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Find sum of a[i]%a[j] for all valid pairs
  • Difficulty Level : Easy
  • Last Updated : 30 Jul, 2019

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

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 <bits/stdc++.h>
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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