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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++ 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;
}




// 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




# 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




// 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




<script>
 
    // Javascript 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
    function numPairs(arr, n)
    {
        let ans = 0;
        for (let i = 0; i < n; i++) {
            for (let j = i + 1; j < n; j++) {
                if (arr[j] % arr[i] == 0)
                    ans++;
            }
        }
        return ans;
    }
     
    let arr = [ 1, 1, 2, 2, 3, 3 ];
    let n = arr.length;
   
    document.write(numPairs(arr, n));
 
</script>

Output: 
11

 

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

Below code is the implementation of the above approach:
 




// 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
            // occurrences 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;
}




// Java Program to find
// the number of pairs
// (i, j) such that arr[i]
// is a factor of arr[j]
import java.util.*;
 
class GFG {
 
  static int upper_bound(ArrayList<Integer> arr, int ele)
  {
    for (int i = 0; i < arr.size(); i++)
      if (arr.get(i) > ele)
        return i;
    return arr.size();
  }
 
  // Function to return the
  // count of Pairs
  static int numPairs(int[] arr, int n)
  {
    HashMap<Integer, ArrayList<Integer> > mp
      = new HashMap<Integer, ArrayList<Integer> >();
    int mx = 0;
    for (int i = 0; i < n; i++) {
 
      // Update the maximum
      mx = Math.max(mx, arr[i]);
 
      // Store the indices of
      // every element
      ArrayList<Integer> l1;
      if (!mp.containsKey(arr[i]))
        l1 = new ArrayList<Integer>();
      else
        l1 = mp.get(arr[i]);
      l1.add(i);
      mp.put(arr[i], l1);
    }
 
    int ans = 0;
    for(var i : mp.entrySet())
    {
 
      int ctr = 1;
 
      // Access all indices of i
      for (int j : i.getValue())
      {
 
        // Add the number of
        // occurrences of i
        // after j-th index
        ans += (i.getValue()).size() - ctr;
 
        // Traverse all multiples of i
        for (int k = 2 * i.getKey(); k <= mx;
             k += i.getKey()) {
 
          // Find their occurrences
          // after the j-th index
          int numGreater = 0;
          if (mp.containsKey(k))
            numGreater = mp.get(k).size()
 
            - upper_bound(mp.get(k), j);
          // Add the count
          ans += numGreater;
        }
        ctr++;
      }
    }
 
    return ans;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int[] arr = { 1, 2, 3, 4, 5 };
    int n = arr.length;
 
    System.out.println(numPairs(arr, n));
  }
}
 
// This code is contributed by phasing17.




# Python3 Program to find
# the number of pairs
# (i, j) such that arr[i]
# is a factor of arr[j]
def upper_bound(arr, ele):
 
    for i in range(len(arr)):
        if (arr[i] > ele):
            return i;
    return len(arr)
 
# Function to return the
# count of Pairs
def numPairs(arr, n):
 
    mp = dict();
    mx = 0;
    for i in range(n):
 
        # Update the maximum
        mx = max(mx, arr[i]);
 
        # Store the indices of
        # every element
        if arr[i] not in mp:
            mp[arr[i]] = []
        mp[arr[i]].append(i);
 
    ans = 0;
    for first, second in mp.items():
         
        ctr = 1;
 
        # Access all indices of i
        for j in second:
 
            # Add the number of
            # occurrences of i
            # after j-th index
            ans += len(second) - ctr;
 
            # Traverse all multiples of i
            for k in range(2 * first, mx + 1, first):
 
                # Find their occurrences
                # after the j-th index
                numGreater = 0;
                if k in mp:
                    numGreater  = (len(mp[k]) - upper_bound(mp[k], j));
                # Add the count
                ans += numGreater;
             
            ctr += 1;
 
    return ans;
 
# Driver code
arr = [ 1, 2, 3, 4, 5 ];
n = len(arr);
 
print(numPairs(arr, n))
     
# This code is contributed by phasing17




// C# Program to find
// the number of pairs
// (i, j) such that arr[i]
// is a factor of arr[j]
using System;
using System.Collections.Generic;
 
class GFG {
 
    static int upper_bound(List<int> arr, int ele)
    {
        for (var i = 0; i < arr.Count; i++)
            if (arr[i] > ele)
                return i;
        return arr.Count;
    }
   
    // Function to return the
    // count of Pairs
    static int numPairs(int[] arr, int n)
    {
        Dictionary<int, List<int> > mp
            = new Dictionary<int, List<int> >();
        int mx = 0;
        for (int i = 0; i < n; i++) {
 
            // Update the maximum
            mx = Math.Max(mx, arr[i]);
 
            // Store the indices of
            // every element
            if (!mp.ContainsKey(arr[i]))
                mp[arr[i]] = new List<int>();
            mp[arr[i]].Add(i);
        }
 
        int ans = 0;
        foreach(var i in mp)
        {
 
            int ctr = 1;
 
            // Access all indices of i
            foreach(int j in i.Value)
            {
 
                // Add the number of
                // occurrences of i
                // after j-th index
                ans += (i.Value).Count - ctr;
 
                // Traverse all multiples of i
                for (int k = 2 * i.Key; k <= mx;
                     k += i.Key) {
 
                    // Find their occurrences
                    // after the j-th index
                    int numGreater = 0;
                    if (mp.ContainsKey(k))
                        numGreater
                            = mp[k].Count
                              - upper_bound(mp[k], j);
                    // Add the count
                    ans += numGreater;
                }
                ctr++;
            }
        }
 
        return ans;
    }
   
    // Driver code
    public static void Main(string[] args)
    {
        int[] arr = { 1, 2, 3, 4, 5 };
        int n = arr.Length;
 
        Console.WriteLine(numPairs(arr, n));
    }
}
 
// This code is contributed by phasing17.




// JS Program to find
// the number of pairs
// (i, j) such that arr[i]
// is a factor of arr[j]
function upper_bound(arr, ele)
{
    for (var i = 0; i < arr.length; i++)
        if (arr[i] > ele)
            return i;
    return arr.length;
}
 
// Function to return the
// count of Pairs
function numPairs(arr, n)
{
    let mp = {};
    let mx = 0;
    for (let i = 0; i < n; i++) {
 
        // Update the maximum
        mx = Math.max(mx, arr[i]);
 
        // Store the indices of
        // every element
        if (!mp.hasOwnProperty(arr[i]))
            mp[arr[i]] = []
        mp[arr[i]].push(i);
    }
 
    let ans = 0;
    for (let [first, second] of Object.entries(mp)) {
         
        first = parseInt(first)
         
        let ctr = 1;
 
        // Access all indices of i
        for (let j of second) {
 
            // Add the number of
            // occurrences of i
            // after j-th index
            ans += second.length - ctr;
 
            // Traverse all multiples of i
            for (let k = 2 * first;
                 k <= mx;
                 k += first) {
 
                // Find their occurrences
                // after the j-th index
                let numGreater = 0;
                if (mp.hasOwnProperty(k))
                    numGreater
                        = parseInt(mp[k].length - upper_bound(mp[k], j));
                // Add the count
                ans += numGreater;
            }
            ctr++;
        }
    }
 
    return ans;
}
 
// Driver code
let arr = [ 1, 2, 3, 4, 5 ];
let n = arr.length;
 
console.log(numPairs(arr, n))
     
// This code is contributed by phasing17

Output: 
5

 


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