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Sum of multiples of Array elements within a given range [L, R]

  • Last Updated : 08 Jul, 2021

Given an array arr[] of positive integers and two integers L and R, the task is to find the sum of all multiples of the array elements in the range [L, R].

Examples: 

Input: arr[] = {2, 7, 3, 8}, L = 7, R = 20 
Output: 197 
Explanation: 
In the range 7 to 20: 
Sum of multiples of 2: 8 + 10 + 12 + 14 + 16 + 18 + 20 = 98 
Sum of multiples of 7: 7 + 14 = 21 
Sum of multiples of 3: 9 + 12 + 15 + 18 = 54 
Sum of multiples of 8: 8 + 16 = 24 
Total sum of all multiples = 98 + 21 + 54 + 24 = 197

Input: arr[] = {5, 6, 7, 8, 9}, L = 39, R = 100 
Output: 3278 

Naive Approach: The naive idea is for each element in the given array arr[] find the multiple of the element in the range [L, R] and print the sum of all the multiples.



Time Complexity: O(N*(L-R)) 
Auxiliary Space: O(1)

Efficient Approach: To optimize the above naive approach we will use the concept discussed below: 

  1. For any integer X, the number of multiples of X till any integer Y is given by Y/X.
  2. Let N = Y/X 
    Then, the sum of all the above multiple is given by X*N*(N-1)/2.

For Example: 

For X = 2 and Y = 12 
Sum of multiple is: 
=> 2 + 4 + 6 + 8 + 10 + 12 
=> 2*(1 + 2 + 3 + 4 + 5 + 6) 
=> 2*(6*5)/2 
=> 20. 

Using the above concept the problem can be solved using the below steps: 

  1. Calculate the sum of all multiples of arr[i] up to R using the above-discussed formula.
  2. Calculate the sum of all multiples of arr[i] up to L – 1 using the above-discussed formula.
  3. Subtract the above two values in the above steps to get the sum of all multiples between ranges [L, R].
  4. Repeat the above process for all the elements and print the 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 all
// multiples of N up to K
int calcSum(int k, int n)
{
    // Calculate the sum
    int value = (k * n * (n
                          + 1))
                / 2;
    // Return the sum
    return value;
}
 
// Function to find the total sum
int findSum(int* a, int n, int L, int R)
{
    int sum = 0;
    for (int i = 0; i < n; i++) {
 
        // Calculating sum of multiples
        // for each element
 
        // If L is divisible by a[i]
        if (L % a[i] == 0 && L != 0) {
            sum += calcSum(a[i], R / a[i])
                   - calcSum(a[i],
                             (L - 1) / a[i]);
        }
 
        // Otherwise
        else {
            sum += calcSum(a[i], R / a[i])
                   - calcSum(a[i], L / a[i]);
        }
    }
 
    // Return the final sum
    return sum;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 2, 7, 3, 8 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Given range
    int L = 7;
    int R = 20;
 
    // Function Call
    cout << findSum(arr, N, L, R);
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to find the sum of
// all multiples of N up to K
static int calcSum(int k, int n)
{
     
    // Calculate the sum
    int value = (k * n * (n + 1)) / 2;
     
    // Return the sum
    return value;
}
 
// Function to find the total sum
static int findSum(int[] a, int n,
                   int L, int R)
{
    int sum = 0;
    for(int i = 0; i < n; i++)
    {
        
       // Calculating sum of multiples
       // for each element
        
       // If L is divisible by a[i]
       if (L % a[i] == 0 && L != 0)
       {
           sum += calcSum(a[i], R / a[i]) -
                  calcSum(a[i], (L - 1) / a[i]);
       }
        
       // Otherwise
       else
       {
           sum += calcSum(a[i], R / a[i]) -
                  calcSum(a[i], L / a[i]);
       }
    }
 
    // Return the final sum
    return sum;
}
 
// Driver Code
public static void main (String[] args)
{
     
    // Given array arr[]
    int arr[] = { 2, 7, 3, 8 };
 
    int N = arr.length;
 
    // Given range
    int L = 7;
    int R = 20;
 
    // Function Call
    System.out.println(findSum(arr, N, L, R));
}
}
 
// This code is contributed by shubhamsingh10

Python3




# Python3 program for the above approach
 
# Function to find the sum of
# all multiples of N up to K
def calcSum(k, n):
 
    # Calculate the sum
    value = (k * n * (n + 1)) // 2
     
    # Return the sum
    return value
     
# Function to find the total sum
def findSum(a, n, L, R):
 
    sum = 0
    for i in range(n):
         
        # Calculating sum of multiples
        # for each element
         
        # If L is divisible by a[i]
        if (L % a[i] == 0 and L != 0):
            sum += (calcSum(a[i], R // a[i]) -
                    calcSum(a[i], (L - 1) // a[i]))
         
        # Otherwise
        else:
            sum += (calcSum(a[i], R // a[i]) -
                    calcSum(a[i], L // a[i]))
     
    # Return the final sum
    return sum;
 
# Driver code
if __name__=="__main__":
     
    # Given array arr[]
    arr = [ 2, 7, 3, 8 ]
 
    N = len(arr)
 
    # Given range
    L = 7
    R = 20
 
    # Function call
    print(findSum(arr, N, L, R))    
 
# This code is contributed by rutvik_56

C#




// C# program for the above approach
using System;
class GFG{
      
// Function to find the sum of
// all multiples of N up to K
static int calcSum(int k, int n)
{
      
    // Calculate the sum
    int value = (k * n * (n + 1)) / 2;
      
    // Return the sum
    return value;
}
  
// Function to find the total sum
static int findSum(int[] a, int n,
                   int L, int R)
{
    int sum = 0;
    for(int i = 0; i < n; i++)
    {
         
       // Calculating sum of multiples
       // for each element
         
       // If L is divisible by a[i]
       if (L % a[i] == 0 && L != 0)
       {
           sum += calcSum(a[i], R / a[i]) -
                  calcSum(a[i], (L - 1) / a[i]);
       }
         
       // Otherwise
       else
       {
           sum += calcSum(a[i], R / a[i]) -
                  calcSum(a[i], L / a[i]);
       }
    }
  
    // Return the final sum
    return sum;
}
  
// Driver Code
public static void Main (string[] args)
{
      
    // Given array arr[]
    int []arr = new int[]{ 2, 7, 3, 8 };
  
    int N = arr.Length;
  
    // Given range
    int L = 7;
    int R = 20;
  
    // Function Call
    Console.Write(findSum(arr, N, L, R));
}
}
  
// This code is contributed by Ritik Bansal

Javascript




<script>
// Javascript implementation for the above approach
 
// Function to find the sum of
// all multiples of N up to K
function calcSum(k, n)
{
     
    // Calculate the sum
    let value = (k * n * (n + 1)) / 2;
     
    // Return the sum
    return value;
}
 
// Function to find the total sum
function findSum(a, n, L, R)
{
    let sum = 0;
    for(let i = 0; i < n; i++)
    {
        
       // Calculating sum of multiples
       // for each element
        
       // If L is divisible by a[i]
       if (L % a[i] == 0 && L != 0)
       {
           sum += calcSum(a[i], Math.floor(R / a[i])) -
                  calcSum(a[i], Math.floor((L - 1) / a[i]));
       }
        
       // Otherwise
       else
       {
           sum += calcSum(a[i], Math.floor(R / a[i])) -
                  calcSum(a[i], Math.floor(L / a[i]));
       }
    }
 
    // Return the final sum
    return sum;
}
 
    // Driver Code
     
    // Given array arr[]
    let arr = [ 2, 7, 3, 8 ];
 
    let N = arr.length;
 
    // Given range
    let L = 7;
    let R = 20;
 
    // Function Call
   document.write(findSum(arr, N, L, R));
         
</script>
Output: 
197

 

Time Complexity: O(N), where N is the number of elements in the given array. 
Auxiliary Space: O(1)
 

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