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Count number of even and odd elements in an array
  • Difficulty Level : Easy
  • Last Updated : 05 Mar, 2021

For the given array of integers, count even and odd elements.

Examples: 

Input: 
int arr[5] = {2, 3, 4, 5, 6}
Output: 
Number of even elements = 3    
Number of odd elements = 2  

Input:
int arr[5] = {22, 32, 42, 52, 62}
Output: 
Number of even elements = 5  
Number of odd elements = 0

Solution: 
We can also check if a number is odd or even

  • By doing AND of 1 and that digit, if the result comes out to be 1 then the number is odd otherwise even.
  • By its divisibility by 2. A number is said to be odd if it is not divisible by 2, otherwise its even.
     

Here, we will check if a number is odd, then we will increment the odd counter otherwise we will increment the even counter. 

Below is the implementation of the above approach:



C++




// CPP program to count number of even
// and odd elements in an array
#include <iostream>
using namespace std;
 
void CountingEvenOdd(int arr[], int arr_size)
{
    int even_count = 0;
    int odd_count = 0;
 
    // loop to read all the values in the array
    for (int i = 0; i < arr_size; i++) {
         
          // checking if a number is completely
        // divisible by 2
        if (arr[i] & 1 == 1)
            odd_count++;
        else
            even_count++;
    }
 
    cout << "Number of even elements = " << even_count
         << "\nNumber of odd elements = " << odd_count;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
       
      // Function Call
    CountingEvenOdd(arr, n);
}

Java




// JAVA program to count number of even
// and odd elements in an array
import java.io.*;
 
class GFG {
 
    static void CountingEvenOdd(int arr[], int arr_size)
    {
        int even_count = 0;
        int odd_count = 0;
 
        // loop to read all the values in
        // the array
        for (int i = 0; i < arr_size; i++) {
             
              // checking if a number is
            // completely divisible by 2
            if ((arr[i] & 1) == 1)
                odd_count++;
            else
                even_count++;
        }
 
        System.out.println("Number of even"
                           + " elements = " + even_count
                           + " Number of odd elements = "
                           + odd_count);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 2, 3, 4, 5, 6 };
        int n = arr.length;
           
          // Function Call
        CountingEvenOdd(arr, n);
    }
}
 
// This code is Contributed by anuj_67.

Python3




# Python3 program to count number of
# even and odd elements in an array
 
 
def CountingEvenOdd(arr, arr_size):
    even_count = 0
    odd_count = 0
 
    # loop to read all the values
    # in the array
    for i in range(arr_size):
 
        # checking if a number is
        # completely divisible by 2
        if (arr[i] & 1 == 1):
            odd_count += 1
        else:
            even_count += 1
 
    print("Number of even elements = ",
          even_count)
    print("Number of odd elements = ",
          odd_count)
 
 
# Driver Code
arr = [2, 3, 4, 5, 6]
n = len(arr)
 
# Function Call
CountingEvenOdd(arr, n)
 
# This code is contributed by sahishelangia

C#




// C# program to count number of even
// and odd elements in an array
using System;
 
class GFG {
 
    static void CountingEvenOdd(int[] arr, int arr_size)
    {
        int even_count = 0;
        int odd_count = 0;
 
        // loop to read all the values in
        // the array
        for (int i = 0; i < arr_size; i++) {
             
              // checking if a number is
            // completely divisible by 2
            if ((arr[i] & 1) == 1)
                odd_count++;
            else
                even_count++;
        }
 
        Console.WriteLine("Number of even"
                          + " elements = " + even_count
                          + " Number of odd elements = "
                          + odd_count);
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 2, 3, 4, 5, 6 };
        int n = arr.Length;
       
          // Function Call
        CountingEvenOdd(arr, n);
    }
}
 
// This code is Contributed by anuj_67.

PHP




<?php
// PHP program to count number of even
// and odd elements in an array
 
function CountingEvenOdd( $arr, $arr_size)
{
    $even_count = 0;        
    $odd_count = 0;            
         
    // loop to read all the values in
    // the array
    for( $i = 0 ; $i < $arr_size ; $i++)
    {
        // checking if a number is
        // completely divisible by 2
        if ($arr[$i] & 1 == 1)
            $odd_count ++ ;    
        else               
            $even_count ++ ;        
    }
 
    echo "Number of even elements = " ,
        $even_count," Number of odd " ,
            "elements = " ,$odd_count ;    
}
 
// Driver Code
    $arr = array(2, 3, 4, 5, 6);
    $n = count($arr);
 
    // Function Call
    CountingEvenOdd($arr, $n);
 
// This code is Contributed by anuj_67.
?>

Javascript




<script>
 
// Javascript program to count number of even
// and odd elements in an array
 
function CountingEvenOdd(arr, arr_size)
{
    let even_count = 0;
    let odd_count = 0;
 
    // loop to read all the values in the array
    for (let i = 0; i < arr_size; i++) {
         
        // checking if a number is completely
        // divisible by 2
        if (arr[i] & 1 == 1)
            odd_count++;
        else
            even_count++;
    }
 
    document.write("Number of even elements = " + even_count
        + "<br>" + "Number of odd elements = " + odd_count);
}
 
// Driver Code
 
    let arr = [ 2, 3, 4, 5, 6 ];
    let n = arr.length;
     
    // Function Call
    CountingEvenOdd(arr, n);
 
// This code is contributed by Mayank Tyagi
 
</script>
Output
Number of even elements = 3
Number of odd elements = 2

Time Complexity: O(n)

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