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Count number of even and odd elements in an array

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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




// C program to count number of even
// and odd elements in an array
#include <stdio.h>
  
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++;
    }
  
    printf("Number of even elements = %d \nNumber of odd "
           "elements = %d",
           even_count, 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);
}

C++




// C++ 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)

Auxiliary Space: O(1) because it is using constant space for variables


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Last Updated : 13 Mar, 2023
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