Sum of all the elements in an array divisible by a given number K

Given an array containing N elements and a number K. The task is to find the sum of all such elements which are divisible by K.

Examples: 

Input : arr[] = {15, 16, 10, 9, 6, 7, 17}
        K = 3
Output : 30
Explanation: As 15, 9, 6 are divisible by 3. So, sum of elements divisible by K = 15 + 9 + 6 = 30.

Input : arr[] = {5, 3, 6, 8, 4, 1, 2, 9}
        K = 2
Output : 20

The idea is to traverse the array and check the elements one by one. If an element is divisible by K then add that element’s value with the sum so far and continue this process while the end of the array reached.

Below is the implementation of the above approach:  

C++

// C++ program to find sum of all the elements
// in an array divisible by a given number K

#include <iostream>
using namespace std;

// Function to find sum of all the elements
// in an array divisible by a given number K
int findSum(int arr[], int n, int k)
{
    int sum = 0;

    // Traverse the array
    for (int i = 0; i < n; i++) {

        // If current element is divisible by k
        // add it to sum
        if (arr[i] % k == 0) {
            sum += arr[i];
        }
    }

    // Return calculated sum
    return sum;
}

// Driver code
int main()
{
    int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;

    cout << findSum(arr, n, k);

    return 0;
}

C

// C program to find sum of all the elements
// in an array divisible by a given number K
#include <stdio.h>

// Function to find sum of all the elements
// in an array divisible by a given number K
int findSum(int arr[], int n, int k)
{
    int sum = 0;

    // Traverse the array
    for (int i = 0; i < n; i++) {

        // If current element is divisible by k
        // add it to sum
        if (arr[i] % k == 0) {
            sum += arr[i];
        }
    }

    // Return calculated sum
    return sum;
}

// Driver code
int main()
{
    int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;

    printf("%d",findSum(arr, n, k));

    return 0;
}

// This code is contributed by kothavvsaakash.

Java

// Java program to find sum of all the elements
// in an array divisible by a given number K

import java.io.*;

class GFG {

// Function to find sum of all the elements
// in an array divisible by a given number K
static int findSum(int arr[], int n, int k)
{
    int sum = 0;

    // Traverse the array
    for (int i = 0; i < n; i++) {

        // If current element is divisible by k
        // add it to sum
        if (arr[i] % k == 0) {
            sum += arr[i];
        }
    }

    // Return calculated sum
    return sum;
}

    // Driver code
    public static void main (String[] args) {

    int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = arr.length;
    int k = 3;

    System.out.println( findSum(arr, n, k));
    }
}

// this code is contributed by anuj_67..

Python3

# Python3 program to find sum of
# all the elements in an array
# divisible by a given number K

# Function to find sum of all
# the elements in an array
# divisible by a given number K
def findSum(arr, n, k) :

    sum = 0

    # Traverse the array
    for i in range(n) :

        # If current element is divisible
        # by k add it to sum
        if arr[i] % k == 0 :

            sum += arr[i]

    # Return calculated sum
    return sum

# Driver code
if __name__ == "__main__" :

    arr = [ 15, 16, 10, 9, 6, 7, 17]
    n = len(arr)
    k = 3

    print(findSum(arr, n, k))

# This code is contributed by ANKITRAI1

C#

// C# program to find sum of all the elements
// in an array divisible by a given number K

using System;

public class GFG{

// Function to find sum of all the elements
// in an array divisible by a given number K
static int findSum(int []arr, int n, int k)
{
    int sum = 0;

    // Traverse the array
    for (int i = 0; i < n; i++) {

        // If current element is divisible by k
        // add it to sum
        if (arr[i] % k == 0) {
            sum += arr[i];
        }
    }

    // Return calculated sum
    return sum;
}

    // Driver code

    static public void Main (){

    int []arr = { 15, 16, 10, 9, 6, 7, 17 };
    int n = arr.Length;
    int k = 3;

    Console.WriteLine( findSum(arr, n, k));
    }
}
//This code is contributed by anuj_67..

PHP

<?php
// PHP program to find sum of all
// the elements in an array divisible
// by a given number K

// Function to find sum of all
// the elements in an array
// divisible by a given number K
function findSum($arr, $n, $k)
{
    $sum = 0;

    // Traverse the array
    for ($i = 0; $i < $n; $i++)
    {

        // If current element is divisible
        // by k add it to sum
        if ($arr[$i] % $k == 0)
        {
            $sum += $arr[$i];
        }
    }

    // Return calculated sum
    return $sum;
}

// Driver code
$arr = array(15, 16, 10, 9, 6, 7, 17);
$n = sizeof($arr);
$k = 3;

echo findSum($arr, $n, $k);

// This code is contributed
// by Akanksha Rai(Abby_akku)

Javascript

<script>
    // Javascript program to find sum of all the elements
    // in an array divisible by a given number K

    // Function to find sum of all the elements
    // in an array divisible by a given number K
    function findSum(arr, n, k)
    {
        let sum = 0;

        // Traverse the array
        for (let i = 0; i < n; i++) {

            // If current element is divisible by k
            // add it to sum
            if (arr[i] % k == 0) {
                sum += arr[i];
            }
        }

        // Return calculated sum
        return sum;
    }

    let arr = [ 15, 16, 10, 9, 6, 7, 17 ];
    let n = arr.length;
    let k = 3;

    document.write(findSum(arr, n, k));

</script>

Output

30

complexity Analysis:

• Time Complexity: O(N), where N is the number of elements in the array.
• Auxiliary Space: O(1)

Using Recursion :

Start traversing the array using recursion and check if the current element is divisible by K. If yes then update the answer. Print the answer when all the elements get traversed.

Method 1 : (Pass By Reference)

Implementation :

• Make a variable sum and pass that address of that variable
• We passing address of sum variable so that no change occur in it after every another function call.  
• We just increment our pointer and make another function call.
• Add the value at pointer into our sum and make another function call until the base condition hits. 

C++

#include <iostream>
using namespace std;

void solve(int arr[], int k, int i, int& ans)
{
    if (i < 0) { // boundry condition to stop the recurion
        return;
    }

    if (arr[i] % k == 0) { // current element is divisible
                           // so add that element to our ans
        ans += arr[i];
    }
    solve(arr, k, i - 1,
          ans); // further iterate array using recursion
}

int main()
{
    int n = 6;
    int arr[n] = { 2, 6, 7, 12, 14, 18 };
    int k = 2;
    int sum = 0;

    solve(arr, k, n - 1, sum);

    cout << sum << " ";
    return 0;
}

//this code is contributed by bhardwajji

Java

import java.util.*;

public class Main {
    // function to recursively compute the sum of elements in arr that are divisible by k
    public static void solve(int arr[], int k, int i, int[] ans) {
        if (i < 0) { // base case to stop recursion
            return;
        }
        if (arr[i] % k == 0) { // current element is divisible by k
            ans[0] += arr[i]; // add current element to the sum
        }
        solve(arr, k, i - 1, ans); // recursively call the function with the previous index
    }

    public static void main(String[] args) {
        int n = 6;
        int arr[] = {2, 6, 7, 12, 14, 18};
        int k = 2;
        int[] sum = new int[1]; // store the sum in an array to avoid pass-by-value

        solve(arr, k, n - 1, sum);

        System.out.print(sum[0] + " ");
    }
}

Python3

def solve(arr, k, i, ans):
    # Base case to stop recursion
    if i < 0:
        return

    # If the current element is divisible by k, add it to the sum
    if arr[i] % k == 0:
        ans[0] += arr[i]

    # Recursively call the function with the previous index
    solve(arr, k, i - 1, ans)

# Main function
if __name__ == '__main__':
    n = 6
    arr = [2, 6, 7, 12, 14, 18]
    k = 2

    # Store the sum in an array to avoid pass-by-value
    sum_arr = [0]

    # Call the recursive function
    solve(arr, k, n - 1, sum_arr)

    # Print the sum
    print(sum_arr[0], end=' ')

C#

using System;

public class Program
{
  public static void Solve(int[] arr, int k, int i, ref int ans)
  {
    if (i < 0) // boundry condition to stop the recursion
      return;


    if (arr[i] % k == 0) // current element is divisible
    {
      // so add that element to our ans
      ans += arr[i];
    }

    Solve(arr, k, i - 1, ref ans); // further iterate array using recursion
  }

  public static void Main()
  {
    int n = 6;
    int[] arr = { 2, 6, 7, 12, 14, 18 };
    int k = 2;
    int sum = 0;

    Solve(arr, k, n - 1, ref sum);

    Console.WriteLine(sum);
  }
}

Javascript

function solve(arr, k, i, ans) {
  // Base case to stop recursion
  if (i < 0) {
    return;
  }

  // If the current element is divisible by k, add it to the sum
  if (arr[i] % k === 0) {
    ans[0] += arr[i];
  }

  // Recursively call the function with the previous index
  solve(arr, k, i - 1, ans);
}

// Main function
const n = 6;
const arr = [2, 6, 7, 12, 14, 18];
const k = 2;

// Store the sum in an array to avoid pass-by-value
const sum_arr = [0];

// Call the recursive function
solve(arr, k, n - 1, sum_arr);

// Print the sum
console.log(sum_arr[0]);

Output

52 

Method 2 : (Pass By Value)

Implementation : 

• Take a sum variable and only pass its value into the function solve.
• Function will return int value and we use that value to find our ans.
• If number is divisible by K then return the number ( arr[i] ) and if not divisible then return 0.

C++

#include <iostream>
using namespace std;

int solve(int arr[], int k, int i)
{ // returning the sum of elements divisible by k
    if (i < 0) {
        return 0;
    }

    if (arr[i] % k == 0) {
        return arr[i]
               + solve(arr, k,
                       i - 1); // add current element and
                               // further call the recursion
    }
    else {
        return 0
               + solve(arr, k,
                       i - 1); // nothing to add so just
                               // call the futher recursion
    }
}

int main()
{
    int n = 6;
    int arr[n] = { 2, 6, 7, 12, 14, 18 };
    int k = 2;

    int sum = solve(arr, k, n - 1);

    cout << sum << " ";
    return 0;
}

// this code is contributed by bhardwajji

Java

/*package whatever //do not write package name here */
import java.util.*;

public class Main {

    // Function to calculate the sum of array elements that
    // are divisible by k
    static int solve(int[] arr, int k, int i)
    {
        // Base case: If the index is less than zero, return
        // zero
        if (i < 0) {
            return 0;
        }

        // If the current element is divisible by k, add it
        // to the sum and call the function recursively
        if (arr[i] % k == 0) {
            return arr[i] + solve(arr, k, i - 1);
        }
        // If the current element is not divisible by k,
        // just call the function recursively without adding
        // anything
        else {
            return 0 + solve(arr, k, i - 1);
        }
    }

    // Main function
    public static void main(String[] args)
    {
        int n = 6;
        int[] arr = { 2, 6, 7, 12, 14, 18 };
        int k = 2;

        // Calculate the sum of array elements that are
        // divisible by k
        int sum = solve(arr, k, n - 1);

        // Print the sum
        System.out.println(sum);
    }
}

Python3

# This function returns the sum of elements divisible by k
def solve(arr, k, i):
    if i < 0:
        return 0
    if arr[i] % k == 0:
        # add current element and further call the recursion
        return arr[i] + solve(arr, k, i - 1)
    else:
        # nothing to add so just call the further recursion
        return 0 + solve(arr, k, i - 1)


n = 6
arr = [2, 6, 7, 12, 14, 18]
k = 2
sum = solve(arr, k, n - 1)
print(sum)

C#

using System;

public class MainClass {
  public static int Solve(int[] arr, int k, int i)
  {
    // returning the sum of elements divisible by k
    if (i < 0) {
      return 0;
    }

    if (arr[i] % k == 0) {
      // add current element and further call the
      // recursion
      return arr[i] + Solve(arr, k, i - 1);
    }
    else {
      // nothing to add so just call the futher
      // recursion
      return 0 + Solve(arr, k, i - 1);
    }
  }

  public static void Main()
  {
    int n = 6;
    int[] arr = { 2, 6, 7, 12, 14, 18 };
    int k = 2;

    int sum = Solve(arr, k, n - 1);

    Console.WriteLine(sum);
  }
}

Javascript

// JavaScript program to find the sum of array elements
// divisible by a given number k
function solve(arr, k, i) {
// returning the sum of elements divisible by k
if (i < 0) {
return 0;
}

if (arr[i] % k == 0) {
return (
arr[i] +
solve(arr, k, i - 1) // add current element and further call the recursion
);
} else {
return 0 + solve(arr, k, i - 1); // nothing to add so just call the further recursion
}
}

// Driver code
const arr = [2, 6, 7, 12, 14, 18];
const k = 2;

const sum = solve(arr, k, arr.length - 1);

console.log(sum);

Output

52 

Complexity Analysis:

• Time Complexity: O(n) 
• Auxiliary Space: O(1)