Count pairs whose product contains single distinct prime factor
Given an array arr[] of size N, the task is to count the number of pairs from the given array whose product contains only a single distinct prime factor.
Examples:
Input: arr[] = {1, 2, 3, 4}
Output: 4
Explanation:
Pairs having single distinct prime factor in their product is as follows:
arr[0] * arr[1] = (1 * 2) = 2. Therefore, the single distinct prime factor is 2.
arr[0] * arr[2] = (1 * 3) = 3. Therefore, the single distinct prime factor is 3.
arr[0] * arr[3] = (1 * 4) = 22 Therefore, the single distinct prime factor is 2.
arr[1] * arr[3] = (2 * 4) = 8 23 Therefore, the single distinct prime factor is 2.
Therefore, the required output is 4.Input: arr[] = {2, 4, 6, 8}
Output: 3
Naive Approach: The simplest approach to solve this problem is to traverse the array and generate all possible pairs of the array and for each pair, check if the product of elements contains only a single distinct prime factor or not. If found to be true, then increment the count. Finally, print the count.
Time Complexity: O(N2 * √X), where X is the maximum possible product of a pair in the given array.
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach the idea is to use Hashing. Follow the steps below to solve the problem:
- Initialize a variable, say cntof1 to store count of array elements whose value is 1.
- Create map, say mp to store the count of array elements which contains only a single distinct prime factor.
- Traverse the array and for each array elements, check if the count of distinct prime factors is 1 or not. If found to be true then insert the current element into mp.
- Initialize a variable, say res to store the count of pairs whose product of elements contains only a single distinct prime factor.
- Traverse the map and update the res += cntof1 * (X) + (X *(X- 1)) / 2. Where X stores the count of the array element which contains only a single distinct prime factor i.
- Finally, print the value of res.
Below is the implementation of the above approach
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find a single // distinct prime factor of Nint singlePrimeFactor(int N){ // Stores distinct // prime factors of N unordered_set<int> disPrimeFact; // Calculate prime factor of N for (int i = 2; i * i <= N; ++i) { // Calculate distinct // prime factor while (N % i == 0) { // Insert i into // disPrimeFact disPrimeFact.insert(i); // Update N N /= i; } } // If N is not equal to 1 if (N != 1) { // Insert N into // disPrimeFact disPrimeFact.insert(N); } // If N contains a single // distinct prime factor if (disPrimeFact.size() == 1) { // Return single distinct // prime factor of N return *disPrimeFact.begin(); } // If N contains more than one // distinct prime factor return -1;}// Function to count pairs in the array// whose product contains only// single distinct prime factorint cntsingleFactorPair(int arr[], int N){ // Stores count of 1s // in the array int countOf1 = 0; // mp[i]: Stores count of array elements // whose distinct prime factor is only i unordered_map<int, int> mp; // Traverse the array arr[] for (int i = 0; i < N; i++) { // If current element is 1 if(arr[i] == 1) { countOf1++; continue; } // Store distinct // prime factor of arr[i] int factorValue = singlePrimeFactor(arr[i]); // If arr[i] contains more // than one prime factor if (factorValue == -1) { continue; } // If arr[i] contains // a single prime factor else { mp[factorValue]++; } } // Stores the count of pairs whose // product of elements contains only // a single distinct prime factor int res = 0; // Traverse the map mp[] for (auto it : mp) { // Stores count of array elements // whose prime factor is (it.first) int X = it.second; // Update res res += countOf1 * X + (X * (X - 1) ) / 2; } return res;}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4 }; int N = sizeof(arr) / sizeof(arr[0]); cout << cntsingleFactorPair(arr, N); return 0;} |
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Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to find a single // distinct prime factor of Nstatic int singlePrimeFactor(int N){ // Stores distinct // prime factors of N HashSet<Integer> disPrimeFact = new HashSet<>(); // Calculate prime factor of N for (int i = 2; i * i <= N; ++i) { // Calculate distinct // prime factor while (N % i == 0) { // Insert i into // disPrimeFact disPrimeFact.add(i); // Update N N /= i; } } // If N is not equal to 1 if (N != 1) { // Insert N into // disPrimeFact disPrimeFact.add(N); } // If N contains a single // distinct prime factor if (disPrimeFact.size() == 1) { // Return single distinct // prime factor of N for(int i : disPrimeFact) return i; } // If N contains more than // one distinct prime factor return -1;}// Function to count pairs in // the array whose product // contains only single distinct // prime factorstatic int cntsingleFactorPair(int arr[], int N){ // Stores count of 1s // in the array int countOf1 = 0; // mp[i]: Stores count of array // elements whose distinct prime // factor is only i HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); // Traverse the array arr[] for (int i = 0; i < N; i++) { // If current element is 1 if(arr[i] == 1) { countOf1++; continue; } // Store distinct // prime factor of arr[i] int factorValue = singlePrimeFactor(arr[i]); // If arr[i] contains more // than one prime factor if (factorValue == -1) { continue; } // If arr[i] contains // a single prime factor else { if(mp.containsKey(factorValue)) mp.put(factorValue, mp.get(factorValue) + 1); else mp.put(factorValue, 1); } } // Stores the count of pairs whose // product of elements contains only // a single distinct prime factor int res = 0; // Traverse the map mp[] for (Map.Entry<Integer, Integer> it : mp.entrySet()) { // Stores count of array // elements whose prime // factor is (it.first) int X = it.getValue(); // Update res res += countOf1 * X + (X * (X - 1) ) / 2; } return res;}// Driver Codepublic static void main(String[] args){ int arr[] = {1, 2, 3, 4}; int N = arr.length; System.out.print( cntsingleFactorPair(arr, N));}}// This code is contributed by gauravrajput1 |
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Python3
# Python3 program to implement# the above approach# Function to find a single# distinct prime factor of Ndef singlePrimeFactor(N): # Stores distinct # prime factors of N disPrimeFact = {} # Calculate prime factor of N for i in range(2, N + 1): if i * i > N: break # Calculate distinct # prime factor while (N % i == 0): # Insert i into # disPrimeFact disPrimeFact[i] = 1 # Update N N //= i # If N is not equal to 1 if (N != 1): # Insert N into # disPrimeFact disPrimeFact[N] = 1 # If N contains a single # distinct prime factor if (len(disPrimeFact) == 1): # Return single distinct # prime factor of N return list(disPrimeFact.keys())[0] # If N contains more than one # distinct prime factor return -1# Function to count pairs in the array# whose product contains only# single distinct prime factordef cntsingleFactorPair(arr, N): # Stores count of 1s # in the array countOf1 = 0 # mp[i]: Stores count of array elements # whose distinct prime factor is only i mp = {} # Traverse the array arr[] for i in range(N): # If current element is 1 if (arr[i] == 1): countOf1 += 1 continue # Store distinct # prime factor of arr[i] factorValue = singlePrimeFactor(arr[i]) # If arr[i] contains more # than one prime factor if (factorValue == -1): continue # If arr[i] contains # a single prime factor else: mp[factorValue] = mp.get(factorValue, 0) + 1 # Stores the count of pairs whose # product of elements contains only # a single distinct prime factor res = 0 # Traverse the map mp[] for it in mp: # Stores count of array elements # whose prime factor is (it.first) X = mp[it] # Update res res += countOf1 * X + (X * (X - 1) ) // 2 return res# Driver Codeif __name__ == '__main__': arr = [ 1, 2, 3, 4 ] N = len(arr) print(cntsingleFactorPair(arr, N))# This code is contributed by mohit kumar 29 |
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C#
// C# program to implement// the above approachusing System; using System.Collections.Generic; class GFG{// Function to find a single // distinct prime factor of Nstatic int singlePrimeFactor(int N){ // Stores distinct // prime factors of N HashSet<int> disPrimeFact = new HashSet<int>(); // Calculate prime factor of N for (int i = 2; i * i <= N; ++i) { // Calculate distinct // prime factor while (N % i == 0) { // Insert i into // disPrimeFact disPrimeFact.Add(i); // Update N N /= i; } } // If N is not equal to 1 if(N != 1) { // Insert N into // disPrimeFact disPrimeFact.Add(N); } // If N contains a single // distinct prime factor if (disPrimeFact.Count == 1) { // Return single distinct // prime factor of N foreach(int i in disPrimeFact) return i; } // If N contains more than // one distinct prime factor return -1;}// Function to count pairs in // the array whose product // contains only single distinct // prime factorstatic int cntsingleFactorPair(int []arr, int N){ // Stores count of 1s // in the array int countOf1 = 0; // mp[i]: Stores count of array // elements whose distinct prime // factor is only i Dictionary<int, int> mp = new Dictionary<int, int>(); // Traverse the array arr[] for (int i = 0; i < N; i++) { // If current element is 1 if(arr[i] == 1) { countOf1++; continue; } // Store distinct // prime factor of arr[i] int factorValue = singlePrimeFactor(arr[i]); // If arr[i] contains more // than one prime factor if (factorValue == -1) { continue; } // If arr[i] contains // a single prime factor else { if(mp.ContainsKey(factorValue)) mp[factorValue] = mp[factorValue] + 1; else mp.Add(factorValue, 1); } } // Stores the count of pairs whose // product of elements contains only // a single distinct prime factor int res = 0; // Traverse the map mp[] foreach(KeyValuePair<int, int> ele1 in mp) { // Stores count of array // elements whose prime // factor is (it.first) int X = ele1.Value; // Update res res += countOf1 * X + (X * (X - 1) ) / 2; } return res;}// Driver Codepublic static void Main(){ int []arr = {1, 2, 3, 4}; int N = arr.Length; Console.WriteLine( cntsingleFactorPair(arr, N));}}// This code is contributed by bgangwar59 |
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Output:
4
Time Complexity: O(N√X), where X is the maximum element of the given array.
Auxiliary Space: O(N)
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