Sum of Array (original) (raw)

Last Updated : 31 Mar, 2026

Given an array of integers **arr[], find the **sum of its elements.

**Examples:

**Input : arr[] = [1, 2, 3]
**Output : 6
**Explanation: 1 + 2 + 3 = 6

**Input : arr[] = [15, 12, 13, 10]
**Output : 50

Table of Content

Iterative Solution - O(n) Time and O(1) Space
Recursive Solution - O(n) Time and O(n) Space
Inbuilt Methods - O(n) Time and O(1) Space

Iterative Solution - O(n) Time and O(1) Space

Iterate through each element of the array and add it to a variable called sum. Initialize the sum variable to 0 before starting the iteration. After completing the iteration, return the final value of sum.

**Step-by-step execution:

For arr[] = [12, 3, 4, 15]

Initialize : sum = 0
i = 0, sum = 0 + 12 = 12
i = 1, sum = 12 + 3 = 15
i = 2, sum = 15 + 4 = 19
i = 3, sum = 19 + 15 = 34

Final sum = 34

C++ `

#include #include using namespace std;

int arraySum(vector& arr) { int sum = 0;

// Iterate through all elements and add them to sum
for (int i = 0; i < arr.size(); i++)
    sum += arr[i];

return sum;

}

int main() { vector arr = {12, 3, 4, 15};

cout << arraySum(arr);
return 0;

}

C

#include <stdio.h>

int arraySum(int arr[], int n) { int sum = 0;

// Iterate through all elements and add them to sum
for (int i = 0; i < n; i++)
    sum += arr[i];

return sum;

}

int main() { int arr[] = { 12, 3, 4, 15 }; int n = sizeof(arr) / sizeof(arr[0]);

printf("%d", arraySum(arr, n));
return 0;

}

Java

import java.io.*;

class GFG { static int arr[] = { 12, 3, 4, 15 };

static int arraySum()
{
    int sum = 0;

    // Iterate through all elements and add them to sum
    for (int i = 0; i < arr.length; i++)
        sum += arr[i];

    return sum;
}

public static void main(String[] args)
{
    System.out.println(arraySum());
}

}

Python

def arraySum(arr): sum = 0

# Iterate through all elements and add them to sum
for num in arr:
    sum += num

return sum

if name == "main": arr = [12, 3, 4, 15] print(arraySum(arr))

C#

using System;

class GFG {

static int arraySum(int[] arr, int n)
{
    int sum = 0;

    // Iterate through all elements and add them to sum
    for (int i = 0; i < n; i++)
        sum += arr[i];

    return sum;
}

public static void Main()
{
    int[] arr = { 12, 3, 4, 15 };
    int n = arr.Length;

    Console.Write(arraySum(arr, n));
}

}

JavaScript

function arraySum(arr) {
let sum = 0;

// Iterate through all elements and add them to sum  
for (let i = 0; i < arr.length; i++)  
    sum += arr[i];  

return sum;

}

// Driver code let arr = [12, 3, 4, 15]; console.log(arraySum(arr));

Recursive Solution - O(n) Time and O(n) Space

Use recursion to compute the sum by reducing the problem size in each call. In the base case, when the array size becomes 0, return 0. In the recursive case, return the element at index **n - 1 added to the result of a recursive call with the size reduced by one, continuing until all elements are processed.

dsa1

C++ `

#include #include using namespace std;

int arraySum(vector& arr, int n) { // base case if (n == 0) { return 0; } else {

    // recursively calling the function
    return arr[n - 1] + arraySum(arr, n - 1);
}

}

int main() { vector arr = { 12, 3, 4, 15 }; int n = arr.size();

cout << arraySum(arr, n);
return 0;

}

C

#include <stdio.h> #include <stdlib.h> #include <string.h>

int arraySum(int arr[], int n) { // base case if (n == 0) { return 0; } else {

    // recursively calling the function
    return arr[0] + arraySum(arr + 1, n - 1);
}

}

int main() { int arr[] = { 12, 3, 4, 15 }; int n = sizeof(arr) / sizeof(arr[0]);

printf("%d", arraySum(arr, n));

return 0;

}

Java

import java.io.*;

class GFG {

static int arraySum(int[] arr, int n)
{
    // base case
    if (n <= 0) {
        return 0;
    }

    // recursively calling the function
    return arraySum(arr, n - 1) + arr[n - 1];
}

public static void main(String[] args)
{
    int arr[] = { 12, 3, 4, 15 };
    int s = arraySum(arr, arr.length);

    System.out.println(s);
}

}

Python

def arraySum(arr):

# base case
if len(arr) == 0:
    return 0

# recursively calling the function
return arr[0] + arraySum(arr[1:])

if name == "main": arr = [12, 3, 4, 15] print(arraySum(arr))

C#

using System;

public class GFG {

static int arraySum(int[] arr, int n)
{
    // base case
    if (n <= 0) {
        return 0;
    }

    // recursively calling the function
    return arraySum(arr, n - 1) + arr[n - 1];
}

public static void Main()
{
    int[] arr = { 12, 3, 4, 15 };
    int sum = arraySum(arr, arr.Length);

    Console.Write(sum);
}

}

JavaScript

function arraySum(arr, n) { // base case if (n <= 0) { return 0; }

// recursively calling the function
return arraySum(arr, n - 1) + arr[n - 1];

}

// Driver code let arr = [12, 3, 4, 15]; let sum = arraySum(arr, arr.length);

console.log(sum);

Inbuilt Methods - O(n) Time and O(1) Space

Use built-in functions to compute the sum of elements in a given array. These functions eliminate the need for explicit iteration and improve code simplicity.

C++ `

#include #include #include using namespace std;

int main() { vector arr = { 12, 3, 4, 15 };

// calling accumulate function to compute sum of elements
cout << accumulate(arr.begin(), arr.end(), 0);

return 0;

}

Java

import java.util.Arrays;

public class GFG {

public static void main(String[] args) {
    int[] arr = {12, 3, 4, 15};
    
    // using stream to compute sum of elements
    int sum = Arrays.stream(arr).sum();
    
    System.out.println(sum);
}

}

Python

if name == "main":

arr = [12, 3, 4, 15]

# calling sum function to compute sum of elements
print(sum(arr))

C#

using System; using System.Linq;

class GFG {

public static void Main()
{
    int[] arr = { 12, 3, 4, 15 };

    // calling LINQ Sum method to compute sum of elements
    int sum = arr.Sum();

    Console.Write(sum);
}

}

JavaScript

const arr = [12, 3, 4, 15];

// calling reduce function to compute sum of elements const sum = arr.reduce((accumulator, currentValue) => accumulator + currentValue, 0);

console.log(sum);