Find the sum of the first half and second half elements of an array
Last Updated :
27 Mar, 2023
Given an array arr of size N. The task is to find the sum of the first half (N/2) elements and the second half elements (N - N/2) of an array.
Examples:
Input : arr[] = {20, 30, 60, 10, 25, 15, 40}
Output : 110, 90
Sum of first N/2 elements 20 + 30 + 60 is 110Input : arr[] = {50, 35, 20, 15}
Output : 85, 35
Approach:
- Initialize SumFirst and SumSecond as 0.
- Traverse the given array.
- Now add elements in SumFirst if the current index is less than N/2 otherwise add in SumSecond.
Below is the implementation of the above approach:
C
// C program to find the sum of the first half elements and
// second half elements of the array. Function to find find
// the sum of the first half elements and second half
// elements of the array.
#include <stdio.h>
void sum_of_elements(int arr[], int n)
{
int sum_of_first = 0, sum_of_second = 0;
// adding
for (int i = 0; i < n; i++) {
// adding first half
if (i < n / 2) {
sum_of_first += arr[i];
}
// adding second half
else {
sum_of_second += arr[i];
}
}
// printing answer
printf("Sum of first half elements is %d\n",
sum_of_first);
printf("Sum of second half elements is %d\n",
sum_of_second);
}
// Driver code
int main()
{
int arr[] = { 20, 30, 60, 10, 25, 15, 40 };
int n = sizeof(arr) / sizeof(arr[0]);
// function call
sum_of_elements(arr, n);
return 0;
}
// C# program to count pairs
// whose sum divisible by 'K'
using System;
class GFG {
public static void sum_of_elements(int[] arr, int n)
{
int sumfirst = 0, sumsecond = 0;
for (int i = 0; i < n; i++) {
// Add elements in first half sum
if (i < n / 2) {
sumfirst += arr[i];
}
// Add elements in the second half sum
else {
sumsecond += arr[i];
}
}
Console.WriteLine("Sum of first half elements is "
+ sumfirst);
Console.WriteLine("Sum of second half elements is "
+ sumsecond);
}
// Driver code
static public void Main()
{
int[] arr = { 20, 30, 60, 10, 25, 15, 40 };
int n = arr.Length;
// Function call
sum_of_elements(arr, n);
}
}
// This code is contributed by nidhiva
// C++ program to find the sum of the first half
// elements and second half elements of an array
#include <bits/stdc++.h>
using namespace std;
// Function to find the sum of the first half
// elements and second half elements of an array
void sum_of_elements(int arr[], int n) {
int sumfirst = 0, sumsecond = 0;
for (int i = 0; i < n; i++) {
// Add elements in first half sum
if (i < n / 2)
sumfirst += arr[i];
// Add elements in the second half sum
else
sumsecond += arr[i];
}
cout << "Sum of first half elements is " << sumfirst
<< endl;
cout << "Sum of second half elements is " << sumsecond
<< endl;
}
// Driver Code
int main()
{
int arr[] = { 20, 30, 60, 10, 25, 15, 40 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
sum_of_elements(arr, n);
return 0;
}
// Java program to count pairs
// whose sum divisible by 'K'
import java.util.*;
class GFG {
public static void sum_of_elements(int[] arr, int n)
{
int sumfirst = 0, sumsecond = 0;
for (int i = 0; i < n; i++) {
// Add elements in first half sum
if (i < n / 2) {
sumfirst += arr[i];
}
// Add elements in the second half sum
else {
sumsecond += arr[i];
}
}
System.out.println("Sum of first half elements is "
+ sumfirst);
System.out.println("Sum of second half elements is "
+ sumsecond);
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 20, 30, 60, 10, 25, 15, 40 };
int n = arr.length;
// Function call
sum_of_elements(arr, n);
}
}
// This code is contributed by Princi Singh
# Python3 program to find the sum of
# the first half elements and
# second half elements of an array
# Function to find the sum of
# the first half elements and
# second half elements of an array
def sum_of_elements(arr, n):
sumfirst = 0
sumsecond = 0
for i in range(n):
# Add elements in first half sum
if (i < n // 2):
sumfirst += arr[i]
# Add elements in the second half sum
else:
sumsecond += arr[i]
print("Sum of first half elements is",
sumfirst, end="\n")
print("Sum of second half elements is",
sumsecond, end="\n")
# Driver Code
arr = [20, 30, 60, 10, 25, 15, 40]
n = len(arr)
# Function call
sum_of_elements(arr, n)
# This code is contributed
# by Akanksha Rai
<script>
// Javascript program to count pairs
// whose sum divisible by 'K'
function sum_of_elements(arr , n)
{
var sumfirst = 0, sumsecond = 0;
for (i = 0; i < n; i++) {
// Add elements in first half sum
if (i < parseInt(n / 2))
{
sumfirst += arr[i];
}
// Add elements in the second half sum
else {
sumsecond += arr[i];
}
}
document.write(
"Sum of first half elements is " + sumfirst+"<br/>"
);
document.write(
"Sum of second half elements is " + sumsecond+"<br/>"
);
}
// Driver code
var arr = [ 20, 30, 60, 10, 25, 15, 40 ];
var n = arr.length;
// Function call
sum_of_elements(arr, n);
// This code contributed by umadevi9616
</script>
Output: Sum of first half elements is 110
Sum of second half elements is 90
Time complexity: O(N), as we are using a loop to traverse the array.
Auxiliary Space: O(1), as we are not using any extra space.
Approach 2: Traversing half of the array length.
The idea is to traverse half the length of the array and calculate the first sum and second sum simultaneously by
- Initializing firstSum=0 and LastSum=0 and also traversing the array starting from the 0th index to N/2.
- Adding values of arr[i] to the firstSum.
- Adding values of arr[i+n/2] to the secondSum.
- An edge case occurs when the length of the array is odd. In this case, the first sum has the sum of the first N/2 elements while secondSum has the sum of the remaining elements excluding the last element (since the loop runs for n/2 times the last element is excluded) Hence we take care of this exclusively.
Below is the code for the same.
C#
// c# code to find the sum of first n/2 and last n- n/2
// elements of an array by traversing the array length/2
// times
using System;
public class GFG {
public static void Main()
{
int[] arr = { 20, 30, 60, 10, 25, 15, 40 };
int n = arr.Length;
// Function call
sum(arr, n);
}
// function definition
public static void sum(int[] arr, int n)
{
int firstSum = 0, secondSum = 0;
// initializing the firstSum and secondSum variables
for (int i = 0; i < n / 2; i++) {
// adding elements of the first half to firstSum
firstSum += arr[i];
// adding elements of the second half to
// secondSum
secondSum += arr[i + n / 2];
}
// checking for the odd case length
if (n % 2 != 0)
secondSum += arr[n - 1];
// printing the sums
Console.WriteLine("Sum of first half elements is "
+ firstSum);
Console.WriteLine("Sum of second half elements is "
+ secondSum);
}
}
// c++ code to find the sum of first n/2 and last n- n/2
// elements of an array by traversing the array length/2
// times
#include <bits/stdc++.h>
using namespace std;
// Function definition
void sum(int arr[], int n)
{
int firstSum = 0, secondSum = 0;
// initializing the firstSum and secondSum variables
for (int i = 0; i < n / 2; i++) {
// adding elements of the first half to firstSum
firstSum += arr[i];
// adding elements of the second half to secondSum
secondSum += arr[i + n / 2];
}
// checking for the odd case length
if (n % 2 != 0)
secondSum += arr[n - 1];
// printing the sums
cout << "Sum of first half elements is " << firstSum
<< endl;
cout << "Sum of second half elements is " << secondSum
<< endl;
}
// Driver Code
int main()
{
int arr[] = { 20, 30, 60, 10, 25, 15, 40 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
sum(arr, n);
return 0;
}
// Java code to find the sum of first n/2 and last n- n/2
// elements of an array by traversing the array length/2
// times
import java.util.*;
public class Main {
// Function definition
static void sum(int arr[], int n)
{
int firstSum = 0, secondSum = 0;
// initializing the firstSum and secondSum variables
for (int i = 0; i < n / 2; i++) {
// adding elements of the first half to firstSum
firstSum += arr[i];
// adding elements of the second half to
// secondSum
secondSum += arr[i + n / 2];
}
// checking for the odd case length
if (n % 2 != 0)
secondSum += arr[n - 1];
// printing the sums
System.out.println("Sum of first half elements is "
+ firstSum);
System.out.println("Sum of second half elements is "
+ secondSum);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 20, 30, 60, 10, 25, 15, 40 };
int n = arr.length;
// Function call
sum(arr, n);
}
}
# python code to find the sum of first n/2 and last n- n/2
# elements of an array by traversing the array length/2
# times
# function definition
def sum(arr, n):
first_sum, second_sum = 0, 0
# initializing the first_sum and second_sum variables
for i in range(n // 2):
# adding elements of the first half to first_sum
first_sum += arr[i]
# adding elements of the second half to second_sum
second_sum += arr[i + n // 2]
# checking for the odd case length
if n % 2 != 0:
second_sum += arr[n - 1]
# printing the sums
print('Sum of first half elements is', first_sum)
print("Sum of second half elements is", second_sum)
# Driver Code
arr = [20, 30, 60, 10, 25, 15, 40]
n = len(arr)
# Function call
sum(arr, n)
// JavaScript code to find the sum of first n/2 and last n- n/2
// elements of an array by traversing the array length/2
// times
// Function definition
function sum(arr, n) {
let firstSum = 0, secondSum = 0;
// initializing the firstSum and secondSum variables
for (let i = 0; i < Math.floor(n / 2); i++) {
// adding elements of the first half to firstSum
firstSum += arr[i];
// adding elements of the second half to secondSum
secondSum += arr[i + Math.floor(n / 2)];
}
// checking for the odd case length
if (n % 2 !== 0)
secondSum += arr[n - 1];
// printing the sums
console.log(`Sum of first half elements is ${firstSum}`);
console.log(`Sum of second half elements is ${secondSum}`);
}
// Driver Code
const arr = [20, 30, 60, 10, 25, 15, 40];
const n = arr.length;
// Function call
sum(arr, n);
Output:
Sum of first half elements is 110
Sum of second half elements is 90
Time complexity: O(N/2)
Auxiliary Space: O(1)
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