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Count pairs whose products exist in array

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<iostream>
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

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)

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. Finarlly 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 */

Output:

3

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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