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)