# Java Program to Find the Row having Maximum 0’s in a Binary Matrix

In the previous article, we have seen Java Program to Find the Row having Maximum 1’s in a Binary Matrix

In this article we are going to see how we can write a program to count maximum number of zero’s in a binary matrix in JAVA language.

## Java Program to Find the Row having Maximum Number of 0 in a Binary Matrix

A 3*3 Matrix is having 3 rows and 3 columns where this 3*3 represents the dimension of the matrix. Means there are 3*3 i.e. total 9 elements in a 3*3 Matrix.

Let’s understand it in more simpler way.

                   | A00   A01   A02 |
Matrix A =  | A10   A11   A12 |
| A20   A21   A22 | 3*3
• Matrix A represents a 3*3 matrix.
• A‘ represents the matrix element
• Aij‘ represents the matrix element at it’s matrix position/index.
• i‘ represents the row index
• j‘ represents the column index
• Means A00=Aij  where i=0 and j=0,  A01=aij where i=0 and j=1 and like this.
• Here we have started row value from 0 and column value from 0.

A binary matrix is a matrix that has only 0 or 1 as its elements.

For example:

1 0 0
1 1 1
0 1 0


Let’s see different ways to find the Row having Maximum Number of 0 in a Binary Matrix

### Method-1: Java Program to Find the Row having Maximum Number of 0 in a Binary Matrix By Static Initialization of Array Elements

Approach:

• Initialize an binary array of size 3×3, with elements.
• Use two for loops to iterate the rows and columns .
• Inside the for loops count all zeros using a counter.
• Store the count and index if the count is greater than the previous row count.
• Print the result.

Program:

import java.util.Scanner;
public class Matrix
{
public static void main(String args[])
{
//Scanner class to take input
Scanner scan = new Scanner(System.in);

// Initializing the 3X3 matrix i.e. 2D array
int arr[][] = {{0,0,0},{0,1,0},{1,1,0}};
int row, col ;

System.out.print("The matrix elements are:");
printMatrix(arr);
maxOne(arr);
}

// Method to print the matrix
static void printMatrix(int arr[][])
{
int row, col;
// Loop to print the elements
for(row=0;row<3;row++)
{
// Used for formatting
System.out.print("\n");
for(col=0;col<3;col++)
{
System.out.print(arr[row][col]+" ");
}
}
}

// Looks for the highest occurence of 0
static void maxOne(int arr[][])
{
int counter=0,prev[]={0,-1}, row,col;
// Counts and stores the occurence of 0 and the row index
for(row=0;row<3;row++)
{
counter =0;
for(col=0;col<3;col++)
{
if(arr[row][col]==0)
counter++;

}
if(prev[0]<counter)
{
prev[0]=counter;
prev[1]=row;
}
}
if(prev[0]==0)
System.out.println("\nNo rows have 0's");
else
System.out.println("\nThe "+(prev[1]+1)+" row has the highest occurrence of 0: "+prev[0]);
}

}
Output:

The matrix elements are:
0 0 0
0 1 0
1 1 0
The 1 row has the highest occurrence of 0 : 3

### Method-2: Java Program to Find the Row having Maximum Number of 0 in a Binary Matrix By Dynamic Initialization of Array Elements

Approach:

• Declare one array of size 3×3.
• Ask the user for input of binary array elements and store them in the array using two for loops.
• Use two for loops to iterate the rows and columns .
• Inside the for loops count all zeros in a row using a counter.
• Store the count and index if the count is greater than the previous row count.
• Print the result.

Program:

import java.util.Scanner;
public class matrix
{
public static void main(String args[])
{
//Scanner class to take input
Scanner scan = new Scanner(System.in);

// Declare the 3X3 matrix i.e. 2D array
int arr[][] = new int[3][3];
int row, col ;

// Taking matrix input
System.out.println("Enter matrix elements : ");
for(row=0;row<3;row++)
for(col=0;col<3;col++)
arr[row][col] = scan.nextInt();

System.out.print("The matrix elements are:");
printMatrix(arr);
maxOne(arr);
}

// Method to print the matrix
static void printMatrix(int arr[][])
{
int row, col;
// Loop to print the elements
for(row=0;row<3;row++)
{
// Used for formatting
System.out.print("\n");
for(col=0;col<3;col++)
{
System.out.print(arr[row][col]+" ");
}
}
}

// Looks for the highet occurence of 0
static void maxOne(int arr[][])
{
int counter=0,prev[]={0,-1}, row,col;
// Counts and stores the occurence of 0 and the row index
for(row=0;row<3;row++)
{
counter =0;
for(col=0;col<3;col++)
{
if(arr[row][col]==0)
counter++;

}
if(prev[0]<counter)
{
prev[0]=counter;
prev[1]=row;
}
}
if(prev[0]==0)
System.out.println("\nNo rows have 1's");
else
System.out.println("\nThe "+(prev[1]+1)+" row has the highest occurrence of 0: "+prev[0]);
}

}
Output:

Enter matrix elements : 1 1 1 0 0 1 1 1 0
The matrix elements are:
1 1 1
0 0 1
1 1 0
The 2 row has the highest occurrence of 0: 2

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