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
#include <stdio.h>
void CountingEvenOdd( int arr[], int arr_size)
{
int even_count = 0;
int odd_count = 0;
for ( int i = 0; i < arr_size; i++) {
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);
}
int main()
{
int arr[] = { 2, 3, 4, 5, 6 };
int n = sizeof (arr) / sizeof (arr[0]);
CountingEvenOdd(arr, n);
}
|
C++
#include <iostream>
using namespace std;
void CountingEvenOdd( int arr[], int arr_size)
{
int even_count = 0;
int odd_count = 0;
for ( int i = 0; i < arr_size; i++) {
if (arr[i] & 1 == 1)
odd_count++;
else
even_count++;
}
cout << "Number of even elements = " << even_count
<< "\nNumber of odd elements = " << odd_count;
}
int main()
{
int arr[] = { 2, 3, 4, 5, 6 };
int n = sizeof (arr) / sizeof (arr[0]);
CountingEvenOdd(arr, n);
}
|
Java
import java.io.*;
class GFG {
static void CountingEvenOdd( int arr[], int arr_size)
{
int even_count = 0 ;
int odd_count = 0 ;
for ( int i = 0 ; i < arr_size; i++) {
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);
}
public static void main(String[] args)
{
int arr[] = { 2 , 3 , 4 , 5 , 6 };
int n = arr.length;
CountingEvenOdd(arr, n);
}
}
|
Python3
def CountingEvenOdd(arr, arr_size):
even_count = 0
odd_count = 0
for i in range (arr_size):
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)
arr = [ 2 , 3 , 4 , 5 , 6 ]
n = len (arr)
CountingEvenOdd(arr, n)
|
C#
using System;
class GFG {
static void CountingEvenOdd( int [] arr, int arr_size)
{
int even_count = 0;
int odd_count = 0;
for ( int i = 0; i < arr_size; i++) {
if ((arr[i] & 1) == 1)
odd_count++;
else
even_count++;
}
Console.WriteLine( "Number of even"
+ " elements = " + even_count
+ " Number of odd elements = "
+ odd_count);
}
public static void Main()
{
int [] arr = { 2, 3, 4, 5, 6 };
int n = arr.Length;
CountingEvenOdd(arr, n);
}
}
|
PHP
<?php
function CountingEvenOdd( $arr , $arr_size )
{
$even_count = 0;
$odd_count = 0;
for ( $i = 0 ; $i < $arr_size ; $i ++)
{
if ( $arr [ $i ] & 1 == 1)
$odd_count ++ ;
else
$even_count ++ ;
}
echo "Number of even elements = " ,
$even_count , " Number of odd " ,
"elements = " , $odd_count ;
}
$arr = array (2, 3, 4, 5, 6);
$n = count ( $arr );
CountingEvenOdd( $arr , $n );
?>
|
Javascript
<script>
function CountingEvenOdd(arr, arr_size)
{
let even_count = 0;
let odd_count = 0;
for (let i = 0; i < arr_size; i++) {
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);
}
let arr = [ 2, 3, 4, 5, 6 ];
let n = arr.length;
CountingEvenOdd(arr, n);
</script>
|
OutputNumber 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