Given a positive integers ‘q’ and ‘n’. For each query ‘q’ find whether a number ‘n’ have exactly four distinct divisors or not. If the number have exactly four divisors then print ‘Yes’ else ‘No’.1 <= q, n <= 106
Input: 2 10 12 Output: Yes No Explanation: For 1st query, n = 10 has exactly four divisor i.e., 1, 2, 5, 10. For 2nd query, n = 12 has exactly six divisor i.e., 1, 2, 3, 4, 6, 12.
Simple approach is to count factors by generating all divisors of a number by using this approach, after that check whether the count of all factors are equal to ‘4’ or not. Time complexity of this approach is O(sqrt(n)).
Better approach is to use Number theory. For a number to be have four factors, it must satisfy the following conditions:-
- If number is a product of exactly two prime numbers(say p, q). Thus we can assure that it will have four factors i.e, 1, p, q, n.
- If a number is cube of a prime number (or cube root of the number is prime). For example, let’s say n = 8, cube root = 2 that means ‘8’ can be written as 2*2*2 hence four factors are:- 1, 2, 4 and 8.
We can use sieve of Eratosthenes such that we will pre-calculate all the prime factor from 1 to 106. Now we will mark all numbers which are the product of two prime number by using two ‘for loops’ i.e, mark[p * q] =true. Meanwhile we will also mark all numbers(cube root) by taking cube of number i.e, mark[p * p * p] = true.
After that we can easily answer each query in O(1) time.
Below is pseudo code, have a look for better understanding
C++
// C++ program to check whether number has// exactly four distinct factors or not#include <bits/stdc++.h>using namespace std;// Initialize global variable according// to given condition so that it can be// accessible to all functionconst int N = 1e6;bool fourDiv[N + 1];// Function to calculate all number having// four distinct distinct factorsvoid fourDistinctFactors(){ // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // not a prime, else true. bool primeAll[N + 1]; memset(primeAll, true, sizeof(primeAll)); for (int p = 2; p * p <= N; p++) { // If prime[p] is not changed, then it // is a prime if (primeAll[p] == true) { // Update all multiples of p for (int i = p * 2; i <= N; i += p) primeAll[i] = false; } } // Initialize prime[] array which will // contains all the primes from 1-N vector<int> prime; for (int p = 2; p <= N; p++) if (primeAll[p]) prime.push_back(p); // Set the marking of all primes to false memset(fourDiv, false, sizeof(fourDiv)); // Iterate over all the prime numbers for (int i = 0; i < prime.size(); ++i) { int p = prime[i]; // Mark cube root of prime numbers if (1LL * p * p * p <= N) fourDiv[p * p * p] = true; for (int j = i + 1; j < prime.size(); ++j) { int q = prime[j]; if (1LL * p * q > N) break; // Mark product of prime numbers fourDiv[p * q] = true; } }}// Driver programint main(){ fourDistinctFactors(); int num = 10; if (fourDiv[num]) cout << "Yes\n"; else cout << "No\n"; num = 12; if (fourDiv[num]) cout << "Yes\n"; else cout << "No\n"; return 0;} |
Java
// Java program to check whether number has// exactly four distinct factors or notimport java.util.*;class GFG{// Initialize global variable according// to given condition so that it can be// accessible to all functionstatic int N = (int)1E6;static boolean[] fourDiv=new boolean[N + 1];// Function to calculate all number having// four distinct distinct factorsstatic void fourDistinctFactors(){ // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // not a prime, else true. boolean[] primeAll=new boolean[N + 1]; for (int p = 2; p * p <= N; p++) { // If prime[p] is not changed, then it // is a prime if (primeAll[p] == false) { // Update all multiples of p for (int i = p * 2; i <= N; i += p) primeAll[i] = true; } } // Initialize prime[] array which will // contains all the primes from 1-N ArrayList<Integer> prime=new ArrayList<Integer>(); for (int p = 2; p <= N; p++) if (!primeAll[p]) prime.add(p); // Iterate over all the prime numbers for (int i = 0; i < prime.size(); ++i) { int p = prime.get(i); // Mark cube root of prime numbers if (1L * p * p * p <= N) fourDiv[p * p * p] = true; for (int j = i + 1; j < prime.size(); ++j) { int q = prime.get(j); if (1L * p * q > N) break; // Mark product of prime numbers fourDiv[p * q] = true; } }}// Driver programpublic static void main(String[] args){ fourDistinctFactors(); int num = 10; if (fourDiv[num]) System.out.println("Yes"); else System.out.println("No"); num = 12; if (fourDiv[num]) System.out.println("Yes"); else System.out.println("No");}}// This code is contributed by mits |
Python3
# Python3 program to check whether number # has exactly four distinct factors or not# Initialize global variable according to # given condition so that it can be # accessible to all functionN = 1000001;fourDiv = [False] * (N + 1);# Function to calculate all number # having four distinct factorsdef fourDistinctFactors(): # Create a boolean array "prime[0..n]" # and initialize all entries it as true. # A value in prime[i] will finally be # false if i is not a prime, else true. primeAll = [True] * (N + 1); p = 2; while (p * p <= N): # If prime[p] is not changed, then it # is a prime if (primeAll[p] == True): # Update all multiples of p i = p * 2; while (i <= N): primeAll[i] = False; i += p; p += 1; # Initialize prime[] array which will # contain all the primes from 1-N prime = []; for p in range(2, N + 1): if (primeAll[p]): prime.append(p); # Iterate over all the prime numbers for i in range(len(prime)): p = prime[i]; # Mark cube root of prime numbers if (1 * p * p * p <= N): fourDiv[p * p * p] = True; for j in range(i + 1, len(prime)): q = prime[j]; if (1 * p * q > N): break; # Mark product of prime numbers fourDiv[p * q] = True;# Driver CodefourDistinctFactors();num = 10;if (fourDiv[num]): print("Yes");else: print("No");num = 12;if (fourDiv[num]): print("Yes");else: print("No");# This code is contributed by mits |
C#
// C# program to check whether number has// exactly four distinct factors or notusing System;using System.Collections;class GFG{ // Initialize global variable according// to given condition so that it can be// accessible to all functionstatic int N = (int)1E6;static bool[] fourDiv = new bool[N + 1];// Function to calculate all number having// four distinct distinct factorsstatic void fourDistinctFactors(){ // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // not a prime, else true. bool[] primeAll = new bool[N + 1]; for (int p = 2; p * p <= N; p++) { // If prime[p] is not changed, // then it is a prime if (primeAll[p] == false) { // Update all multiples of p for (int i = p * 2; i <= N; i += p) primeAll[i] = true; } } // Initialize prime[] array which will // contains all the primes from 1-N ArrayList prime = new ArrayList(); for (int p = 2; p <= N; p++) if (!primeAll[p]) prime.Add(p); // Iterate over all the prime numbers for (int i = 0; i < prime.Count; ++i) { int p = (int)prime[i]; // Mark cube root of prime numbers if (1L * p * p * p <= N) fourDiv[p * p * p] = true; for (int j = i + 1; j < prime.Count; ++j) { int q = (int)prime[j]; if (1L * p * q > N) break; // Mark product of prime numbers fourDiv[p * q] = true; } }}// Driver Codepublic static void Main(){ fourDistinctFactors(); int num = 10; if (fourDiv[num]) Console.WriteLine("Yes"); else Console.WriteLine("No"); num = 12; if (fourDiv[num]) Console.WriteLine("Yes"); else Console.WriteLine("No");}}// This code is contributed by mits |
PHP
<?php// GFG PHP Compiler not support 64-bit// PHP program to check whether// number has exactly four // distinct factors or not// Initialize global variable // according to given condition // so that it can be accessible// to all function$N = 1000001;$fourDiv = array_fill(0, $N + 1, false);// Function to calculate // all number having four// distinct factorsfunction fourDistinctFactors(){ global $N; global $fourDiv; // Create a boolean array // "prime[0..n]" and initialize // all entries it as true. A // value in prime[i] will finally // be false if i is not a prime, // else true. $primeAll = array_fill(0, $N + 1, true); for ($p = 2; $p * $p <= $N; $p++) { // If prime[p] is not // changed, then it // is a prime if ($primeAll[$p] == true) { // Update all multiples of p for ($i = $p * 2; $i <= $N; $i += $p) $primeAll[$i] = false; } } // Initialize prime[] array // which will contains all // the primes from 1-N $prime; $x = 0; for ($p = 2; $p <= $N; $p++) if ($primeAll[$p]) $prime[$x++] = $p; // Iterate over all // the prime numbers for ($i = 0; $i < $x; ++$i) { $p = $prime[$i]; // Mark cube root // of prime numbers if (1 * $p * $p * $p <= $N) $fourDiv[$p * $p * $p] = true; for ($j = $i + 1; $j < $x; ++$j) { $q = $prime[$j]; if (1 * $p * $q > $N) break; // Mark product of // prime numbers $fourDiv[$p * $q] = true; } }}// Driver CodefourDistinctFactors();$num = 10;if ($fourDiv[$num]) echo "Yes\n";else echo "No\n";$num = 12;if ($fourDiv[$num]) echo "Yes\n";else echo "No\n";// This code is contributed by mits?> |
Javascript
<script>// JavaScript program to check whether number has// exactly four distinct factors or not // Initialize global variable according// to given condition so that it can be// accessible to all functionvar N = 1000000;var fourDiv = Array(N+1).fill(false);// Function to calculate all number having// four distinct distinct factorsfunction fourDistinctFactors(){ // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // not a prime, else true. var primeAll = Array(N+1).fill(false); for (var p = 2; p * p <= N; p++) { // If prime[p] is not changed, // then it is a prime if (primeAll[p] == false) { // Update all multiples of p for (var i = p * 2; i <= N; i += p) primeAll[i] = true; } } // Initialize prime[] array which will // contains all the primes from 1-N var prime = []; for (var p = 2; p <= N; p++) if (!primeAll[p]) prime.push(p); // Iterate over all the prime numbers for (var i = 0; i < prime.length; ++i) { var p = prime[i]; // Mark cube root of prime numbers if (p * p * p <= N) fourDiv[p * p * p] = true; for(var j = i + 1; j < prime.length; ++j) { var q = prime[j]; if (p * q > N) break; // Mark product of prime numbers fourDiv[p * q] = true; } }}// Driver CodefourDistinctFactors();var num = 10;if (fourDiv[num]) document.write("Yes<br>");else document.write("No<br>");num = 12;if (fourDiv[num]) document.write("Yes");else document.write("No");</script> |
Output:
Yes No
Time Complexity: O(n log(log n))
Auxiliary Space: O(n)
This article is contributed by Aarti_Rathi and Shubham Bansal. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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