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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

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:

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.

Count pairs whose products exist in array

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python 3

C#

PHP

C++

Java

Python3

C#

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