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++ 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 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 // 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
// 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;
} |
// 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> |
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:
#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;
} |
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(); |
Output:
3
Time complexity :O(n log n)
Auxiliary Space: O(n)