# C++ program to count numbers of even and odd elements in an array

## Count Even and Odd Elements in C++

Here, in this page we will discuss the count even and odd elements in C++ programming language. We are given with an array and need to print the count of the even elements and odd elements in the given input array. ## Method discussed :

• Method 1 : Using Modulo Operator
• Method 2 : Using Bit-wise AND operator.

Let’s go through with above two methods one by one,

## Method 1 :

• Declare two variables say count_even=0, count_odd=0.
• Start iterating over the array,
• Check if(arr[i]%2==0) set even_count++.
• Else set odd_count++
• Print the value of even_count and odd_count.

### Time and Space Complexity :

• Time Complexity : O(n)
• Space Complexity : O(1)

### Method 1  : Code in C++

Run

```#include<bits/stdc++.h>
using namespace std;

int main(){

int arr[] = {1, 7, 8, 4, 5, 16, 8};
int n = sizeof(arr)/sizeof(arr);

int even_count=0, odd_count=0;

for(int i=0; i<n; i++){
if(arr[i]%2==0)
even_count++;

else
odd_count++;
}
cout<<"Even Elements count : " <<even_count<< "\nOdd Elements count : " << odd_count;
}```

### Output :

```Even Elements count : 4
Odd Elements count : 3```

## Method 2 :

In this method we will use bit-wise AND operator. By doing AND of 1 with array element, if the result comes out to be 1 then the number is odd otherwise even.

### Method 2  : Code in C++

Run

```#include<bits/stdc++.h>
using namespace std;

int main(){

int arr[] = {1, 7, 8, 4, 5, 16, 8};
int n = sizeof(arr)/sizeof(arr);

int even_count=0, odd_count=0;

for(int i=0; i<n; i++){
if(arr[i]&1==0)
even_count++;

else
odd_count++;
}
cout<<"Even Elements count :" <<even_count<< "\nOdd Elements count : " << odd_count;
}```

### Output :

```Even Elements count : 4
Odd Elements count : 3```