# Count of all values of N in [L, R] such that count of primes upto N is also prime

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:
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:
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:

1. Find all prime numbers up to R using a sieve.
2. Maintain a frequency array to store the count of primes up to for all values up to R.
3. 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.
4. 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 ` `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 = prime = 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

 `// Java implementation of the approach ` `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     `

## 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 = 1;` `    ``prime = 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`

Output:

```5

```

Time Complexity: O(R*log(log(R)))
Auxillary Space: O(R) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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