Count of Double Prime numbers in a given range L to R

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Given two integers L and R, the task to find the number of Double Prime numbers in the range.

A number N is called double prime when the count of prime numbers in the range 1 to N (excluding 1 and including N) is also prime.

Examples:

Input: L = 3, R = 10
Output: 4
Explanation:
For 3, we have the range 1, 2, 3, and count of prime number is 2 (which is also a prime no.)
For 4, we have the range 1, 2, 3, 4, and count of a prime number is 2 (which is also a prime no.)
For 5, we have the range 1, 2, 3, 4, 5, and count of a prime number is 3 (which is also a prime no.)
For 6, we have the range 1, 2, 3, 4, 5, 6, and count of prime numbers is 3 (which is also a prime no.)
For 7, we have the range 1, 2, 3, 4, 5, 6, 7, and count of prime numbers is 4 which is nonprime.
Similarly for other numbers till R = 10, the count of prime numbers is nonprime. Hence the count of double prime numbers is 4.
Input: L = 4, R = 12
Output: 5
Explanation:
For the given range there are total 5 double prime numbers.

Approach:
To solve the problem mentioned above we will use the concept of Sieve to generate prime numbers.

Generate all prime numbers for 0 to 10⁶ and store in an array.
Initialize a variable count to keep a track of prime numbers from 1 to some ith position.
Then for every prime number we will increment the count and also set dp[count] = 1 (where dp is the array which stores a double prime number) indicating the number of numbers from 1 to some ith position that are prime.
Lastly, find the cumulative sum of dp array so the answer will be dp[R] – dp[L – 1].

Below is the implementation of the above approach:

C++

// C++ program to find the count
// of Double Prime numbers
// in the range L to R
 
#include <bits/stdc++.h>
using namespace std;
 
// Array to make Sieve
// where arr[i]=0 indicates
// non prime and arr[i] = 1
// indicates prime
int arr[1000001];
 
// Array to find double prime
int dp[1000001];
 
// Function to find the number
// double prime numbers in range
void count()
{
    int maxN = 1000000, i, j;
 
    // Assume all numbers as prime
    for (i = 0; i < maxN; i++)
        arr[i] = 1;
 
    arr[0] = 0;
    arr[1] = 0;
 
    for (i = 2; i * i <= maxN; i++) {
 
        // Check if the number is prime
        if (arr[i] == 1) {
 
            // check for multiples of i
            for (j = 2 * i; j <= maxN; j += i) {
 
                // Make all multiples of
                // ith prime as non-prime
                arr[j] = 0;
            }
        }
    }
 
    int cnt = 0;
 
    for (i = 0; i <= maxN; i++) {
        // Check if number at ith position
        // is prime then increment count
        if (arr[i] == 1)
            cnt++;
 
        if (arr[cnt] == 1)
 
            // Indicates count of numbers
            // from 1 to i that are
            // also prime and
            // hence double prime
            dp[i] = 1;
 
        else
            // If number is not a double prime
            dp[i] = 0;
    }
    for (i = 1; i <= maxN; i++)
        // finding cumulative sum
        dp[i] += dp[i - 1];
}
 
// Driver code
int main()
{
    int L = 4, R = 12;
    count();
    cout << dp[R] - dp[L - 1];
 
    return 0;
}

Java

// Java program to find the count
// of Double Prime numbers
// in the range L to R
import java.util.*;
import java.lang.*;
class GFG{
     
// Array to make Sieve
// where arr[i]=0 indicates
// non prime and arr[i] = 1
// indicates prime
static int[] arr = new int[1000001];
 
// Array to find double prime
static int[] dp = new int[1000001];
 
// Function to find the number
// double prime numbers in range
static void count()
{
    int maxN = 1000000, i, j;
 
    // Assume all numbers as prime
    for (i = 0; i < maxN; i++)
        arr[i] = 1;
 
    arr[0] = 0;
    arr[1] = 0;
 
    for (i = 2; i * i <= maxN; i++) 
    {
 
        // Check if the number is prime
        if (arr[i] == 1) 
        {
 
            // check for multiples of i
            for (j = 2 * i; j <= maxN; j += i) 
            {
 
                // Make all multiples of
                // ith prime as non-prime
                arr[j] = 0;
            }
        }
    }
 
    int cnt = 0;
 
    for (i = 0; i <= maxN; i++) 
    {
        // Check if number at ith position
        // is prime then increment count
        if (arr[i] == 1)
            cnt++;
 
        if (arr[cnt] == 1)
 
            // Indicates count of numbers
            // from 1 to i that are
            // also prime and
            // hence double prime
            dp[i] = 1;
 
        else
            // If number is not a double prime
            dp[i] = 0;
    }
     
    for (i = 1; i <= maxN; i++)
     
        // finding cumulative sum
        dp[i] += dp[i - 1];
}
 
// Driver code
public static void main(String[] args)
{
    int L = 4, R = 12;
    count();
    System.out.println(dp[R] - dp[L - 1]);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to find the count
# of Double Prime numbers
# in the range L to R
 
# Array to make Sieve
# where arr[i]=0 indicates
# non prime and arr[i] = 1
# indicates prime
arr = [0] * 1000001
 
# Array to find double prime
dp = [0] * 1000001
 
# Function to find the number
# double prime numbers in range
def count():
     
    maxN = 1000000
 
    # Assume all numbers as prime
    for i in range(0, maxN):
        arr[i] = 1
 
    arr[0] = 0
    arr[1] = 0
     
    i = 2
    while(i * i <= maxN):
 
        # Check if the number is prime
        if (arr[i] == 1):
 
            # Check for multiples of i
            for j in range(2 * i, maxN + 1, i):
 
                # Make all multiples of
                # ith prime as non-prime
                arr[j] = 0
                 
        i += 1
 
    cnt = 0
 
    for i in range(0, maxN + 1):
         
        # Check if number at ith position
        # is prime then increment count
        if (arr[i] == 1):
            cnt += 1
 
        if (arr[cnt] == 1):
 
            # Indicates count of numbers
            # from 1 to i that are
            # also prime and
            # hence double prime
            dp[i] = 1
 
        else:
             
            # If number is not a double prime
            dp[i] = 0
     
    for i in range(0, maxN + 1):
         
        # Finding cumulative sum
        dp[i] += dp[i - 1]
 
# Driver code
L = 4
R = 12
     
count()
 
print(dp[R] - dp[L - 1])
 
# This code is contributed by sanjoy_62

C#

// C# program to find the count
// of Double Prime numbers
// in the range L to R
using System;
class GFG{
     
// Array to make Sieve
// where arr[i]=0 indicates
// non prime and arr[i] = 1
// indicates prime
static int[] arr = new int[1000001];
 
// Array to find double prime
static int[] dp = new int[1000001];
 
// Function to find the number
// double prime numbers in range
static void count()
{
    int maxN = 1000000, i, j;
 
    // Assume all numbers as prime
    for (i = 0; i < maxN; i++)
        arr[i] = 1;
 
    arr[0] = 0;
    arr[1] = 0;
 
    for (i = 2; i * i <= maxN; i++) 
    {
 
        // Check if the number is prime
        if (arr[i] == 1) 
        {
 
            // check for multiples of i
            for (j = 2 * i; j <= maxN; j += i) 
            {
 
                // Make all multiples of
                // ith prime as non-prime
                arr[j] = 0;
            }
        }
    }
 
    int cnt = 0;
 
    for (i = 0; i <= maxN; i++) 
    {
        // Check if number at ith position
        // is prime then increment count
        if (arr[i] == 1)
            cnt++;
 
        if (arr[cnt] == 1)
 
            // Indicates count of numbers
            // from 1 to i that are
            // also prime and
            // hence double prime
            dp[i] = 1;
 
        else
            // If number is not a double prime
            dp[i] = 0;
    }
     
    for (i = 1; i <= maxN; i++)
     
        // finding cumulative sum
        dp[i] += dp[i - 1];
}
 
// Driver code
public static void Main()
{
    int L = 4, R = 12;
    count();
    Console.Write(dp[R] - dp[L - 1]);
}
}
 
// This code is contributed by Code_Mech

Javascript

<script>
     
// Javascript program to find the count
// of Double Prime numbers
// in the range L to R
 
// Array to make Sieve
// where arr[i]=0 indicates
// non prime and arr[i] = 1
// indicates prime
let arr = [];
   
// Array to find double prime
let dp = [];
   
// Function to find the number
// double prime numbers in range
function count()
{
    let maxN = 1000000, i, j;
   
    // Assume all numbers as prime
    for (i = 0; i < maxN; i++)
        arr[i] = 1;
   
    arr[0] = 0;
    arr[1] = 0;
   
    for (i = 2; i * i <= maxN; i++) 
    {
   
        // Check if the number is prime
        if (arr[i] == 1) 
        {
   
            // check for multiples of i
            for (j = 2 * i; j <= maxN; j += i) 
            {
   
                // Make all multiples of
                // ith prime as non-prime
                arr[j] = 0;
            }
        }
    }
   
    let cnt = 0;
   
    for (i = 0; i <= maxN; i++) 
    {
        // Check if number at ith position
        // is prime then increment count
        if (arr[i] == 1)
            cnt++;
   
        if (arr[cnt] == 1)
   
            // Indicates count of numbers
            // from 1 to i that are
            // also prime and
            // hence double prime
            dp[i] = 1;
   
        else
            // If number is not a double prime
            dp[i] = 0;
    }
       
    for (i = 1; i <= maxN; i++)
       
        // finding cumulative sum
        dp[i] += dp[i - 1];
}
     
// Driver code
 
    let L = 4, R = 12;
    count();
    document.write(dp[R] - dp[L - 1]);
    
   // This code is contributed by susmitakundugoaldanga.
</script>

Output:

Time Complexity: O((R-L)*log(log(R)))

Auxiliary Space: O(R)

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