Program to print Sum of even and odd elements in an array

Prerequisite – Array Basics
Given an array, write a program to find the sum of values of even and odd index positions separately.

Examples: 

Input : arr[] = {1, 2, 3, 4, 5, 6}
Output :Even index positions sum 9
        Odd index positions sum 12
Explanation: Here, n = 6 so there will be 3 even 
index positions and 3 odd index positions in an array
Even = 1 + 3 + 5 = 9
Odd =  2 + 4 + 6 = 12

Input : arr[] = {10, 20, 30, 40, 50, 60, 70}
Output : Even index positions sum 160
        Odd index positions sum 120
Explanation: Here, n = 7 so there will be 3 odd
index positions and 4 even index positions in an array
Even = 10 + 30 + 50 + 70 = 160
Odd = 20 + 40 + 60 = 120 

Implementation:

C++

// CPP program to find out
// Sum of elements at even and
// odd index positions separately
#include <iostream>

using namespace std;

// Function to calculate sum
void EvenOddSum(int arr[], int n)
{
    int even = 0;
    int odd = 0;
    for (int i = 0; i < n; i++) {
        // Loop to find even, odd sum
        if (i % 2 == 0)
            even += arr[i];
        else
            odd += arr[i];
    }

    cout << "Even index positions sum " << even;
    cout << "\nOdd index positions sum " << odd;
}

// Driver function
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);

    EvenOddSum(arr, n);

    return 0;
}

Java

// Java program to find out
// Sum of elements at even and
// odd index positions separately
import java.io.*;

class EvenOddSum {
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 3, 4, 5, 6 };
        int even = 0, odd = 0;

        // Loop to find even, odd sum
        for (int i = 0; i < arr.length; i++) {
            if (i % 2 == 0)
                even += arr[i];
            else
                odd += arr[i];
        }

        System.out.println("Even index positions sum: " + even);
        System.out.println("Odd index positions sum: " + odd);
    }
}

Python3

# Python program to find out
# Sum of elements at even and
# odd index positions separately

# Function to calculate Sum
def EvenOddSum(a, n):
    even = 0
    odd = 0
    for i in range(n):

        # Loop to find even, odd Sum
        if i % 2 == 0:
            even += a[i]
        else:
            odd += a[i]

    print ("Even index positions sum ", even)
    print ("nOdd index positions sum ", odd)

# Driver Function

arr = [1, 2, 3, 4, 5, 6]
n = len(arr)

EvenOddSum(arr, n)

# This code is contributed by Sachin Bisht

C#

// C# program to find out
// Sum of elements at even and
// odd index positions separately
using System;

public class GFG {

    public static void Main()
    {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int even = 0, odd = 0;

        // Loop to find even, odd sum
        for (int i = 0; i < arr.Length; i++)
        {
            if (i % 2 == 0)
                even += arr[i];
            else
                odd += arr[i];
        }

        Console.WriteLine("Even index positions"
                             + " sum: " + even);

        Console.WriteLine("Odd index positions "
                               + "sum: " + odd);
    }
}

// This code is contributed by Sam007.

PHP

<?php
// PHP program to find out
// Sum of elements at even and
// odd index positions separately

// Function to calculate sum
function EvenOddSum($arr, $n)
{
    $even = 0;
    $odd = 0;
    for ($i = 0; $i < $n; $i++)
    {

        // Loop to find even, odd sum
        if ($i % 2 == 0)
            $even += $arr[$i];
        else
            $odd += $arr[$i];
    }

    echo("Even index positions sum " . $even);
    echo("\nOdd index positions sum " . $odd);
}

// Driver Code
$arr = array( 1, 2, 3, 4, 5, 6 );
$n = sizeof($arr);

EvenOddSum($arr, $n);

// This code is contributed by Ajit.
?>

Javascript

<script>

// Javascript program to find out
// Sum of elements at even and
// odd index positions separately

// Function to calculate sum
function EvenOddSum(arr, n)
{
    let even = 0;
    let odd = 0;
    for (let i = 0; i < n; i++)
    {

        // Loop to find even, odd sum
        if (i % 2 == 0)
            even += arr[i];
        else
            odd += arr[i];
    }

    document.write("Even index positions sum " + even);
    document.write("<br>" + "Odd index positions sum " + odd);
}

// Driver function
    let arr = [ 1, 2, 3, 4, 5, 6 ];
    let n = arr.length;

    EvenOddSum(arr, n);

// This code is contributed by Mayank Tyagi

</script>

Output

Even index positions sum 9
Odd index positions sum 12

Time Complexity: O(length(arr))
Auxiliary Space: 0(1)

Method 2: Using bitwise & operator

C++

// CPP program to find out
// Sum of elements at even and
// odd index positions separately  using bitwise & operator
#include <iostream>

using namespace std;

// Function to calculate sum
void EvenOddSum(int arr[], int n)
{
    int even = 0;
    int odd = 0;
    for (int i = 0; i < n; i++) {
        // Loop to find even, odd sum
        if (i & 1 != 0)
            odd += arr[i];
        else
            even += arr[i];
    }

    cout << "Even index positions sum " << even;
    cout << "\nOdd index positions sum " << odd;
}

// Driver function
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);

    EvenOddSum(arr, n);

    return 0;
}

//This code is contributed by Vinay Pinjala.

Java

// Java program to find out
// Sum of elements at even and
// odd index positions separately using bitwise & operator
public class Main {
  public static void main(String[] args)
  {
    int[] arr = { 1, 2, 3, 4, 5, 6 };
    int n = arr.length;
    int even = 0;
    int odd = 0;
    for (int i = 0; i < n; i++)
    {

      // Loop to find even, odd sum
      if ((i & 1) != 0) {
        odd += arr[i];
      }
      else {
        even += arr[i];
      }
    }
    System.out.println("Even index positions sum: "
                       + even);
    System.out.println("Odd index positions sum: "
                       + odd);
  }
}

// This code is contributed by divyansh2212

Python3

# Python program to find out
# Sum of elements at even and
# odd index positions separately using bitwise & operator

# Function to calculate Sum
def EvenOddSum(a, n):
    even = 0
    odd = 0
    for i in range(n):

        # Loop to find even, odd Sum
        if i &1:
            odd += a[i]
        else:
            even += a[i]

    print ("Even index positions sum ", even)
    print ("Odd index positions sum ", odd)

# Driver Function

arr = [1, 2, 3, 4, 5, 6]
n = len(arr)

EvenOddSum(arr, n)

# This code is contributed by tvsk

C#

using System;

namespace TestApplication
{
  class Program
  {
    static void Main(string[] args)
    {
      int[] arr = { 1, 2, 3, 4, 5, 6 };
      int n = arr.Length;
      int even = 0;
      int odd = 0;
      for (int i = 0; i < n; i++)
      {
        // Loop to find even, odd sum
        if ((i & 1) != 0)
        {
          odd += arr[i];
        }
        else
        {
          even += arr[i];
        }
      }
      Console.WriteLine("Even index positions sum: " + even);
      Console.WriteLine("Odd index positions sum: " + odd);
    }
  }
}

Javascript

// JS program to find out
// Sum of elements at even and
// odd index positions separately  using bitwise & operator

// Function to calculate sum
function EvenOddSum(arr, n)
{
    let even = 0;
    let odd = 0;
    for (let i = 0; i < n; i++)
    {

        // Loop to find even, odd sum
        if (i & 1 != 0)
            odd += arr[i];
        else
            even += arr[i];
    }

    console.log("Even index positions sum " + even);
    console.log("\nOdd index positions sum " + odd);
}

// Driver function
let arr = [ 1, 2, 3, 4, 5, 6 ];
let n = arr.length;

EvenOddSum(arr, n);

Output

Even index positions sum 9
Odd index positions sum 12

Time Complexity: O(length(arr))
Auxiliary Space: 0(1)

Method 3: Using slicing in python:

Calculate sum of all even indices using slicing and repeat the same with odd indices and print sum.

Below is the implementation:

C++

#include <iostream>
using namespace std;

// Function to calculate sum
int EvenOddSum(int arr[] , int n)
{
  int even = 0;
  int odd = 0;

  // Loop to find even, odd sum
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0)
      even += arr[i];
    else
      odd += arr[i];
  }
  cout<<"Even index positions sum "<<even<<"\n" ;
  cout<<"Odd index positions sum " <<odd;
}

int main()
{
  int arr[] = { 1, 2, 3, 4, 5, 6 };
  int n =sizeof(arr)/sizeof(arr[0]);

  EvenOddSum(arr, n);
}

// This code is contributed by ksrikanth0498.

Java

// Java program to find out sum of elements at even
// and odd index positions separately

import java.io.*;

class GFG {

    // Function to calculate sum
    static void EvenOddSum(int[] arr, int n)
    {
        int even = 0;
        int odd = 0;

        // Loop to find even, odd sum
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0)
                even += arr[i];
            else
                odd += arr[i];
        }
        System.out.println("Even index positions sum "
                           + even);
        System.out.print("Odd index positions sum " + odd);
    }

    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int n = arr.length;

        EvenOddSum(arr, n);
    }
}

// This code is contributed by lokeshmvs21.

Python3

# Python program to find out
# Sum of elements at even and
# odd index positions separately

# Function to calculate Sum


def EvenOddSum(a, n):
    even_sum = sum(a[::2])
    odd_sum = sum(a[1::2])
    print("Even index positions sum", even_sum)
    print("Odd index positions sum", odd_sum)


# Driver Function

arr = [1, 2, 3, 4, 5, 6]
n = len(arr)

EvenOddSum(arr, n)

# This code is contributed by vikkycirus

C#

// C# program to find out sum of elements at even
// and odd index positions separately
using System;

public class GFG {

    // Function to calculate sum
    static void EvenOddSum(int[] arr, int n)
    {
        int even = 0;
        int odd = 0;

        // Loop to find even, odd sum
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0)
                even += arr[i];
            else
                odd += arr[i];
        }
        Console.WriteLine("Even index positions sum "
                          + even);
        Console.WriteLine("Odd index positions sum " + odd);
    }

    static public void Main()
    {

        // Code
        int[] arr = { 1, 2, 3, 4, 5, 6 };
        int n = arr.Length;

        EvenOddSum(arr, n);
    }
}

// This code is contributed by lokeshmvs21.

Javascript

<script>

// Javascript program to find out
// Sum of elements at even and
// odd index positions separately

// Function to calculate sum
function EvenOddSum(arr, n)
{
    let even = 0;
    let odd = 0;
    for (let i = 0; i < n; i++)
    {

        // Loop to find even, odd sum
        if (i % 2 == 0)
            even += arr[i];
        else
            odd += arr[i];
    }

    document.write("Even index positions sum " + even);
    document.write("<br>" + "Odd index positions sum " + odd);
}

// Driver function
    let arr = [ 1, 2, 3, 4, 5, 6 ];
    let n = arr.length;

    EvenOddSum(arr, n);

    // This code is contributed by avijitmondal1998.
</script>

Output

Even index positions sum 9
Odd index positions sum 12

Time Complexity: O(n) where n is no of elements in given array
Auxiliary Space: 0(1)

Method 2: Using bitwise | operator

C++14

// C++ program to find out
// Sum of elements at even and
// odd index positions separately using bitwise | operator

#include<bits/stdc++.h>
using namespace std;
// Function to calculate Sum
void EvenOddSum(vector<int>a, int n)
{
    int even = 0;
    int odd = 0;
    for(int i=0; i<n; i++)
    {
        // Loop to find even, odd Sum
        if ((i|1)==i)
          odd+=a[i];
        else
          even+=a[i];
    }
    cout<< "Even index positions sum "<< even<<endl;
    cout<< "Odd index positions sum "<< odd<<endl;
}

// Driver Function
int main()
{
    vector<int>arr = {1, 2, 3, 4, 5, 6};
    int n = arr.size();

    EvenOddSum(arr, n);
}

Python3

# Python program to find out
# Sum of elements at even and
# odd index positions separately using bitwise | operator

# Function to calculate Sum
def EvenOddSum(a, n):
    even = 0
    odd = 0
    for i in range(n):

        # Loop to find even, odd Sum
        if i|1==i:
          odd+=a[i]
        else:
          even+=a[i]
    print ("Even index positions sum ", even)
    print ("Odd index positions sum ", odd)

# Driver Function

arr = [1, 2, 3, 4, 5, 6]
n = len(arr)

EvenOddSum(arr, n)

Javascript

// JavaScript program to find out
// Sum of elements at even and
// odd index positions separately using bitwise | operator

// Function to calculate Sum
function EvenOddSum(a, n) {
  let even = 0;
  let odd = 0;
  for(let i = 0; i < n; i++) {
    // Loop to find even, odd Sum
    if ((i|1) === i)
      odd += a[i];
    else
      even += a[i];
  }
  console.log("Even index positions sum ", even);
  console.log("Odd index positions sum ", odd);
}

// Driver Function

  const arr = [1, 2, 3, 4, 5, 6];
  const n = arr.length;

  EvenOddSum(arr, n);

Java

import java.util.*;

class Main {
    // Function to calculate Sum of elements at even and odd
    // index positions separately
    static void EvenOddSum(List<Integer> a, int n)
    {
        int even = 0;
        int odd = 0;
        for (int i = 0; i < n; i++) {
            // Loop to find even, odd Sum
            if ((i | 1) == i) {
                odd += a.get(i);
            }
            else {
                even += a.get(i);
            }
        }
        // Print the sum of elements at even and odd index
        // positions
        System.out.println("Even index positions sum "
                           + even);
        System.out.println("Odd index positions sum "
                           + odd);
    }

    // Driver Function
    public static void main(String[] args)
    {
        // Create a list of integers
        List<Integer> arr = new ArrayList<>(
            Arrays.asList(1, 2, 3, 4, 5, 6));
        // Get the size of the list
        int n = arr.size();
        // Call the EvenOddSum function
        EvenOddSum(arr, n);
    }
}

C#

// C# program to find out
// Sum of elements at even and
// odd index positions separately using bitwise | operator

using System;
using System.Collections.Generic;

class MainClass {
    // Function to calculate Sum
    static void EvenOddSum(List<int> a, int n)
    {
        int even = 0;
        int odd = 0;
        for (int i = 0; i < n; i++) {
            // Loop to find even, odd Sum
            if ((i | 1) == i)
                odd += a[i];
            else
                even += a[i];
        }
        Console.WriteLine("Even index positions sum "
                          + even);
        Console.WriteLine("Odd index positions sum " + odd);
    }

    // Driver Function
    public static void Main(string[] args)
    {
        List<int> arr = new List<int>{ 1, 2, 3, 4, 5, 6 };
        int n = arr.Count;

        EvenOddSum(arr, n);
    }
}

Output

Even index positions sum  9
Odd index positions sum  12

Time Complexity: O(length(arr))
Auxiliary Space: 0(1)

This article is contributed by Rishabh Jain. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.