Given an array arr[] of N integers, the task is to find the number of pairs (arr[i], arr[j]) such that arr[i]*arr[j] is a perfect square.
Examples:
Input: arr[] = { 1, 2, 4, 8, 5, 6}
Output: 2
Explanation:
The pairs such that the product of an element is perfectly square are (1, 4) and (8, 2).Input: arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }
Output: 4
Explanation:
The pairs such that the product of an element is perfectly square are (1, 4), (1, 9), (2, 8) and (4, 9).
Naive Approach:
Run two loops from 1 to n and count all the pairs (i, j) where arr[i]*arr[j] is a perfect square. The time complexity of this approach will be O(N2).
C++
// C++ code for above approach.#include <bits/stdc++.h>using namespace std;// Function to check if number// is perfect square or notbool checkperfectsquare(int n){ // If ceil and floor are equal // the number is a perfect // square if (ceil((double)sqrt(n)) == floor((double)sqrt(n))) { return true; } else { return false; }}// Function that return total count// of pairs with perfect square productint countPairs(int arr[], int n){ int count = 0; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // Checking the pair with // arr[i]*arr[j] as perfect square if (checkperfectsquare(arr[i] * arr[j])) { count++; } } } // Returning the count return count;}// Driver codeint main(){ int arr[] = { 1, 2, 4, 8, 5, 6 }; // Size of arr[] int n = sizeof(arr) / sizeof(int); cout << countPairs(arr, n) << endl; return 0;}// This code is contributed by Utkarsh Kumar. |
Java
// Java code for above approach.import java.io.*;class GFG { // Function to check if number // is perfect square or not static boolean checkperfectsquare(int n) { // If ceil and floor are equal // the number is a perfect // square if (Math.ceil((double)Math.sqrt(n)) == Math.floor((double)Math.sqrt(n))) { return true; } else { return false; } } // Function that return total count // of pairs with perfect square product static int countPairs(int arr[], int n) { int count = 0; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // Checking the pair with // arr[i]*arr[j] as perfect square if (checkperfectsquare(arr[i] * arr[j])) { count++; } } } // Returning the count return count; } // Driver code public static void main(String[] args) { int arr[] = { 1, 2, 4, 8, 5, 6 }; // Size of arr[] int n = arr.length; System.out.println(countPairs(arr, n)); }}// This code is contributed by Pushpesh Raj. |
Python3
import math# Function to check if number# is perfect square or notdef checkperfectsquare(n): # If ceil and floor are equal # the number is a perfect # square if math.ceil(math.sqrt(n)) == math.floor(math.sqrt(n)): return True else: return False# Function that return total count# of pairs with perfect square productdef countPairs(arr, n): count = 0 for i in range(n): for j in range(i + 1, n): # Checking the pair with # arr[i]*arr[j] as perfect square if checkperfectsquare(arr[i] * arr[j]): count += 1 # Returning the count return count# Driver codeif __name__ == '__main__': arr = [1, 2, 4, 8, 5, 6] # Size of arr[] n = len(arr) print(countPairs(arr, n)) |
Javascript
// JavaScript code for above approach.// Function to check if number// is perfect square or notfunction checkperfectsquare(n) { // If ceil and floor are equal // the number is a perfect // square if (Math.ceil(Math.sqrt(n)) == Math.floor(Math.sqrt(n))) { return true; } else { return false; }}// Function that return total count// of pairs with perfect square productfunction countPairs(arr, n) { let count = 0; for (let i = 0; i < n; i++) { for (let j = i + 1; j < n; j++) { // Checking the pair with // arr[i]*arr[j] as perfect square if (checkperfectsquare(arr[i] * arr[j])) { count++; } } } // Returning the count return count;}// Driver codelet arr = [1, 2, 4, 8, 5, 6];// Size of arr[]let n = arr.length;console.log(countPairs(arr, n));// This code is contributed prasad264 |
C#
using System;public class MainClass { public static bool CheckPerfectSquare(int n) { // If ceil and floor are equal // the number is a perfect // square if (Math.Ceiling(Math.Sqrt(n)) == Math.Floor(Math.Sqrt(n))) { return true; } else { return false; } } public static int CountPairs(int[] arr, int n) { int count = 0; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // Checking the pair with // arr[i]*arr[j] as perfect square if (CheckPerfectSquare(arr[i] * arr[j])) { count += 1; } } } // Returning the count return count; } public static void Main(string[] args) { int[] arr = { 1, 2, 4, 8, 5, 6 }; // Size of arr[] int n = arr.Length; Console.WriteLine(CountPairs(arr, n)); }}// This code is contributed by shivhack999 |
Output
2
Time Complexity : O(n^2) // since two nested loops are used the time taken by the algorithm to complete all operation is quadratic.
Space Complexity : O(1) // since no extra array is used so the space taken by the algorithm is constant
Efficient Approach:
Each integer in arr[] can be represented in the following form:
arr[i] = k*x ..............(1) where k is not divisible by any perfect square other than 1, and x = perfect square,
Steps:
- Represent every element in the form of equation(1).
- Then, for every pair (arr[i], arr[j]) in arr[] can be represented as:
arr[i] = ki*x; arr[j] = kj*y; where x and y are perfect square
- For pairs (arr[i], arr[j]), the product of arr[i] and arr[j] can be perfectly square if and only if ki = kj
- Use Sieve of Eratosthenes to pre-compute the value of k for every element in array arr[].
- Store the frequency of k for every element in arr[] in map.
- Therefore, the total number of pairs is given by the number of pairs formed by elements with a frequency greater than 1.
- The total number of pairs formed by n elements is given by:
Number of Pairs = (f*(f-1))/2 where f is the frequency of an element.
Below is the implementation of the above approach:
C++
// C++ program to calculate the number of// pairs with product is perfect square#include <bits/stdc++.h>using namespace std;// Prime[] array to calculate Prime Numberint prime[100001] = { 0 };// Array k[] to store the value of k for// each element in arr[]int k[100001] = { 0 };// For value of k, Sieve function is// implementedvoid Sieve(){ // Initialize k[i] to i for (int i = 1; i < 100001; i++) k[i] = i; // Prime Sieve for (int i = 2; i < 100001; i++) { // If i is prime then remove all // factors of prime from it if (prime[i] == 0) for (int j = i; j < 100001; j += i) { // Update that j is not // prime prime[j] = 1; // Remove all square divisors // i.e. if k[j] is divisible // by i*i then divide it by i*i while (k[j] % (i * i) == 0) k[j] /= (i * i); } }}// Function that return total count// of pairs with perfect square productint countPairs(int arr[], int n){ // Map used to store the frequency of k unordered_map<int, int> freq; // Store the frequency of k for (int i = 0; i < n; i++) { freq[k[arr[i]]]++; } int sum = 0; // The total number of pairs is the // summation of (fi * (fi - 1))/2 for (auto i : freq) { sum += ((i.second - 1) * i.second) / 2; } return sum;}// Driver codeint main(){ int arr[] = { 1, 2, 4, 8, 5, 6 }; // Size of arr[] int n = sizeof(arr) / sizeof(int); // To pre-compute the value of k Sieve(); // Function that return total count // of pairs with perfect square product cout << countPairs(arr, n) << endl; return 0;} |
Java
// Java program to calculate the number of// pairs with product is perfect squareimport java.util.*;class GFG{ // Prime[] array to calculate Prime Numberstatic int []prime = new int[100001]; // Array k[] to store the value of k for// each element in arr[]static int []k = new int[100001]; // For value of k, Sieve function is// implementedstatic void Sieve(){ // Initialize k[i] to i for (int i = 1; i < 100001; i++) k[i] = i; // Prime Sieve for (int i = 2; i < 100001; i++) { // If i is prime then remove all // factors of prime from it if (prime[i] == 0) for (int j = i; j < 100001; j += i) { // Update that j is not // prime prime[j] = 1; // Remove all square divisors // i.e. if k[j] is divisible // by i*i then divide it by i*i while (k[j] % (i * i) == 0) k[j] /= (i * i); } }} // Function that return total count// of pairs with perfect square productstatic int countPairs(int arr[], int n){ // Map used to store the frequency of k HashMap<Integer,Integer> freq = new HashMap<Integer,Integer>(); // Store the frequency of k for (int i = 0; i < n; i++) { if(freq.containsKey(k[arr[i]])) { freq.put(k[arr[i]], freq.get(k[arr[i]])+1); } else freq.put(k[arr[i]], 1); } int sum = 0; // The total number of pairs is the // summation of (fi * (fi - 1))/2 for (Map.Entry<Integer,Integer> i : freq.entrySet()){ sum += ((i.getValue() - 1) * i.getValue()) / 2; } return sum;} // Driver codepublic static void main(String[] args){ int arr[] = { 1, 2, 4, 8, 5, 6 }; // Size of arr[] int n = arr.length; // To pre-compute the value of k Sieve(); // Function that return total count // of pairs with perfect square product System.out.print(countPairs(arr, n) +"\n"); }}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to calculate the number # of pairs with product is perfect square# prime[] array to calculate Prime Numberprime = [0] * 100001# Array to store the value of k # for each element in arr[]k = [0] * 100001# For value of k, Sieve implementeddef Sieve(): # Initialize k[i] to i for i in range(1, 100001): k[i] = i # Prime sieve for i in range(2, 100001): # If i is prime then remove all # factors of prime from it if (prime[i] == 0): for j in range(i, 100001, i): # Update that j is not prime prime[j] = 1 # Remove all square divisors # i.e if k[j] is divisible by # i*i then divide it by i*i while (k[j] % (i * i) == 0): k[j] /= (i * i)# Function that return total count of# pairs with perfect square productdef countPairs (arr, n): # Store the frequency of k freq = dict() for i in range(n): if k[arr[i]] in freq.keys(): freq[k[arr[i]]] += 1 else: freq[k[arr[i]]] = 1 Sum = 0 # The total number of pairs is the # summation of (fi * (fi - 1))/2 for i in freq: Sum += (freq[i] * (freq[i] - 1)) / 2 return Sum# Driver code arr = [ 1, 2, 4, 8, 5, 6 ]# Length of arrn = len(arr) # To pre-compute the value of kSieve()# Function that return total count # of pairs with perfect square product print(int(countPairs(arr, n)))# This code is contributed by himanshu77 |
C#
// C# program to calculate the number of// pairs with product is perfect squareusing System;using System.Collections.Generic;class GFG{ // Prime[] array to calculate Prime Numberstatic int []prime = new int[100001]; // Array k[] to store the value of k for// each element in []arrstatic int []k = new int[100001]; // For value of k, Sieve function is// implementedstatic void Sieve(){ // Initialize k[i] to i for (int i = 1; i < 100001; i++) k[i] = i; // Prime Sieve for (int i = 2; i < 100001; i++) { // If i is prime then remove all // factors of prime from it if (prime[i] == 0) for (int j = i; j < 100001; j += i) { // Update that j is not // prime prime[j] = 1; // Remove all square divisors // i.e. if k[j] is divisible // by i*i then divide it by i*i while (k[j] % (i * i) == 0) k[j] /= (i * i); } }} // Function that return total count// of pairs with perfect square productstatic int countPairs(int []arr, int n){ // Map used to store the frequency of k Dictionary<int,int> freq = new Dictionary<int,int>(); // Store the frequency of k for (int i = 0; i < n; i++) { if(freq.ContainsKey(k[arr[i]])) { freq[k[arr[i]]] = freq[k[arr[i]]]+1; } else freq.Add(k[arr[i]], 1); } int sum = 0; // The total number of pairs is the // summation of (fi * (fi - 1))/2 foreach (KeyValuePair<int,int> i in freq){ sum += ((i.Value - 1) * i.Value) / 2; } return sum;} // Driver codepublic static void Main(String[] args){ int []arr = { 1, 2, 4, 8, 5, 6 }; // Size of []arr int n = arr.Length; // To pre-compute the value of k Sieve(); // Function that return total count // of pairs with perfect square product Console.Write(countPairs(arr, n) +"\n"); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program to calculate the number of// pairs with product is perfect square// Prime[] array to calculate Prime Numberlet prime = new Array(100001).fill(0);// Array k[] to store the value of k for// each element in arr[]let k = new Array(100001).fill(0);// For value of k, Sieve function is// implementedfunction Sieve(){ // Initialize k[i] to i for (let i = 1; i < 100001; i++) k[i] = i; // Prime Sieve for (let i = 2; i < 100001; i++) { // If i is prime then remove all // factors of prime from it if (prime[i] == 0) for (let j = i; j < 100001; j += i) { // Update that j is not // prime prime[j] = 1; // Remove all square divisors // i.e. if k[j] is divisible // by i*i then divide it by i*i while (k[j] % (i * i) == 0) k[j] /= (i * i); } }}// Function that return total count// of pairs with perfect square productfunction countPairs(arr, n){ // Map used to store the frequency of k let freq = new Map(); // Store the frequency of k for (let i = 0; i < n; i++) { if(freq.has(k[arr[i]])) { freq.set(k[arr[i]], freq.get(k[arr[i]])+1); } else freq.set(k[arr[i]], 1); } let sum = 0; // The total number of pairs is the // summation of (fi * (fi - 1))/2 for (let i of freq) { sum += ((i[1] - 1) * i[1]) / 2; } return sum;}// Driver codelet arr = [ 1, 2, 4, 8, 5, 6 ];// Size of arr[]let n = arr.length;// To pre-compute the value of kSieve();// Function that return total count// of pairs with perfect square productdocument.write(countPairs(arr, n) + "<br>");// This code is contributed by _saurabh_jaiswal</script> |
Output:
2
Time Complexity: O(N*log(log N)), since sieve of Eratosthenes takes N *log(log N) time to execute
Auxiliary Space: O(N + 105)
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