Given two positive integers L and R, the task is to find out the total number of values between range [L, R] such that the count of prime numbers from 1 to N is also prime.
Examples:
Input: L = 3, R = 10
Output: 4
Explanation:
Number of primes upto 3, 4, 5, 6, 7, 8, 9 and 10 are 2, 2, 3, 3, 4, 4, 4 and 4 respectively. So, there are a total 4 such numbers {3, 4, 5, 6}[3, 10].
Input: L = 4, R = 12
Output: 5
Explanation:
Number of primes upto 4, 5, 6, 7, 8, 9, 10, 11 and 12 are 2, 3, 3, 4, 4, 4, 4, 5 and 5 respectively. So, there are total 5 such numbers {4, 5, 6, 11, 12} which satisfy the condition in the range [4, 12].
Naive Approach:
The simplest approach to solve the problem is to traverse for all values in the range [1, L – 1] count the number of primes in that range. Once, calculated, check if the count is prime or not. Now, start traversing the values in the range [L, R] one by one and check if the number is prime or not and increase the count accordingly. For every updated count check if it is prime or not and accordingly update the count of required numbers from the given range.
Time Complexity: O(R2)
Efficient Approach:
The above approach can be further optimized by the Sieve of Eratosthenes. Follow the steps below to solve the problem:
- Find all prime numbers up to R using a sieve.
- Maintain a frequency array to store the count of primes up to for all values up to R.
- Create another count array(say freqPrime[]) and add 1 at position i if the cumulative count of total primes up to i is itself a prime number.
- Now for any range L to R, then the number of crazy primes can be calculated by freqPrime[R] – freqPrime[L – 1].
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count the number of// crazy primes in the given range [L, R]int count_crazy_primes(int L, int R){ // Stores all primes int prime[R + 1] = { 0 }; // Stores count of primes int countPrime[R + 1] = { 0 }; // Stores if frequency of // primes is a prime or not // upto each index int freqPrime[R + 1] = { 0 }; prime[0] = prime[1] = 1; // Sieve of Eratosthenes for (int p = 2; p * p <= R; p++) { if (prime[p] == 0) { for (int i = p * p; i <= R; i += p) prime[i] = 1; } } // Count primes for (int i = 1; i <= R; i++) { countPrime[i] = countPrime[i - 1]; // If i is a prime if (!prime[i]) { countPrime[i]++; } } // Stores frequency of primes for (int i = 1; i <= R; i++) { freqPrime[i] = freqPrime[i - 1]; // If the frequency of primes // is a prime if (!prime[countPrime[i]]) { // Increase count of // required numbers freqPrime[i]++; } } // Return the required count return (freqPrime[R] - freqPrime[L - 1]);}// Driver Codeint main(){ // Given Range int L = 4, R = 12; // Function Call cout << count_crazy_primes(L, R); return 0;} |
Java
// Java implementation of the approach import java.io.*;public class GFG{ // Function to count the number of// crazy primes in the given range [L, R]static int count_crazy_primes(int L, int R){ // Stores all primes int prime[] = new int[R + 1]; // Stores count of primes int countPrime[] = new int[R + 1]; // Stores if frequency of // primes is a prime or not // upto each index int freqPrime[] = new int[R + 1]; prime[0] = 1; prime[1] = 1; // Sieve of Eratosthenes for(int p = 2; p * p <= R; p++) { if (prime[p] == 0) { for(int i = p * p; i <= R; i += p) prime[i] = 1; } } // Count primes for(int i = 1; i <= R; i++) { countPrime[i] = countPrime[i - 1]; // If i is a prime if (prime[i] != 0) { countPrime[i]++; } } // Stores frequency of primes for(int i = 1; i <= R; i++) { freqPrime[i] = freqPrime[i - 1]; // If the frequency of primes // is a prime if (prime[countPrime[i]] != 0) { // Increase count of // required numbers freqPrime[i]++; } } // Return the required count return (freqPrime[R] - freqPrime[L - 1]);}// Driver code public static void main (String[] args) { // Given range int L = 4, R = 12; // Function call System.out.println(count_crazy_primes(L, R));} } // This code is contributed by Pratima Pandey |
Python3
# Python3 program for the above approach# Function to count the number of# crazy primes in the given range [L, R]def count_crazy_primes(L, R): # Stores all primes prime = [0] * (R + 1) # Stores count of primes countPrime = [0] * (R + 1) # Stores if frequency of # primes is a prime or not # upto each index freqPrime = [0] * (R + 1) prime[0] = prime[1] = 1 # Sieve of Eratosthenes p = 2 while p * p <= R: if (prime[p] == 0): for i in range (p * p, R + 1 , p): prime[i] = 1 p += 1 # Count primes for i in range (1 , R + 1): countPrime[i] = countPrime[i - 1] # If i is a prime if (not prime[i]): countPrime[i] += 1 # Stores frequency of primes for i in range (1, R + 1): freqPrime[i] = freqPrime[i - 1] # If the frequency of primes # is a prime if (not prime[countPrime[i]]): # Increase count of # required numbers freqPrime[i] += 1 # Return the required count return (freqPrime[R] - freqPrime[L - 1])# Driver Codeif __name__ =="__main__": # Given Range L = 4 R = 12 # Function Call print(count_crazy_primes(L, R))# This code is contributed by Chitranayal |
C#
// C# implementation of the approach using System;class GFG{ // Function to count the number of// crazy primes in the given range [L, R]static int count_crazy_primes(int L, int R){ // Stores all primes int []prime = new int[R + 1]; // Stores count of primes int []countPrime = new int[R + 1]; // Stores if frequency of // primes is a prime or not // upto each index int []freqPrime = new int[R + 1]; prime[0] = 1; prime[1] = 1; // Sieve of Eratosthenes for(int p = 2; p * p <= R; p++) { if (prime[p] == 0) { for(int i = p * p; i <= R; i += p) prime[i] = 1; } } // Count primes for(int i = 1; i <= R; i++) { countPrime[i] = countPrime[i - 1]; // If i is a prime if (prime[i] != 0) { countPrime[i]++; } } // Stores frequency of primes for(int i = 1; i <= R; i++) { freqPrime[i] = freqPrime[i - 1]; // If the frequency of primes // is a prime if (prime[countPrime[i]] != 0) { // Increase count of // required numbers freqPrime[i]++; } } // Return the required count return (freqPrime[R] - freqPrime[L - 1]);} // Driver code public static void Main(String[] args) { // Given range int L = 4, R = 12; // Function call Console.WriteLine( count_crazy_primes(L, R));} } // This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approach// Function to count the number of// crazy primes in the given range [L, R]function count_crazy_primes(L, R){ // Stores all primes let prime = Array.from({length: R+1}, (_, i) => 0); // Stores count of primes let countPrime = Array.from({length: R+1}, (_, i) => -1); // Stores if frequency of // primes is a prime or not // upto each index let freqPrime = Array.from({length: R+1}, (_, i) => 0); prime[0] = 1; prime[1] = 1; // Sieve of Eratosthenes for(let p = 2; p * p <= R; p++) { if (prime[p] == 0) { for(let i = p * p; i <= R; i += p) prime[i] = 1; } } // Count primes for(let i = 1; i <= R; i++) { countPrime[i] = countPrime[i - 1]; // If i is a prime if (prime[i] != 0) { countPrime[i]++; } } // Stores frequency of primes for (let i = 1; i <= R; i++) { freqPrime[i] = freqPrime[i - 1]; // If the frequency of primes // is a prime if (!prime[countPrime[i]]) { // Increase count of // required numbers freqPrime[i]++; } } // Return the required count return (freqPrime[R] - freqPrime[L - 1]);} // Driver Code // Given range let L = 4, R = 12; // Function call document.write(count_crazy_primes(L, R));</script> |
Output:
5
Time Complexity: O(R*log(log(R)))
Auxiliary Space: O(R)
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