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Java Program to Count Even and Odd Numbers in an Array
This article is created to cover a program in Java, to count and print the total number of even and odd numbers available in an array entered by user.
Count Even and Odd Numbers in an Array of 10 Numbers
The question is, write a Java program to count even and odd numbers in an array of 10 elements. The 10 elements for the array must be received by user at run-time. Here is its answer:
import java.util.Scanner; public class fresherearth { public static void main(String[] args) { int[] arr = new int[10]; Scanner s = new Scanner(System.in); System.out.print("Enter 10 Numbers for the Array: "); for(int i=0; i<10; i++) arr[i] = s.nextInt(); int countEven=0, countOdd=0; for(int i=0; i<10; i++) { if(arr[i]%2==0) countEven++; else countOdd++; } System.out.println("\nEven Number: " +countEven); System.out.println("Odd Number: " +countOdd); } }
Here is its sample run with user input 10, 13, 14, 17, 18, 20, 21, 22, 30, 40 as 10 numbers:
Count Even and Odd Numbers in an Array of n Numbers
Since the program given above has a limitation, of only operating with 10 elements. Therefore I've modified that program with the program given below. This program allows user to define the size of array too, along with its elements:
import java.util.Scanner; public class fresherearth { public static void main(String[] args) { Scanner s = new Scanner(System.in); System.out.print("Enter the Size of Array: "); int TOTAL = s.nextInt(); int[] arr = new int[TOTAL]; System.out.print("Enter " +TOTAL+" Numbers for the Array: "); for(int i=0; i<TOTAL; i++) arr[i] = s.nextInt(); int countEven=0, countOdd=0; for(int i=0; i<TOTAL; i++) { if(arr[i]%2==0) countEven++; else countOdd++; } System.out.println("\nEven Number: " +countEven); System.out.println("Odd Number: " +countOdd); } }
Here is its sample run with user input 5 as size of array, and 10, 12, 13, 14, 15 as its five elements:
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