Count all distinct pairs with product equal to K
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; } |
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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 |
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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 |
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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 |
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Output:
1
Time Complexity: O(N2)
Efficient Approach: The idea is to use hashing.
- Initially, insert all the elements from the array into the hashmap. While inserting, ignore if a particular element is already present in the hashmap.
- 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.
- 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; } |
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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)); } } |
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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 |
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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 |
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Output:
1
Time Complexity: O(N * log(N))
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