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

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 below steps:

  1. Calculate the sum of all multiples of arr[i] upto R using the above dicussed formula.
  2. Calculate the sum of all multiples of arr[i] upto L – 1 using the above dicussed formula.
  3. Subtract the above two values in the above steps to get the sum of all multiples between range [L, R].
  4. Repeat the above process for all the elements and print the sum.

Below is the implementation of above approach:

C++

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// 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;
}

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Java

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// 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

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C#

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// 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

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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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Improved By : SHUBHAMSINGH10, btc_148