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 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[0]); 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
<script>// JavaScript program for the above approach// 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);}// 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:
9
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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