# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program for the above approach ` `#include ` `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); ` ` `  `    ``// 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 `

## 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 `

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