// C++ program to count pairs whose product exist in array #include   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 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 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# 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 

<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 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 

// A hashing based C++ program to count pairs whose product // exists in arr[] #include   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; } 

// 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 

# 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 

// 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 */ 

<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> 

#include    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; } 

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);  } } 

# 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) 

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);  } } 

// 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();