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Given an array, count those pair whose product value is present in array. 
Examples: 
 

Input : arr[] = {6, 2, 4, 12, 5, 3}
Output : 3
All pairs whose product exist in array
(6 , 2) (2, 3) (4, 3)
Input : arr[] = {3, 5, 2, 4, 15, 8}
Output : 2

 

A Simple solution is to generate all pairs of given array and check if product exists in the array. If exists, then increment count. Finally return count.
Below is implementation of above idea 
 

C++




// C++ program to count pairs whose product exist in array
#include<bits/stdc++.h>
using namespace std;
 
// Returns count of pairs whose product exists in arr[]
int countPairs( int arr[] ,int n)
{
    int result = 0;
    for (int i = 0; i < n ; i++)
    {
        for (int j = i+1 ; j < n ; j++)
        {
            int product = arr[i] * arr[j] ;
 
            // find product in an array
            for (int k = 0; k < n; k++)
            {
                // if product found increment counter
                if (arr[k] == product)
                {
                    result++;
                    break;
                }
            }
        }
    }
 
    // return Count of all pair whose product exist in array
    return result;
}
 
//Driver program
int main()
{
    int arr[] = {6 ,2 ,4 ,12 ,5 ,3} ;
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << countPairs(arr, n);
    return 0;
}


Java




// Java program to count pairs
// whose product exist in array
import java.io.*;
 
class GFG
{
     
// Returns count of pairs
// whose product exists in arr[]
static int countPairs(int arr[],
                      int n)
{
    int result = 0;
    for (int i = 0; i < n ; i++)
    {
        for (int j = i + 1 ; j < n ; j++)
        {
            int product = arr[i] * arr[j] ;
 
            // find product
            // in an array
            for (int k = 0; k < n; k++)
            {
                // if product found
                // increment counter
                if (arr[k] == product)
                {
                    result++;
                    break;
                }
            }
        }
    }
 
    // return Count of all pair
    // whose product exist in array
    return result;
}
 
// Driver Code
public static void main (String[] args)
{
int arr[] = {6, 2, 4, 12, 5, 3} ;
int n = arr.length;
System.out.println(countPairs(arr, n));
}
}
 
// This code is contributed by anuj_67.


Python 3




# Python program to count pairs whose
# product exist in array
 
# Returns count of pairs whose
# product exists in arr[]
def countPairs(arr, n):
 
    result = 0;
    for i in range (0, n):
 
        for j in range(i + 1, n):
             
            product = arr[i] * arr[j] ;
 
            # find product in an array
            for k in range (0, n):
         
                # if product found increment counter
                if (arr[k] == product):
                    result = result + 1;
                    break;
 
    # return Count of all pair whose
    # product exist in array
    return result;
 
# Driver program
arr = [6, 2, 4, 12, 5, 3] ;
n = len(arr);
print(countPairs(arr, n));
     
# This code is contributed
# by Shivi_Aggarwal


C#




// C# program to count pairs
// whose product exist in array
using System;
 
class GFG
{
 
// Returns count of pairs
// whose product exists in arr[]
public static int countPairs(int[] arr,
                             int n)
{
    int result = 0;
    for (int i = 0; i < n ; i++)
    {
        for (int j = i + 1 ; j < n ; j++)
        {
            int product = arr[i] * arr[j];
 
            // find product in an array
            for (int k = 0; k < n; k++)
            {
                // if product found
                // increment counter
                if (arr[k] == product)
                {
                    result++;
                    break;
                }
            }
        }
    }
 
    // return Count of all pair
    // whose product exist in array
    return result;
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] arr = new int[] {6, 2, 4, 12, 5, 3};
    int n = arr.Length;
    Console.WriteLine(countPairs(arr, n));
}
}
 
// This code is contributed by Shrikant13


Javascript




<script>
// javascript program to count pairs
// whose product exist in array
 
    // Returns count of pairs
    // whose product exists in arr
    function countPairs(arr, n)
    {
        var result = 0;
        for (i = 0; i < n; i++)
        {
            for (j = i + 1; j < n; j++)
            {
                var product = arr[i] * arr[j];
 
                // find product
                // in an array
                for (k = 0; k < n; k++)
                {
                 
                    // if product found
                    // increment counter
                    if (arr[k] == product)
                    {
                        result++;
                        break;
                    }
                }
            }
        }
 
        // return Count of all pair
        // whose product exist in array
        return result;
    }
 
    // Driver Code
        var arr = [ 6, 2, 4, 12, 5, 3 ];
        var n = arr.length;
        document.write(countPairs(arr, n));
 
// This code is contributed by Rajput-Ji
</script>


PHP




<?php
// PHP program to count pairs
// whose product exist in array
 
// Returns count of pairs whose
// product exists in arr[]
function countPairs($arr, $n)
{
    $result = 0;
    for ($i = 0; $i < $n ; $i++)
    {
        for ($j = $i + 1 ; $j < $n ; $j++)
        {
            $product = $arr[$i] * $arr[$j] ;
 
            // find product in an array
            for ($k = 0; $k < $n; $k++)
            {
                // if product found increment counter
                if ($arr[$k] == $product)
                {
                    $result++;
                    break;
                }
            }
        }
    }
 
    // return Count of all pair whose
    // product exist in array
    return $result;
}
 
// Driver Code
$arr = array(6, 2, 4, 12, 5, 3);
$n = sizeof($arr);
echo countPairs($arr, $n);
 
// This code is contributed
// by Akanksha Rai


Output: 
 

3

Time complexity: O(n3)

Auxiliary Space: O(1)
An Efficient solution is to use ‘hash’ that stores all array element. Generate all possible pair of given array ‘arr’ and check product of each pair is in ‘hash’. If exists, then increment count. Finally return count. 
Below is implementation of above idea 
 

C++




// A hashing based C++ program to count pairs whose product
// exists in arr[]
#include<bits/stdc++.h>
using namespace std;
 
// Returns count of pairs whose product exists in arr[]
int countPairs(int arr[] , int n)
{
    int result = 0;
 
    // Create an empty hash-set that store all array element
    set< int > Hash;
 
    // Insert all array element into set
    for (int i = 0 ; i < n; i++)
        Hash.insert(arr[i]);
 
    // Generate all pairs and check is exist in 'Hash' or not
    for (int i = 0 ; i < n; i++)
    {
        for (int j = i + 1; j<n ; j++)
        {
            int product = arr[i]*arr[j];
 
            // if product exists in set then we increment
            // count by 1
            if (Hash.find(product) != Hash.end())
                result++;
        }
    }
 
    // return count of pairs whose product exist in array
    return result;
}
 
// Driver program
int main()
{
    int arr[] = {6 ,2 ,4 ,12 ,5 ,3};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << countPairs(arr, n) ;
    return 0;
}


Java




// A hashing based Java program to count pairs whose product
// exists in arr[]
import java.util.*;
 
class GFG
{
 
    // Returns count of pairs whose product exists in arr[]
    static int countPairs(int arr[], int n) {
        int result = 0;
 
        // Create an empty hash-set that store all array element
        HashSet< Integer> Hash = new HashSet<>();
 
        // Insert all array element into set
        for (int i = 0; i < n; i++)
        {
            Hash.add(arr[i]);
        }
 
        // Generate all pairs and check is exist in 'Hash' or not
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
                int product = arr[i] * arr[j];
 
                // if product exists in set then we increment
                // count by 1
                if (Hash.contains(product))
                {
                    result++;
                }
            }
        }
 
        // return count of pairs whose product exist in array
        return result;
    }
 
    // Driver program
    public static void main(String[] args)
    {
        int arr[] = {6, 2, 4, 12, 5, 3};
        int n = arr.length;
        System.out.println(countPairs(arr, n));
    }
}
 
// This code has been contributed by 29AjayKumar


Python3




# A hashing based C++ program to count
# pairs whose product exists in arr[]
 
# Returns count of pairs whose product
# exists in arr[]
def countPairs(arr, n):
    result = 0
 
    # Create an empty hash-set that
    # store all array element
    Hash = set()
 
    # Insert all array element into set
    for i in range(n):
        Hash.add(arr[i])
 
    # Generate all pairs and check is
    # exist in 'Hash' or not
    for i in range(n):
        for j in range(i + 1, n):
            product = arr[i] * arr[j]
 
            # if product exists in set then
            # we increment count by 1
            if product in(Hash):
                result += 1
     
    # return count of pairs whose
    # product exist in array
    return result
 
# Driver Code
if __name__ == '__main__':
    arr = [6, 2, 4, 12, 5, 3]
    n = len(arr)
    print(countPairs(arr, n))
     
# This code is contributed by
# Sanjit_Prasad


C#




// A hashing based C# program to count pairs whose product
// exists in arr[]
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Returns count of pairs whose product exists in arr[]
    static int countPairs(int []arr, int n)
    {
        int result = 0;
 
        // Create an empty hash-set that store all array element
        HashSet<int> Hash = new HashSet<int>();
 
        // Insert all array element into set
        for (int i = 0; i < n; i++)
        {
            Hash.Add(arr[i]);
        }
 
        // Generate all pairs and check is exist in 'Hash' or not
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
                int product = arr[i] * arr[j];
 
                // if product exists in set then we increment
                // count by 1
                if (Hash.Contains(product))
                {
                    result++;
                }
            }
        }
 
        // return count of pairs whose product exist in array
        return result;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {6, 2, 4, 12, 5, 3};
        int n = arr.Length;
        Console.WriteLine(countPairs(arr, n));
    }
}
 
/* This code contributed by PrinciRaj1992 */


Javascript




<script>
// A hashing based javascript program to count pairs whose product
// exists in arr
 
    // Returns count of pairs whose product exists in arr
    function countPairs(arr , n) {
        var result = 0;
 
        // Create an empty hash-set that store all array element
        var Hash = new Set();
 
        // Insert all array element into set
        for (i = 0; i < n; i++) {
            Hash.add(arr[i]);
        }
 
        // Generate all pairs and check is exist in 'Hash' or not
        for (i = 0; i < n; i++) {
            for (j = i + 1; j < n; j++) {
                var product = arr[i] * arr[j];
 
                // if product exists in set then we increment
                // count by 1
                if (Hash.has(product)) {
                    result++;
                }
            }
        }
 
        // return count of pairs whose product exist in array
        return result;
    }
 
    // Driver program
     
        var arr = [ 6, 2, 4, 12, 5, 3 ];
        var n = arr.length;
        document.write(countPairs(arr, n));
 
// This code contributed by Rajput-Ji
</script>


Output: 
 

3

Time complexity : O(n2) ‘Under the assumption insert, find operation take O(1) Time ‘

Auxiliary Space: O(n)

Method 3:Using Unordered Map

Approach:

1.Create an empty map to store the elements of the array and their frequencies.
2.Traverse the array and insert each element into the map along with its frequency.
3.Initialize a count variable to 0 to keep track of the number of pairs.
4.Traverse the array again and for each element, check if it has any factor (other than itself) that is present in the map.
5.If both factors are present in the map, then increment the count of pairs.
6.Return the count of pairs.

Implementation:

C++




#include <bits/stdc++.h>
using namespace std;
 
// Function to count pairs whose product value is present in array
int count_Pairs(int arr[], int n) {
    map<int, int> mp;   // Create a map to store the elements of the array and their frequencies
     
    // Initialize the map with the frequencies of the elements in the array
    for (int i = 0; i < n; i++) {
        mp[arr[i]]++;
    }
     
    int count = 0;   // Initialize the count of pairs to zero
     
    // Traverse the array and check if arr[i] has a factor in the map
    for (int i = 0; i < n; i++) {
        for (int j = 1; j*j <= arr[i]; j++) {
            if (arr[i] % j == 0) {
                int factor1 = j;
                int factor2 = arr[i] / j;
                 
                // If both factors are present in the map, then increment the count of pairs
                if (mp.count(factor1) && mp.count(factor2)) {
                    if (factor1 == factor2 && mp[factor1] < 2) {
                        continue;
                    }
                    count++;
                }
            }
        }
    }
     
    // Return the count of pairs
    return count;
}
 
// Driver code
int main() {
    // Example input
    int arr[] = {6, 2, 4, 12, 5, 3};
    int n = sizeof(arr) / sizeof(arr[0]);
     
    // Count pairs whose product value is present in array
    int count = count_Pairs(arr, n);
     
    // Print the count
    cout << count << endl;
     
    return 0;
}


Java




import java.util.HashMap;
import java.util.Map;
 
public class Main {
    // Function to count pairs whose product value is
    // present in the array
    static int countPairs(int[] arr)
    {
        Map<Integer, Integer> frequencyMap
            = new HashMap<>();
 
        // Initialize the map with the frequencies of the
        // elements in the array
        for (int num : arr) {
            frequencyMap.put(
                num, frequencyMap.getOrDefault(num, 0) + 1);
        }
 
        int count
            = 0; // Initialize the count of pairs to zero
 
        // Traverse the array and check if arr[i] has a
        // factor in the map
        for (int num : arr) {
            for (int j = 1; j * j <= num; j++) {
                if (num % j == 0) {
                    int factor1 = j;
                    int factor2 = num / j;
 
                    // If both factors are present in the
                    // map, then increment the count of
                    // pairs
                    if (frequencyMap.containsKey(factor1)
                        && frequencyMap.containsKey(
                            factor2)) {
                        if (factor1 == factor2
                            && frequencyMap.get(factor1)
                                   < 2) {
                            continue;
                        }
                        count++;
                    }
                }
            }
        }
 
        // Return the count of pairs
        return count;
    }
 
    public static void main(String[] args)
    {
        // Example input
        int[] arr = { 6, 2, 4, 12, 5, 3 };
 
        // Count pairs whose product value is present in the
        // array
        int count = countPairs(arr);
 
        // Print the count
        System.out.println(count);
    }
}


Python




# Function to count pairs whose product value is present in the array
def count_pairs(arr):
    # Create a dictionary to store the elements of the array and their frequencies
    mp = {}
 
    # Initialize the dictionary with the frequencies of the elements in the array
    for num in arr:
        if num in mp:
            mp[num] += 1
        else:
            mp[num] = 1
 
    count = 0  # Initialize the count of pairs to zero
 
    # Traverse the array and check if arr[i] has a factor in the dictionary
    for num in arr:
        for j in range(1, int(num ** 0.5) + 1):
            if num % j == 0:
                factor1 = j
                factor2 = num // j
 
                # If both factors are present in the dictionary,
                # then increment the count of pairs
                if factor1 in mp and factor2 in mp:
                    if factor1 == factor2 and mp[factor1] < 2:
                        continue
                    count += 1
 
    return count
 
 
# Driver code
if __name__ == "__main__":
    # Example input
    arr = [6, 2, 4, 12, 5, 3]
 
    # Count pairs whose product value is present in the array
    count = count_pairs(arr)
 
    # Print the count
    print(count)


C#




using System;
using System.Collections.Generic;
 
class GFG {
    // Function to count pairs whose product value is
    // present in array
    static int CountPairs(int[] arr, int n)
    {
        Dictionary<int, int> mp = new Dictionary<
            int, int>(); // Create a dictionary to store the
                         // elements of the array and their
                         // frequencies
 
        // Initialize the dictionary with the frequencies of
        // the elements in the array
        for (int i = 0; i < n; i++) {
            if (!mp.ContainsKey(arr[i]))
                mp[arr[i]] = 1;
            else
                mp[arr[i]]++;
        }
 
        int count
            = 0; // Initialize the count of pairs to zero
 
        // Traverse the array and check if arr[i] has a
        // factor in the dictionary
        for (int i = 0; i < n; i++) {
            for (int j = 1; j * j <= arr[i]; j++) {
                if (arr[i] % j == 0) {
                    int factor1 = j;
                    int factor2 = arr[i] / j;
 
                    // If both factors are present in the
                    // dictionary, then increment the count
                    // of pairs
                    if (mp.ContainsKey(factor1)
                        && mp.ContainsKey(factor2)) {
                        if (factor1 == factor2
                            && mp[factor1] < 2) {
                            continue;
                        }
                        count++;
                    }
                }
            }
        }
 
        // Return the count of pairs
        return count;
    }
 
    // Driver code
    static void Main(string[] args)
    {
        // Example input
        int[] arr = { 6, 2, 4, 12, 5, 3 };
        int n = arr.Length;
 
        // Count pairs whose product value is present in
        // array
        int count = CountPairs(arr, n);
 
        // Print the count
        Console.WriteLine(count);
    }
}


Javascript




// Function to count pairs whose product value is present in the array
function GFG(arr) {
    // Create a map to store the elements of the array
    // and their frequencies
    const mp = new Map();
    // Initialize the map with the frequencies of the elements
    // in the array
    for (let i = 0; i < arr.length; i++) {
        if (!mp.has(arr[i])) {
            mp.set(arr[i], 0);
        }
        mp.set(arr[i], mp.get(arr[i]) + 1);
    }
    let count = 0; // Initialize the count of pairs to zero
    // Traverse the array and check if arr[i] has a factor in the map
    for (let i = 0; i < arr.length; i++) {
        for (let j = 1; j * j <= arr[i]; j++) {
            if (arr[i] % j === 0) {
                const factor1 = j;
                const factor2 = arr[i] / j;
                // If both factors are present in the map
                // then increment the count of pairs
                if (mp.has(factor1) && mp.has(factor2)) {
                    if (factor1 === factor2 && mp.get(factor1) < 2) {
                        continue;
                    }
                    count++;
                }
            }
        }
    }
    // Return the count of pairs
    return count;
}
// Driver code
function main() {
    // Example input
    const arr = [6, 2, 4, 12, 5, 3];
    // Count pairs whose product value is present in the array
    const count = GFG(arr);
    // Print the count
    console.log(count);
}
main();


Output:

3

Time complexity :O(n log n)

Auxiliary Space: O(n)



 



Last Updated : 16 Oct, 2023
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