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Count all distinct pairs with product equal to K

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  • Last Updated : 20 Aug, 2022
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Given an integer array arr[] of size N and a positive integer K, the task is to count all the distinct pairs in the array with product equal to K.
Examples: 
 

Input: arr[] = {1, 5, 3, 4, 2}, K = 3 
Output:
Explanation: 
There is only one pair (1, 3) with product = K = 3.
Input: arr[] = {1, 2, 16, 4, 4}, K = 16 
Output:
Explanation: 
There are two pairs (1, 16) and (4, 4) with product = K = 16. 
 

 

Efficient Approach: The idea is to use hashing.
 

  1. Initially, insert all the elements from the array into the hashmap. While inserting, ignore if a particular element is already present in the hashmap.
  2. Now, we have all the unique elements in the hash. So, for every element in the array arr[i], we check if arr[i] / K is present in the hashmap or not.
  3. If the value is present, then we increment the count and remove that particular element from the hash( To get the unique pairs).

Below is the implementation of the above approach:
 

C++




// C++ program to count the number of pairs
// whose product is equal to K
 
#include <algorithm>
#include <iostream>
using namespace std;
#define MAX 100000
 
// Function to count the number of pairs
// whose product is equal to K
int countPairsWithProductK(int arr[], int n, int k)
{
    // Initialize the count
    int count = 0;
 
    // Initialize empty hashmap.
    bool hashmap[MAX] = { false };
 
    // Insert array elements to hashmap
    for (int i = 0; i < n; i++)
        hashmap[arr[i]] = true;
 
    for (int i = 0; i < n; i++) {
        int x = arr[i];
 
        double index = 1.0 * k / arr[i];
 
        // Checking if the index is a whole number
        // and present in the hashmap
        if (index >= 0
            && ((index - (int)(index)) == 0)
            && hashmap[k / x])
 
            count++;
        hashmap[x] = false;
    }
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 5, 3, 4, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 3;
 
    cout << countPairsWithProductK(arr, N, K);
    return 0;
}

Java




// Java program to count the number of pairs
// whose product is equal to K
     
class GFG
{
    static int MAX = 100000;
     
    // Function to count the number of pairs
    // whose product is equal to K
    static int countPairsWithProductK(int arr[], int n, int k)
    {
        // Initialize the count
        int count = 0;
        int i;
 
        // Initialize empty hashmap.
        boolean hashmap[] = new boolean[MAX];
         
        // Insert array elements to hashmap
        for (i = 0; i < n; i++)
            hashmap[arr[i]] = true;
         
        for (i = 0; i < n; i++) {
            int x = arr[i];
         
            double index = 1.0 * k / arr[i];
         
            // Checking if the index is a whole number
            // and present in the hashmap
            if (index >= 0
                && ((index - (int)(index)) == 0)
                && hashmap[k / x])
         
                count++;
            hashmap[x] = false;
        }
        return count;
    }
     
    // Driver code
    public static void main(String []args)
    {
        int arr[] = { 1, 5, 3, 4, 2 };
        int N = arr.length;
        int K = 3;
         
        System.out.print(countPairsWithProductK(arr, N, K));
         
    }
}

Python3




# Python3 program to count the number of pairs
# whose product is equal to K
MAX = 100000;
 
# Function to count the number of pairs
# whose product is equal to K
def countPairsWithProductK(arr, n, k) :
 
    # Initialize the count
    count = 0;
 
    # Initialize empty hashmap.
    hashmap = [False]*MAX ;
 
    # Insert array elements to hashmap
    for i in range(n) :
        hashmap[arr[i]] = True;
 
    for i in range(n) :
        x = arr[i];
 
        index = 1.0 * k / arr[i];
 
        # Checking if the index is a whole number
        # and present in the hashmap
        if (index >= 0
            and ((index - int(index)) == 0)
            and hashmap[k // x]) :
 
                count += 1;
         
        hashmap[x] = False;
     
    return count;
 
# Driver code
if __name__ == "__main__" :
    arr = [ 1, 5, 3, 4, 2 ];
    N = len(arr);
    K = 3;
 
    print(countPairsWithProductK(arr, N, K));
 
# This code is contributed by AnkitRai01

C#




// C# program to count the number of pairs
// whose product is equal to K    
using System;
 
class GFG
{
    static int MAX = 100000;
      
    // Function to count the number of pairs
    // whose product is equal to K
    static int countPairsWithProductK(int []arr, int n, int k)
    {
        // Initialize the count
        int count = 0;
        int i;
  
        // Initialize empty hashmap.
        bool []hashmap = new bool[MAX];
          
        // Insert array elements to hashmap
        for (i = 0; i < n; i++)
            hashmap[arr[i]] = true;
          
        for (i = 0; i < n; i++) {
            int x = arr[i];
          
            double index = 1.0 * k / arr[i];
          
            // Checking if the index is a whole number
            // and present in the hashmap
            if (index >= 0
                && ((index - (int)(index)) == 0)
                && hashmap[k / x])
          
                count++;
            hashmap[x] = false;
        }
        return count;
    }
      
    // Driver code
    public static void Main(String []args)
    {
        int []arr = { 1, 5, 3, 4, 2 };
        int N = arr.Length;
        int K = 3;
          
        Console.Write(countPairsWithProductK(arr, N, K));        
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript program to count the number of pairs
// whose product is equal to K
 
    let MAX = 100000;
     
    // Function to count the number of pairs
    // whose product is equal to K
    function countPairsWithProductK(arr,n,k)
    {
        // Initialize the count
        let count = 0;
        let i;
  
        // Initialize empty hashmap.
        let hashmap = new Array(MAX);
          
        // Insert array elements to hashmap
        for (i = 0; i < n; i++)
            hashmap[arr[i]] = true;
          
        for (i = 0; i < n; i++) {
            let x = arr[i];
          
            let index = 1.0 * k / arr[i];
          
            // Checking if the index is a whole number
            // and present in the hashmap
            if (index >= 0
                && ((index - Math.floor(index)) == 0)
                && hashmap[k / x])
          
                count++;
            hashmap[x] = false;
        }
        return count;
    }
     
    // Driver code
    let arr=[1, 5, 3, 4, 2];
    let N = arr.length;
    let K = 3;
    document.write(countPairsWithProductK(arr, N, K));
 
// This code is contributed by rag2127
 
</script>

Output: 

1

 

Time Complexity: O(N)

Auxiliary Space: O(MAX), since MAX space has been taken.
 


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