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.
leave a comment
0 Comments