Find the sum of the first half and second half elements of an array

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 110

Input : arr[] = {50, 35, 20, 15} 
Output : 85, 35

Approach:

1. Initialize SumFirst and SumSecond as 0. 
2. Traverse the given array. 
3. 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#

// 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++

// 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

// 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

# 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

Javascript

<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 

1. Initializing firstSum=0 and LastSum=0 and also traversing the array starting from the 0th index to N/2. 
2. Adding values of arr[i] to the firstSum.
3. Adding values of arr[i+n/2] to the secondSum.
4. 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++

// 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

// 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

# 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

// 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)