Problem:
Score : 700 points

Problem Statement

Takahashi is hosting an sports meet.
There are N people who will participate. These people are conveniently numbered 1 through N.
Also, there are M options of sports for this event. These sports are numbered 1 through M.
Among these options, Takahashi will select one or more sports (possibly all) to be played in the event.

Takahashi knows that Person i's j-th favorite sport is Sport A_{ij}.
Each person will only participate in his/her most favorite sport among the ones that are actually played in the event, and will not participate in the other sports.

Takahashi is worried that one of the sports will attract too many people.
Therefore, he would like to carefully select sports to be played so that the number of the participants in the sport with the largest number of participants is minimized.
Find the minimum possible number of the participants in the sport with the largest number of participants.

Constraints

1 \leq N \leq 300

1 \leq M \leq 300

A_{i1} , A_{i2} , ... , A_{iM} is a permutation of the integers from 1 to M.

Input

Input is given from Standard Input in the following format:

N M
A_{11} A_{12} ... A_{1M}
A_{21} A_{22} ... A_{2M}
:
A_{N1} A_{N2} ... A_{NM}

Output

Print the minimum possible number of the participants in the sport with the largest number of participants.

Sample Input 1

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

Sample Output 1

2

Assume that Sports 1, 3 and 4 are selected to be played. In this case, Person 1 will participate in Sport 1, Person 2 in Sport 3, Person 3 in Sport 3 and Person 4 in Sport 4.
Here, the sport with the largest number of participants is Sport 3, with two participants.
There is no way to reduce the number of participants in the sport with the largest number of participants to 1. Therefore, the answer is 2.

Sample Input 2

3 3
2 1 3
2 1 3
2 1 3

Sample Output 2

3

Since all the people have the same taste in sports, there will be a sport with three participants, no matter what sports are selected.
Therefore, the answer is 3.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.*;

public class Main {

	public static void main(String[] args) {
		Input in = new StdIn();
		int n=in.nextInt(), m=in.nextInt();
		int[][] mat = new int[n][m];
		for(int i=0; i<n; ++i)
			for(int j=0; j<m; ++j)
				mat[i][in.nextInt()-1]=j;
		//System.out.println(Arrays.deepToString(mat));
		int minppl=n;
		boolean[] excludeColumn = new boolean[m];
		for(int i=0; i<m-1; ++i) {
			//Select column / sport to remove
			int[] colFreqs = new int[m];
			for(int j=0; j<n; ++j) {
				int minCol=-1;
				for(int k=0; k<m; ++k)
					if(!excludeColumn[k]&&(minCol==-1||mat[j][minCol]>mat[j][k]))
						minCol=k;
				++colFreqs[minCol];
			}
			int maxCol=-1;
			for(int j=0; j<m; ++j)
				if(maxCol==-1||colFreqs[j]>colFreqs[maxCol])
					maxCol=j;
			minppl=Math.min(colFreqs[maxCol], minppl);
			excludeColumn[maxCol]=true;
		}
		System.out.println(minppl);
	}
	
	interface Input {
		public String next();
		public String nextLine();
		public int nextInt();
		public long nextLong();
		public double nextDouble();
	}
	static class StdIn implements Input {
		final private int BUFFER_SIZE = 1 << 16;
		final private int STRING_SIZE = 1 << 6;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;

		public StdIn() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}
		
		public StdIn(String filename) {
	        try{
	            din = new DataInputStream(new FileInputStream(filename));
	        } catch(Exception e) {
	            throw new RuntimeException();
	        }
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}
		
		public String next() {
			byte[] buf = new byte[STRING_SIZE]; // string length
			int cnt = 0, c;
			while((c=read())!=-1&&(c==' '||c=='\n'||c=='\r'));
			while (c != -1)
			{
				if (c == ' ' || c == '\n'||c=='\r')
					break;
				buf[cnt++] = (byte) c;
				c=read();
			}
			return new String(buf, 0, cnt);
		}

		public String nextLine() {
			byte[] buf = new byte[STRING_SIZE]; // line length
			int cnt = 0, c;
			while((c=read())!=-1&&(c==' '||c=='\n'||c=='\r'));
			while (c != -1)
			{
				if (c == '\n'||c=='\r')
					break;
				buf[cnt++] = (byte) c;
				c = read();
			}
			return new String(buf, 0, cnt);
		}

		public int nextInt() {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do
			{
				ret = ret * 10 + c - '0';
			}  while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}
		
		public int[] readIntArray(int n) {
			int[] ar = new int[n];
			for(int i=0; i<n; ++i)
				ar[i]=nextInt();
			return ar;
		}

		public long nextLong() {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			}
			while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nextDouble() {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			}
			while ((c = read()) >= '0' && c <= '9');

			if (c == '.')
			{
				while ((c = read()) >= '0' && c <= '9')
				{
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() {
			try{
				if (bufferPointer == bytesRead)
					fillBuffer();
				return buffer[bufferPointer++];
			} catch(IOException e) {
				throw new RuntimeException();
			}
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}
	}
}