# Find sum of a[i]%a[j] for all valid pairs

Given an array arr[] of size N. The task is to find the sum of arr[i] % arr[j] for all valid pairs. Answer can be large. So, output answer modulo 1000000007

Examples:

Input: arr[] = {1, 2, 3}
Output: 5
(1 % 1) + (1 % 2) + (1 % 3) + (2 % 1) + (2 % 2)
+ (2 % 3) + (3 % 1) + (3 % 2) + (3 % 3) = 5

Input: arr[] = {1, 2, 4, 4, 4}
Output: 10

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

Approach: Store the frequency of each element and run a nested loop on the frequency array and find the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `#define mod (int)(1e9 + 7) ` ` `  `// Function to return the sum of (a[i] % a[j]) ` `// for all valid pairs ` `int` `Sum_Modulo(``int` `a[], ``int` `n) ` `{ ` `    ``int` `max = *max_element(a, a + n); ` ` `  `    ``// To store the frequency of each element ` `    ``int` `cnt[max + 1] = { 0 }; ` ` `  `    ``// Store the frequency of each element ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cnt[a[i]]++; ` ` `  `    ``// To store the required answer ` `    ``long` `long` `ans = 0; ` ` `  `    ``// For all valid pairs ` `    ``for` `(``int` `i = 1; i <= max; i++) { ` `        ``for` `(``int` `j = 1; j <= max; j++) { ` ` `  `            ``// Update the count ` `            ``ans = ans + cnt[i] * cnt[j] * (i % j); ` `            ``ans = ans % mod; ` `        ``} ` `    ``} ` ` `  `    ``return` `(``int``)(ans); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 1, 2, 3 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` ` `  `    ``cout << Sum_Modulo(a, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` `     `  `static` `int` `mod = (``int``)(1e9 + ``7``); ` ` `  `// Function to return the sum of (a[i] % a[j]) ` `// for all valid pairs ` `static` `int` `Sum_Modulo(``int` `a[], ``int` `n) ` `{ ` `    ``int` `max = Arrays.stream(a).max().getAsInt(); ` ` `  `    ``// To store the frequency of each element ` `    ``int` `[]cnt=``new` `int``[max + ``1``]; ` ` `  `    ``// Store the frequency of each element ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``cnt[a[i]]++; ` ` `  `    ``// To store the required answer ` `    ``long` `ans = ``0``; ` ` `  `    ``// For all valid pairs ` `    ``for` `(``int` `i = ``1``; i <= max; i++)  ` `    ``{ ` `        ``for` `(``int` `j = ``1``; j <= max; j++) ` `        ``{ ` ` `  `            ``// Update the count ` `            ``ans = ans + cnt[i] *  ` `                        ``cnt[j] * (i % j); ` `            ``ans = ans % mod; ` `        ``} ` `    ``} ` ` `  `    ``return` `(``int``)(ans); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `a[] = { ``1``, ``2``, ``3` `}; ` `    ``int` `n = a.length; ` ` `  `    ``System.out.println(Sum_Modulo(a, n)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by PrinciRaj1992  `

## Python3

 `# Python3 implementation of the approach ` `mod ``=` `10``*``*``9` `+` `7` ` `  `# Function to return the sum of  ` `# (a[i] % a[j]) for all valid pairs ` `def` `Sum_Modulo(a, n): ` ` `  `    ``Max` `=` `max``(a) ` ` `  `    ``# To store the frequency of each element ` `    ``cnt ``=` `[``0` `for` `i ``in` `range``(``Max` `+` `1``)] ` ` `  `    ``# Store the frequency of each element ` `    ``for` `i ``in` `a: ` `        ``cnt[i] ``+``=` `1` ` `  `    ``# To store the required answer ` `    ``ans ``=` `0` ` `  `    ``# For all valid pairs ` `    ``for` `i ``in` `range``(``1``, ``Max` `+` `1``): ` `        ``for` `j ``in` `range``(``1``, ``Max` `+` `1``): ` ` `  `            ``# Update the count ` `            ``ans ``=` `ans ``+` `cnt[i] ``*` `\ ` `                        ``cnt[j] ``*` `(i ``%` `j) ` `            ``ans ``=` `ans ``%` `mod ` ` `  `    ``return` `ans ` ` `  `# Driver code ` `a ``=` `[``1``, ``2``, ``3``] ` `n ``=` `len``(a) ` ` `  `print``(Sum_Modulo(a, n)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Linq; ` `class` `GFG  ` `{ ` `     `  `static` `int` `mod = (``int``)(1e9 + 7); ` ` `  `// Function to return the sum of (a[i] % a[j]) ` `// for all valid pairs ` `static` `int` `Sum_Modulo(``int` `[]a, ``int` `n) ` `{ ` `    ``int` `max = a.Max(); ` ` `  `    ``// To store the frequency of each element ` `    ``int` `[]cnt = ``new` `int``[max + 1]; ` ` `  `    ``// Store the frequency of each element ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cnt[a[i]]++; ` ` `  `    ``// To store the required answer ` `    ``long` `ans = 0; ` ` `  `    ``// For all valid pairs ` `    ``for` `(``int` `i = 1; i <= max; i++)  ` `    ``{ ` `        ``for` `(``int` `j = 1; j <= max; j++) ` `        ``{ ` ` `  `            ``// Update the count ` `            ``ans = ans + cnt[i] *  ` `                        ``cnt[j] * (i % j); ` `            ``ans = ans % mod; ` `        ``} ` `    ``} ` `    ``return` `(``int``)(ans); ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]a = { 1, 2, 3 }; ` `    ``int` `n = a.Length; ` ` `  `    ``Console.WriteLine(Sum_Modulo(a, n)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```5
```

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