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++ 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 Code int main()
{ // Given Range
int L = 4, R = 12;
// Function Call
cout << count_crazy_primes(L, R);
return 0;
} |
// 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 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 Code if __name__ = = "__main__" :
# Given Range
L = 4
R = 12
# Function Call
print (count_crazy_primes(L, R))
# This code is contributed by Chitranayal |
// 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 |
<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> |
5
Time Complexity: O(R*log(log(R)))
Auxiliary Space: O(R)