;
    }
    if(dif == 0){
        if(a[0] == n - 1 || 2 * a[0] <= n){
            cout << "YES\n";
        }
        else{
            cout << "NO\n";
        }
    }
    else{   // dif == 1
        int p = 0, q = 0;
        for(int i = 0; i < n; ++i){
            if(a[i] == min_e)   ++p;
            else                ++q;
        }
        if(p < max_e && 2 * (max_e - p) <= q){
            cout << "YES\n";
        }
        else{
            cout << "NO\n";
        }
    }
}