Problem:
Score : 300 points

Problem Statement

Takahashi has N balls. Initially, an integer A_i is written on the i-th ball.

He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls.

Find the minimum number of balls that Takahashi needs to rewrite the integers on them.

Constraints

1 \leq K \leq N \leq 200000

1 \leq A_i \leq N

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 ... A_N

Output

Print the minimum number of balls that Takahashi needs to rewrite the integers on them.

Sample Input 1

5 2
1 1 2 2 5

Sample Output 1

1

For example, if we rewrite the integer on the fifth ball to 2, there are two different integers written on the balls: 1 and 2.
On the other hand, it is not possible to rewrite the integers on zero balls so that there are at most two different integers written on the balls, so we should print 1.

Sample Input 2

4 4
1 1 2 2

Sample Output 2

0

Already in the beginning, there are two different integers written on the balls, so we do not need to rewrite anything.

Sample Input 3

10 3
5 1 3 2 4 1 1 2 3 4

Sample Output 3

3

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
 
//long long
using ll = long long;
 
// pair<int, int>
using PII = pair<int, int>;
 
 
//最大値、mod
const int MOD = 1000000007;
const int mod = 1000000007;
const int INF = 1000000000;
const long long LINF = 1e18;
const int MAX = 510000;
 
//出力系
#define print(x) cout << x << endl
#define prints(x) cout << fixed << setprecision(20) << x << endl
#define printc(x) cout << setw(2) << setfill('0') << x << endl;
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
 
//配列入力
vector<long long>vecin(ll n){
    vector<long long>res(n);
    for(int i = 0; i < n; i++) cin >> res[i];
    return res;
}

// begin() end()
#define all(x) (x).begin(),(x).end()
 
//for
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define rrep(i,a,b) for(int i=(a);i>(b);i--)
#define rep(i,a,b) for(int i=(a);i<(b);i++)
 
//最大公約数 
 ll gcd(ll x, ll y) { return y ? gcd(y,x%y) : x;}

// 最小公倍数
unsigned lcm(unsigned a, unsigned b){
    return a / gcd(a, b) * b; 
}
 
// a = max(a, b), a = min(a, b)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
 
// num ^ pow(mod取る)
ll pow_mod(ll num, ll pow, ll mod) {
    ll prod = 1;
    num %= mod;
    while (pow > 0) {
        if (pow & 1) prod = prod * num % mod;
        num = num * num % mod;
        pow >>= 1;
    }
    return prod;
}
 
 
 
// 二項係数(MODとる、1 ≦ k ≦ n ≦ 10^7 程度)
// COMinit()
// COM(x, y)
// とコンビで使う
 
// テーブルを作る前処理
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 

 
 
// UnionFind
struct UnionFind {
    vector<int> par;
    vector<int> rank;
    vector<ll> Size;
    UnionFind(int n = 1) {
        init(n);
    }
 
    void init(int n = 1) {
        par.resize(n + 1); rank.resize(n + 1); Size.resize(n + 1);
        for (int i = 0; i <= n; ++i) par[i] = i, rank[i] = 0, Size[i] = 1;
    }
 
    int root(int x) {
        if (par[x] == x) {
            return x;
        }
        else {
            int r = root(par[x]);
            return par[x] = r;
        }
    }
 
    bool issame(int x, int y) {
        return root(x) == root(y);
    }
 
    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (rank[x] < rank[y]) swap(x, y);
        if (rank[x] == rank[y]) ++rank[x];
        par[y] = x;
        Size[x] += Size[y];
        return true;
    }
 
    ll size(int x){
        return Size[root(x)];
    }
};

 
//modint構造体 
struct Mint {
    int val;
    Mint inv() const{
        int tmp,a=val,b=mod,x=1,y=0;
        while(b)tmp=a/b,a-=tmp*b,swap(a,b),x-=tmp*y,swap(x,y);
        return Mint(x);
    }
public:
    Mint():val(0){}
    Mint(ll x){if((val=x%mod)<0)val+=mod;}
    Mint pow(ll t){Mint res=1,b=*this; while(t){if(t&1)res*=b;b*=b;t>>=1;}return res;}
    Mint& operator+=(const Mint& x){if((val+=x.val)>=mod)val-=mod;return *this;}
    Mint& operator-=(const Mint& x){if((val+=mod-x.val)>=mod)val-=mod; return *this;}
    Mint& operator*=(const Mint& x){val=(ll)val*x.val%mod; return *this;}
    Mint& operator/=(const Mint& x){return *this*=x.inv();}
    bool operator==(const Mint& x) const{return val==x.val;}
    bool operator!=(const Mint& x) const{return val!=x.val;}
    bool operator<(const Mint& x) const{return val<x.val;}
    bool operator<=(const Mint& x) const{return val<=x.val;}
    bool operator>(const Mint& x) const{return val>x.val;}
    bool operator>=(const Mint& x) const{return val>=x.val;}
    Mint operator+(const Mint& x) const{return Mint(*this)+=x;}
    Mint operator-(const Mint& x) const{return Mint(*this)-=x;}
    Mint operator*(const Mint& x) const{return Mint(*this)*=x;}
    Mint operator/(const Mint& x) const{return Mint(*this)/=x;}
};
struct factorial {
    vector<Mint> Fact, Finv;
public:
   //factorial fact(10000010);
   //fact.nCr(a, b)
  //「fact」の部分は自由に名前変更可能
    factorial(int maxx){
        Fact.resize(maxx+1),Finv.resize(maxx+1); Fact[0]=Mint(1); rep(i,0,maxx)Fact[i+1]=Fact[i]*(i+1);
        Finv[maxx]=Mint(1)/Fact[maxx]; rrep(i,maxx,0)Finv[i-1]=Finv[i]*i;
    }
    Mint fact(int n,bool inv=0){if(inv)return Finv[n];else return Fact[n];}
    Mint nPr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[n-r];}
    Mint nCr(int n,int r){if(n<0||n<r||r<0)return Mint(0);else return Fact[n]*Finv[r]*Finv[n-r];}
};


// 1 * 2 * 3 .... * n (mod)
ll modfact(ll n) { 
    if (n <= 1) return 1;
    return (n * modfact(n - 1)) % MOD;
}


// kが角度だった場合:cos(k * (PI / 180));
//const double PI = acos(-1);のまま使うと円周率(M_PIもあるよ)
const double PI = acos(-1);


// 多次元 vector 生成   例: auto dp = make_vec<long long>(N+1, 5, 5, 5); 
template<class T>
vector<T> make_vec(size_t a){
    return vector<T>(a);
}
template<class T, class... Ts>
auto make_vec(size_t a, Ts... ts){
  return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}

//素因数分解
vector<pair<long long, int>>factorize(long long n){
  vector<pair<long long, int>> res;
  for(long long i = 2; i * i <= n; ++i){
    if(n % i) continue;
    res.emplace_back(i, 0);
    while(n % i == 0){
      n /= i;
      res.back().second++;
    }
  }
if(n != 1) res.emplace_back(n, 1);
return res;
}

// 素数判定
bool primejudge(long long a){
    if(a <= 1) return false;
    for(long long i = 2; i * i <= a; i++){
        if(a % i == 0) return false;
    }
    return true;
}

//　累積和
// vector<long long>sums(vector<int>n){
//     vector<long long>res(n.size() + 1, 0);
//     for(int i = 0; i < n.size(); i++) res[i + 1] = n[i] + res[i];
//     return res;
// }

int dy[4] = {0, 1, 0, -1}, dx[4] = {1, 0, -1, 0};





int main() {
    int n, k;
    cin >> n >> k;
    map<int, int>mp;
    REP(i, n){
        int x;
        cin >> x;
        mp[x]++;
    }
    int tmp = mp.size() - k;
    int ans = 0;
    if(tmp > 0){
        priority_queue<int, vector<int>, greater<int>>que;
        for(auto x : mp) que.push(x.second);
        REP(i, tmp) ans += que.top(), que.pop();
    }
    print(ans);
    return 0;
}