Problem description

Elimination game is a popular game. The game is played on a chessboard with n rows and m rows. A colored chessboard is placed on the square of each row and column of the chessboard. When there are three or more chessboards of the same color in a row or column, these chessboards are eliminated. When there are many places that can be eliminated, the chess pieces in these places will be eliminated at the same time.
Now give you a chessboard with n rows and m rows. Each square in the chessboard has a chessboard. Please give a chessboard that has been eliminated once.
Note: A chess piece may be eliminated at the same time in a row and a column.

Input format

The first line of input contains two integers n, m, separated by spaces, representing the number of rows and columns on the chessboard, respectively.
Next, n rows, m integers per row, separated by spaces, represent the color of the pieces in each square. Colors are numbered 1 to 9.

Output format

Output n rows, m integers per row, adjacent integers are separated by a space, indicating the chessboard after one elimination. If the chess pieces in a square are eliminated, the corresponding square outputs 0, otherwise the color number of the chess pieces is output.

sample input

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

sample output

2 2 3 0 2
3 4 5 0 4
2 3 2 0 3
0 0 0 4 4

Sample explanation

The first and second rows of the fourth column and the fourth row in the chessboard can be eliminated, while the pieces in the other squares are retained.

sample input

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

sample output

2 2 3 0 2
3 0 0 0 0
2 3 2 0 3
2 2 0 0 0

Sample explanation

All 1 in the board and 3 in the last line can be eliminated at the same time, while the pieces in other squares are retained.

Assessment of use case size and conventions

All evaluation cases satisfy: 1 < n, m < 30.

import java.util.*;
public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		
		int[][] a = new int[n][m];
		int[][] b = new int[n][m];//b Special for operation
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				a[i][j] = sc.nextInt();
				b[i][j] = a[i][j];
			}
		}
		
//		First check whether there are more than three consecutive transverse
		for (int i = 0; i < n; i++) {
			for (int j = 1; j < m-1; j++) {			
				if(a[i][j-1]==a[i][j] && a[i][j]==a[i][j+1])//There are three consecutive identical
				{
					int point = j+1;		//Point to three identical third
					if(j+2==m)	//j+1 is the last
					{
						b[i][j-1] = 0;b[i][j] = 0;b[i][j+1] = 0;
						break;
					}	
					else
					{
					for (int j2 = j+2; j2 < m; j2++) {		//Are there three identical ones behind them?
						if(a[i][j2]==a[i][j+1] && j2==point+1)
							b[i][j2] = 0;
						else
							point = -30;
						point++;
						}
					b[i][j-1] = 0;b[i][j] = 0;b[i][j+1] = 0;
					}					
				}
			}
		}		
		
//		Re-test whether there are more than 3 consecutive vertical tests
		for (int i = 0; i < m; i++) {
			for (int j = 1; j < n-1; j++) {			
				if(a[j-1][i]==a[j][i] && a[j][i]==a[j+1][i])//There are three consecutive identical
				{
					int point = j+1;	//Point to three identical third
					if(j+2==n)	//j+1 is the last
					{
						b[j-1][i] = 0;b[j][i] = 0;b[j+1][i] = 0;
						break;
					}	
					else
					{
					for (int j2 = j+2; j2 < n; j2++) {		//Are there three identical ones behind them?
						if(a[j2][i]==a[j+1][i] && j2==point+1)
							b[j2][i] = 0;
						else
							point = -30;
						point++;
						}
					b[j-1][i] = 0;b[j][i] = 0;b[j+1][i] = 0;
					}					
				}
			}
		}
				
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				System.out.print(b[i][j]+" ");
			}
			System.out.println();
		}
	}

}