Count of pairs (i, j) in the array such that arr[i] is a factor of arr[j]

Given an array of integers arr, the task is to calculate the number of pairs (i, j) where i < j such that arr[j] % arr[i] = 0.

Examples:

Input: arr[] = {1, 2, 3, 4, 5}
Output: 5

Input: arr[] = {1, 1, 2, 2, 3, 3}
Output: 11

Approach 1:
Iterate over all pairs of the array and keep incrementing the count of pairs that satisfy the required condition.



Below code is the implementation of the above approach:

C++

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// C++ Program to find
// the number of pairs
// (i, j) such that arr[i]
// is a factor of arr[j]
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the
// count of Pairs
int numPairs(int arr[], int n)
{
    int ans = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (arr[j] % arr[i] == 0)
                ans++;
        }
    }
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 1, 2, 2, 3, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << numPairs(arr, n) << endl;
    return 0;
}

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Java














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// Java Program to find the number of pairs 


// (i, j) such that arr[i] is a factor of arr[j] 


import java.util.*; 


import java.lang.*; 


class GFG{ 


  


// Function to return the 


// count of Pairs 


static int numPairs(int arr[], int n) 


{ 


    int ans = 0; 


    for (int i = 0; i < n; i++) 


    { 


        for (int j = i + 1; j < n; j++)  


        { 


            if (arr[j] % arr[i] == 0) 


                ans++; 


        } 


    } 


    return ans; 


} 


  


// Driver code 


public static void main(String[] args) 


{ 


    int arr[] = { 1, 1, 2, 2, 3, 3 }; 


    int n = arr.length; 


  


    System.out.println(numPairs(arr, n)); 


} 


} 


  


// This code is contributed by offbeat 



















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Python3














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# Python3 program to find the number 


# of pairs (i, j) such that arr[i]  


# is a factor of arr[j]  


  


# Function to return the  


# count of Pairs  


def numPairs(arr, n):  


  


    ans = 0


    for i in range(n):  


        for j in range(i + 1, n):  


              


            if arr[j] % arr[i] == 0:  


                ans += 1


  


    return ans  


  


# Driver code  


arr = [ 1, 1, 2, 2, 3, 3 ]  


n = len(arr)  


  


print(numPairs(arr, n))  


  


# This code is contributed by divyamohan123 



















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C#














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// C# Program to find the number of pairs 


// (i, j) such that arr[i] is a factor of arr[j] 


using System; 


  


class GFG{ 


  


// Function to return the 


// count of Pairs 


static int numPairs(int []arr, int n) 


{ 


    int ans = 0; 


    for(int i = 0; i < n; i++) 


    { 


       for(int j = i + 1; j < n; j++)  


       { 


          if (arr[j] % arr[i] == 0) 


              ans++; 


       } 


    } 


    return ans; 


} 


  


// Driver code 


public static void Main() 


{ 


    int []arr = { 1, 1, 2, 2, 3, 3 }; 


    int n = arr.Length; 


  


    Console.Write(numPairs(arr, n)); 


} 


} 


  


// This code is contributed by Code_Mech 



















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Output:


11





Approach 2: Store the indices of all array elements in a map. Traverse the map and for every occurrence of an element:


    

  • Add the occurrences of the same element after the current occurence.
    • 

  • Add the occurrences of all its multiples after the current occurrence using upper_bound
    • 



Below code is the implementation of the above approach:




C++














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// C++ Program to find 


// the number of pairs 


// (i, j) such that arr[i] 


// is a factor of arr[j] 


  


#include <bits/stdc++.h> 


using namespace std; 


  


// Function to return the 


// count of Pairs 


int numPairs(int arr[], int n) 


{ 


    map<int, vector<int> > mp; 


    int mx = 0; 


    for (int i = 0; i < n; i++) { 


  


        // Update the maximum 


        mx = max(mx, arr[i]); 


  


        // Store the indices of 


        // every element 


        mp[arr[i]].push_back(i); 


    } 


  


    int ans = 0; 


    for (auto i : mp) { 


  


        int ctr = 1; 


  


        // Access all indices of i 


        for (int j : i.second) { 


  


            // Add the number of 


            // occurences of i 


            // after j-th index 


            ans += i.second.size() - ctr; 


  


            // Traverse all multiples of i 


            for (int k = 2 * i.first; 


                 k <= mx; 


                 k += i.first) { 


  


                // Find their occurrences 


                // after the j-th index 


                int numGreater = 0; 


                if (mp.find(k) != mp.end()) 


                    numGreater 


                        = int( 


                            mp[k] 


                                .end() 


                            - upper_bound( 


                                  mp[k].begin(), 


                                  mp[k].end(), j)); 


                // Add the count 


                ans += numGreater; 


            } 


            ctr++; 


        } 


    } 


  


    return ans; 


} 


// Driver code 


int main() 


{ 


    int arr[] = { 1, 2, 3, 4, 5 }; 


    int n = sizeof(arr) / sizeof(arr[0]); 


  


    cout << numPairs(arr, n) << endl; 


    return 0; 


} 



















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Output:


5







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