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Count all distinct pairs with product equal to K

  • Last Updated : 02 Jun, 2021

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

 

Naive Approach: The naive approach for this problem is to consider all pairs one by one and check the product between every pair. If the product is equal to K, then increment the count and finally print it. However, the limitation is that this approach doesn’t work if there are duplicates in the array because we only need to consider distinct pairs. 
Below is the implementation of the above approach:
 

CPP




// C++ program to count the number of pairs
// whose product is equal to K
 
#include <iostream>
using namespace std;
 
// Function to count the number of pairs
// whose product is equal to K
int countPairsWithProdK(int arr[], int n, int k)
{
    int count = 0;
 
    // Pick all elements one by one
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++)
 
            // Check if the product of this pair
            // is equal to K
            if (arr[i] * arr[j] == k)
                count++;
    }
 
    return count;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 5, 3, 4, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 3;
 
    cout << countPairsWithProdK(arr, N, K);
    return 0;
}
Java



// Java program to count the number of pairs
// whose product is equal to K
class GFG {
 
     // Function to count the number of pairs
    // whose product is equal to K
    static int countPairsWithProdK(int arr[], int n, int k)
    {
        int count = 0;
     
        // Pick all elements one by one
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++)
     
                // Check if the product of this pair
                // is equal to K
                if (arr[i] * arr[j] == k)
                    count++;
        }
     
        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.println(countPairsWithProdK(arr, N, K));
    }
}
 
// This code is contributed by AnkitRai01
Python3



# Python3 program to count the number of pairs
# whose product is equal to K
 
# Function to count the number of pairs
# whose product is equal to K
def countPairsWithProdK(arr, n, k):
    count = 0
 
    # Pick all elements one by one
    for i in range(n):
        for j in range(i + 1, n):
 
            # Check if the product of this pair
            # is equal to K
            if (arr[i] * arr[j] == k):
                count += 1
 
    return count
 
# Driver code
arr = [1, 5, 3, 4, 2]
N = len(arr)
K = 3
 
print(countPairsWithProdK(arr, N, K))
 
# This code is contributed by mohit kumar 29
C#



// C# program to count the number of pairs
// whose product is equal to K
using System;
 
class GFG {
 
     // Function to count the number of pairs
    // whose product is equal to K
    static int countPairsWithProdK(int []arr, int n, int k)
    {
        int count = 0;
     
        // Pick all elements one by one
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++)
     
                // Check if the product of this pair
                // is equal to K
                if (arr[i] * arr[j] == k)
                    count++;
        }
     
        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.WriteLine(countPairsWithProdK(arr, N, K));
    }
}
 
// This code is contributed by AnkitRai01
Javascript



<script>
 
// Javascript program to count the number of pairs
// whose product is equal to K
 
// Function to count the number of pairs
// whose product is equal to K
function countPairsWithProdK(arr, n, k)
{
    var count = 0;
 
    // Pick all elements one by one
    for (var i = 0; i < n; i++) {
        for (var j = i + 1; j < n; j++)
 
            // Check if the product of this pair
            // is equal to K
            if (arr[i] * arr[j] == k)
                count++;
    }
 
    return count;
}
 
// Driver code
var arr = [ 1, 5, 3, 4, 2 ];
var N = arr.length;
var K = 3;
document.write(countPairsWithProdK(arr, N, K));
 
// This code is contributed by rutvik_56.
</script>
Output: 


1
 
Time Complexity: O(N2)
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 * log(N))
 
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.



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