Java Program to Find the Row having Maximum Number of 1 in a Binary Matrix

In the previous article, we have seen Java Program to Find Average of Matrix Elements

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

Java Program to Find the Row having Maximum Number of 1 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 1 in a Binary Matrix

By Static Initialization of Array Elements
By Dynamic Initialization of Array Elements

Method-1: Java Program to Find the Row having Maximum Number of 1 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 ones’ 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[][] = {{1,0,0},{0,0,0},{0,0,0}};
        int row, col ;

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

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

}

Output:

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

Method-2: Java Program to Find the Row having Maximum Number of 1 in a Binary Matrix By Dynamic Initialization of Array Elements

Approach:

Declare one array of size 3×3.
Ask the user for input of array binary 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 ones’ 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 1
    static void maxOne(int arr[][])
    {
        int counter=0,prev[]={0,-1}, row,col;
        // Counts and stores the occurence of 1 and the row index
        for(row=0;row<3;row++)
        {
            counter =0;
            for(col=0;col<3;col++)
            {
                if(arr[row][col]==1)
                    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 occurence of 1: "+prev[0]);
    }

}

Output:

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

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