C++ Program to Print Even Numbers in an Array

Given an array of numbers, the task is to print all the even elements of the array. Let us understand it with few examples mentioned below.

Example:

Input:  num1 = [2,7,6,5,4]
Output: 2, 6, 4

Input:  num2 = [1,4,7,8,3]
Output: 4, 8

1. Using for loop 

Approach: Iterate each element in the given using for loop and check if num % 2 == 0 that means if the remainder when divided by 2 is 0 it is even else it is odd. If the condition is satisfied, then only print the number.

Below is the implementation of the above approach:

C++

// C++ program to print even numbers in an array
// using for loop
#include <iostream>
using namespace std;

int main()
{
    // array of numbers
    int num1[] = { 2, 7, 6, 5, 4 };
    // size of an array
    int n = 5;
    for (int i = 0; i < n; i++) {
        // to check the number is even or not
        if (num1[i] % 2 == 0) {
            cout << num1[i] << " ";
        }
    }

    return 0;
}

Output

2 6 4 

Time Complexity: O(N)
Auxiliary Space: O(1), As constant extra space is used.

2. Using while loop

Approach: Iterate each element in the given using while loop and check if num % 2 == 0 that means if the remainder when divided by 2 is 0 it is even else it is odd. If the condition is satisfied, then only print the number.

Below is the implementation of the above approach:

C++

// C++ program to print even numbers in an array
// using while loop
#include <iostream>
using namespace std;

int main()
{
    // array of numbers
    int num1[] = { 7, 5, 9, 6, 2, 3, 4 };

    // size of an array
    int n = 7;
    int i = 0;
    while (i < n) {
        // to check the number even or not
        if (num1[i] % 2 == 0) {
            cout << num1[i] << " ";
        }
        i++;
    }

    return 0;
}

Output

6 2 4 

Time Complexity: O(N)

Auxiliary Space: O(1), As constant extra space is used

3.Using  Bitwise & operator

Approach: Iterate each element in the given array using for loop and check if num & 1 == 0 , as we know AND of 1 and num will give 0 , if num rightmost bit is not set and in even numbers rightmost bit isn’t set. If the condition is satisfied, then only print the number.

Below is the implementation of the above approach:

C++

// C++ program to print all even numbers in the array

#include <iostream>
using namespace std;

int main() {

    // array of numbers
    int num1[] = { 2, 7, 6, 5, 4 };
    // size of an array
    int n = 5;

    for (int i = 0; i < n; i++) {

        int x=num1[i]&1; // x will store Bitwise AND of number and 1

        // as we know , bitwise AND of even number and 1 is 0
        // check if it is 0 , then print number
        if (x==0) {
            cout << num1[i] << " ";
        }
    }
    return 0;
}

// This code is contributed by nikhilsainiofficial546

Output

2 6 4 

Time Complexity: O(N) , As we are iterating the whole array

Auxiliary Space: O(1), As constant extra space is used

4.Using  Bitwise | operator

Approach: Iterate each element in the given array using for loop and check if num | 1 ==num+1, if the condition satisfies then it is even number as we know OR of 1 and num will give num+1 in the case of even numbers, then only print the number.

Below is the implementation of the above approach:

C++

// C++ program to print all even numbers in the array

#include <iostream>
using namespace std;

int main() {

    // array of numbers
    int num1[] = { 2, 7, 6, 5, 4 };
    // size of an array
    int n = 5;

    for (int i = 0; i < n; i++) {


        // as we know , bitwise OR of even number and 1 is even number+1
        // check if it is even , then print number
        if ((num1[i] | 1)==num1[i]+1) {
            cout << num1[i] << " ";
        }
    }
    return 0;
}

// This code is contributed by tvsk

Output

2 6 4 

Time Complexity: O(N) , As we are iterating the whole array

Auxiliary Space: O(1), As constant extra space is used

5. Using recursion

Approach: Pass the array to the recursive function, which will traverse till the last element and print the element if it is even.

Below is the implementation of above approach:

C++

// C++ program to print all even numbers in the array
#include <iostream>
using namespace std;

// Recursive function 'printEven' which takes:
// integer array 'arr'
// array size 'n'
// current index
void printEven(int* arr, int n, int index) {
    // halting recursion when index exceeds size of array
    if(index >= n) return;

    // if current element & 1 results 0,
    // and !(0) is 1
    // thus, every even element will be printed
    if(!(arr[index] & 1)) cout << arr[index] << " ";

    // recursing to next index
    printEven(arr, n, ++index);
}

int main() {
    // array of numbers
    int arr[] = { 2, 7, 6, 5, 4 };
    // size of an array
    int n = 5;
    printEven(arr, n, 0);
    return 0;
}

// This code is contributed by Om Mishra (om_mishra)

Output

2 6 4 

Time Complexity: O(N), as we are traversing entire array

Auxiliary Space: O(1), as constant extra space is used