Sum of array elements which are multiples of a given number

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Given an array arr[] consisting of positive integers and an integer N, the task is to find the sum of all array elements which are multiples of N

Examples:

Input: arr[] = {1, 2, 3, 5, 6}, N = 3
Output: 9
Explanation: From the given array, 3 and 6 are multiples of 3. Therefore, sum = 3 + 6 = 9.

Input: arr[] = {1, 2, 3, 5, 7, 11, 13}, N = 5
Output: 5

Approach: The idea is to traverse the array and for each array element, check if it is a multiple of N or not and add those elements. Follow the steps below to solve the problem:

Initialize a variable, say sum, to store the required sum.
Traverse the given array and for each array element, perform the following operations.
Check whether the array element is a multiple of N or not.
If the element is a multiple of N, then add the element to sum.
Finally, print the value of sum.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the sum of array
// elements which are multiples of N
void mulsum(int arr[], int n, int N)
{
 
    // Stores the sum
    int sum = 0;
 
    // Traverse the given array
    for (int i = 0; i < n; i++) {
 
        // If current element
        // is a multiple of N
        if (arr[i] % N == 0) {
            sum = sum + arr[i];
        }
    }
 
    // Print total sum
    cout << sum;
}
 
// Driver Code
int main()
{
 
    // Given arr[]
    int arr[] = { 1, 2, 3, 5, 6 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    int N = 3;
 
    mulsum(arr, n, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG{
 
// Function to find the sum of array
// elements which are multiples of N
static void mulsum(int arr[], int n, int N)
{
 
    // Stores the sum
    int sum = 0;
 
    // Traverse the given array
    for (int i = 0; i < n; i++)
    {
 
        // If current element
        // is a multiple of N
        if (arr[i] % N == 0) 
        {
            sum = sum + arr[i];
        }
    }
 
    // Print total sum
    System.out.println(sum);
}
 
 
// Driver Code
public static void main(String[] args)
{
     
    // Given arr[]
    int arr[] = { 1, 2, 3, 5, 6 };
    int n = arr.length;
    int N = 3;
    mulsum(arr, n, N);
}
}
 
// This code is contributed by jana_sayantan.

Python

# Python3 program for the above approach
  
# Function to find the sum of array
# elements which are multiples of N
def mulsum(arr, n, N):
      
    # Stores the sum
    sums = 0
  
    # Traverse the array
    for i in range(0, n):
        if arr[i] % N == 0:
              sums = sums + arr[i]
  
    # Print total sum
    print(sums)
  
# Driver Code
if __name__ == "__main__":
  
    # Given arr[]
    arr = [ 1, 2, 3, 5, 6 ]
  
    n = len(arr)
     
    N = 3
  
    # Function call
    mulsum(arr, n, N)

C#

// C# program for the above approach
using System;
public class GFG
{
 
// Function to find the sum of array
// elements which are multiples of N
static void mulsum(int[] arr, int n, int N)
{
 
    // Stores the sum
    int sum = 0;
 
    // Traverse the given array
    for (int i = 0; i < n; i++)
    {
 
        // If current element
        // is a multiple of N
        if (arr[i] % N == 0) 
        {
            sum = sum + arr[i];
        }
    }
 
    // Print total sum
    Console.Write(sum);
}
 
// Driver Code
static public void Main ()
{
    // Given arr[]
    int[] arr = { 1, 2, 3, 5, 6 };
    int n = arr.Length;
    int N = 3;
    mulsum(arr, n, N);
}
}
 
// This code is contributed by Dharanendra L V.

Javascript

<script>
 
// JavaScript program for the above approach
 
// Function to find the sum of array
// elements which are multiples of N
function mulsum(arr, n, N) 
{
     
    // Stores the sum
    var sum = 0;
     
    // Traverse the given array
    for(var i = 0; i < n; i++)
    {
         
        // If current element
        // is a multiple of N
        if (arr[i] % N == 0) 
        {
            sum = sum + arr[i];
        }
    }
     
    // Print total sum
    document.write(sum);
}
 
// Driver Code
 
// Given arr[]
var arr = [ 1, 2, 3, 5, 6 ];
var n = arr.length;
var N = 3;
 
mulsum(arr, n, N);
 
// This code is contributed by rdtank
 
</script>

Output:

Time Complexity: O(N) since one traversal of the array is required to complete all operations hence the overall time required by the algorithm is linear
Auxiliary Space: O(1) since no extra array is used so the space taken by the algorithm is constant

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