Java Program to Check horizontal and vertical symmetry in binary matrix
Last Updated :
18 Feb, 2023
Given a 2D binary matrix of N rows and M columns. The task is to check whether the matrix is horizontally symmetric, vertically symmetric, or both. The matrix is said to be horizontally symmetric if the first row is the same as the last row, the second row is the same as the second last row, and so on. And the matrix is said to be vertically symmetric if the first column is the same as the last column, the second column is the same as the second last column, and so on. Print “VERTICAL” if the matrix is vertically symmetric, “HORIZONTAL” if the matrix is vertically symmetric, “BOTH” if the matrix is vertically and horizontally symmetric, and “NO” if not symmetric.
Examples:
Input: N = 3 M = 3
0 1 0
0 0 0
0 1 0
Output: Both
First and third row are same and also second row
is in middle. So Horizontal Symmetric.
Similarly, First and third column are same and
also second column is in middle, so Vertical
Symmetric.
Input:
0 0 1
1 1 0
0 0 1.
Output: Both
The idea is to use pointers indicating two rows (or columns) and compare each cell of both the pointed rows (or columns).
For Horizontal Symmetry, initialize one pointer i = 0 and another pointer j = N – 1.
Now, compare each element of i-th row and j-th row. Increase i by 1 and decrease j by 1 in each loop cycle. If at least one, not an identical element, is found, mark the matrix as not horizontal symmetric.
Similarly, for Vertical Symmetry, initialize one pointer i = 0 and another pointer j = M – 1.
Now, compare each element of i-th column and j-th column. Increase i by 1 and decrease j by 1 in each loop cycle. If at least one, not an identical element, is found, mark the matrix as not vertical symmetric.
Below is the implementation of the above idea:
Java
import java.io.*;
public class GFG {
static void checkHV( int [][] arr, int N,
int M)
{
boolean horizontal = true ;
boolean vertical = true ;
for ( int i = 0 , k = N - 1 ;
i < N / 2 ; i++, k--) {
for ( int j = 0 ; j < M; j++) {
if (arr[i][j] != arr[k][j]) {
horizontal = false ;
break ;
}
}
}
for ( int i = 0 , k = M - 1 ;
i < M / 2 ; i++, k--) {
for ( int j = 0 ; j < N; j++) {
if (arr[i][j] != arr[k][j]) {
horizontal = false ;
break ;
}
}
}
if (!horizontal && !vertical)
System.out.println( "NO" );
else if (horizontal && !vertical)
System.out.println( "HORIZONTAL" );
else if (vertical && !horizontal)
System.out.println( "VERTICAL" );
else
System.out.println( "BOTH" );
}
static public void main(String[] args)
{
int [][] mat = { { 1 , 0 , 1 },
{ 0 , 0 , 0 },
{ 1 , 0 , 1 } };
checkHV(mat, 3 , 3 );
}
}
|
Output:
BOTH
Time Complexity: O(N*M).
Auxiliary Space: O(1) as it is using constant space for variables
Please refer complete article on Check horizontal and vertical symmetry in binary matrix for more details!
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