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++ 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 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. |
# 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# 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. |
<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> |
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