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
C#
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