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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.IOException;
import java.io.InputStream;
import java.util.Arrays;
import java.util.NoSuchElementException;

public class Main {

	public static void main(String[] args) {

		FastScanner sc = new FastScanner();
		int N = sc.nextInt();
		int K = sc.nextInt();
		int[] A = new int[N];
		for (int i = 0; i < N; i++)
			A[i] = sc.nextInt();

		int[] count = new int[200000];
		for (int i = 0; i < N; i++)
			count[A[i] - 1]++;
		int ans = 0;
		Arrays.sort(count);
		for (int i = 0; i < 200000 - K; i++)
			ans += count[i];

		System.out.println(ans);

	}
	//以下、自作ライブラリ

	//最大公約数
	public static long gcd(long a, long b) {
		if (a < b)
			return gcd(b, a);
		long c = 0;
		c = a % b;
		if (c == 0)
			return b;
		return gcd(b, c);
	}

	//最小公倍数
	public static long lcm(long m, long n) {
		return m * n / gcd(m, n);
	}

	//素数判定
	public static boolean isPrime(int n) {
		if (n < 2)
			return false;
		if (n == 2)
			return true;
		if (n % 2 == 0)
			return false;

		for (int i = 3; i * i <= n; i += 2) {
			if (n % i == 0)
				return false;
		}
		return true;
	}
}

//以下、高速スキャナークラス
class FastScanner {
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}

	private int readByte() {
		if (hasNextByte())
			return buffer[ptr++];
		else
			return -1;
	}

	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	public boolean hasNext() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr]))
			ptr++;
		return hasNextByte();
	}

	public String next() {
		if (!hasNext())
			throw new NoSuchElementException();
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}

	public int nextInt() {
		long nl = nextLong();
		if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
			throw new NumberFormatException();
		return (int) nl;
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}
}