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;Â
// Function to find the sum of array// elements which are multiples of Nvoid mulsum(int arr[], int n, int N){Â
    // Stores the sum    int sum = 0;Â
    // Traverse the given array    for (int i = 0; i < n; i++) {Â
        // If current element        // is a multiple of N        if (arr[i] % N == 0) {            sum = sum + arr[i];        }    }Â
    // Print total sum    cout << sum;}Â
// Driver Codeint main(){Â
    // Given arr[]    int arr[] = { 1, 2, 3, 5, 6 };Â
    int n = sizeof(arr) / sizeof(arr[0]);Â
    int N = 3;Â
    mulsum(arr, n, N);Â
    return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG{Â
// Function to find the sum of array// elements which are multiples of Nstatic void mulsum(int arr[], int n, int N){Â
    // Stores the sum    int sum = 0;Â
    // Traverse the given array    for (int i = 0; i < n; i++)    {Â
        // If current element        // is a multiple of N        if (arr[i] % N == 0)         {            sum = sum + arr[i];        }    }Â
    // Print total sum    System.out.println(sum);}Â
Â
// Driver Codepublic static void main(String[] args){Â Â Â Â Â Â Â Â Â // Given arr[]Â Â Â Â int arr[] = { 1, 2, 3, 5, 6 };Â Â Â Â int n = arr.length;Â Â Â Â int N = 3;Â Â Â Â mulsum(arr, n, N);}}Â
// 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{Â
// Function to find the sum of array// elements which are multiples of Nstatic void mulsum(int[] arr, int n, int N){Â
    // Stores the sum    int sum = 0;Â
    // Traverse the given array    for (int i = 0; i < n; i++)    {Â
        // If current element        // is a multiple of N        if (arr[i] % N == 0)         {            sum = sum + arr[i];        }    }Â
    // Print total sum    Console.Write(sum);}Â
// Driver Codestatic public void Main (){Â Â Â Â // Given arr[]Â Â Â Â int[] arr = { 1, 2, 3, 5, 6 };Â Â Â Â int n = arr.Length;Â Â Â Â int N = 3;Â Â Â Â mulsum(arr, n, N);}}Â
// This code is contributed by Dharanendra L V. |
Javascript
<script>Â
// JavaScript program for the above approachÂ
// Function to find the sum of array// elements which are multiples of Nfunction 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);}Â
// Driver CodeÂ
// Given arr[]var arr = [ 1, 2, 3, 5, 6 ];var n = arr.length;var N = 3;Â
mulsum(arr, n, N);Â
// This code is contributed by rdtankÂ
</script> |
Output:Â
9
Â
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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