Sum of array elements which are multiples of a given number

• Difficulty Level : Basic
• Last Updated : 04 May, 2021

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:

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

Below is the implementation of the above approach:

C++

 // C++ program for the above approach#include using namespace std; // Function to find the sum of array// elements which are multiples of Nvoid 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 Codeint main(){     // Given arr[]    int arr[] = { 1, 2, 3, 5, 6 };     int n = sizeof(arr) / sizeof(arr);     int N = 3;     mulsum(arr, n, N);     return 0;}

Java

 // Java program for the above approachimport java.io.*;import java.util.*;class GFG{ // Function to find the sum of array// elements which are multiples of Nstatic 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 Codepublic 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 Ndef 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 Codeif __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 approachusing System;public class GFG{ // Function to find the sum of array// elements which are multiples of Nstatic 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 Codestatic 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


Output:
9

Time Complexity: O(N)
Auxiliary Space: O(1)

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