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 170 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
C/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); } } |
Python
# 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 evem, 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. ?> |
Output:
Even index positions sum 9 Odd index positions sum 12
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