Given two integers L and R, the task is to find the count of total numbers of prime numbers in the range [L, R] whose sum of the digits is also a prime number.
Examples:
Input: L = 1, R = 10
Output: 4
Explanation:
Prime numbers in the range L = 1 to R = 10 are {2, 3, 5, 7}.
Their sum of digits is {2, 3, 5, 7}.
Since all the numbers are prime, hence the answer to the query is 4.Input: L = 5, R = 20
Output: 3
Explanation:
Prime numbers in the range L = 5 to R = 20 are {5, 7, 11, 13, 17, 19}.1
Their sum of digits is {5, 7, 2, 4, 8, 10}.
Only {5, 7, 2} are prime, hence the answer to the query is 3.
Naive Approach: The naive approach is to iterate for each number in the range [L, R] and check if the number is prime or not. If the number is prime, find the sum of its digits and again check whether the sum is prime or not. If the sum is prime, then increment the counter for the current element in the range [L, R].
Time Complexity: O((R – L)*log(log P)) where P is the prime number in the range [L, R].
Efficient Approach:
- Store all the prime numbers ranging from 1 to 106 in an array using Sieve of Eratosthenes.
- Create another array that will store whether the sum of the digits of all the numbers ranging from 1 to 106 which are prime.
- Now, compute a prefix array to store counts till every value before limit.
- Once we have prefix array, the value of prefix[R] – prefix[L-1] gives the count of element in the given range which are prime and whose sum are also prime.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; int maxN = 1000000; // Create an array for storing primes int arr[1000001]; // Create a prefix array that will // contain whether sum is prime or not int prefix[1000001]; // Function to find primes in the range // and check whether the sum of digits // of a prime number is prime or not void findPrimes() { // Initialise Prime array arr[] for (int i = 1; i <= maxN; i++) arr[i] = 1; // Since 0 and 1 are not prime // numbers we mark them as '0' arr[0] = 0, arr[1] = 0; // Using Sieve Of Eratosthenes for (int i = 2; i * i <= maxN; i++) { // if the number is prime if (arr[i] == 1) { // Mark all the multiples // of i starting from sqaure // of i with '0' ie. composite for (int j = i * i; j <= maxN; j += i) { //'0' represents not prime arr[j] = 0; } } } // Initialise a sum variable as 0 int sum = 0; prefix[0] = 0; for (int i = 1; i <= maxN; i++) { // Check if the number is prime if (arr[i] == 1) { // A temporary variable to // store the number int temp = i; sum = 0; // Loop to calculate the // sum of digits while (temp > 0) { int x = temp % 10; sum += x; temp = temp / 10; // Check if the sum of prime // number is prime if (arr[sum] == 1) { // if prime mark 1 prefix[i] = 1; } else { // If not prime mark 0 prefix[i] = 0; } } } } // computing prefix array for (int i = 1; i <= maxN; i++) { prefix[i] += prefix[i - 1]; } } // Function to count the prime numbers // in the range [L, R] void countNumbersInRange(int l, int r) { // Function Call to find primes findPrimes(); int result = prefix[r] - prefix[l - 1]; // Print the result cout << result << endl; } // Driver Code int main() { // Input range int l, r; l = 5, r = 20; // Function Call countNumbersInRange(l, r); return 0; } |
chevron_right
filter_none
Java
// Java program for the above approach class GFG{ static int maxN = 1000000; // Create an array for storing primes static int []arr = new int[1000001]; // Create a prefix array that will // contain whether sum is prime or not static int []prefix = new int[1000001]; // Function to find primes in the range // and check whether the sum of digits // of a prime number is prime or not static void findPrimes() { // Initialise Prime array arr[] for (int i = 1; i <= maxN; i++) arr[i] = 1; // Since 0 and 1 are not prime // numbers we mark them as '0' arr[0] = 0; arr[1] = 0; // Using Sieve Of Eratosthenes for (int i = 2; i * i <= maxN; i++) { // if the number is prime if (arr[i] == 1) { // Mark all the multiples // of i starting from sqaure // of i with '0' ie. composite for (int j = i * i; j <= maxN; j += i) { //'0' represents not prime arr[j] = 0; } } } // Initialise a sum variable as 0 int sum = 0; prefix[0] = 0; for (int i = 1; i <= maxN; i++) { // Check if the number is prime if (arr[i] == 1) { // A temporary variable to // store the number int temp = i; sum = 0; // Loop to calculate the // sum of digits while (temp > 0) { int x = temp % 10; sum += x; temp = temp / 10; // Check if the sum of prime // number is prime if (arr[sum] == 1) { // if prime mark 1 prefix[i] = 1; } else { // If not prime mark 0 prefix[i] = 0; } } } } // computing prefix array for (int i = 1; i <= maxN; i++) { prefix[i] += prefix[i - 1]; } } // Function to count the prime numbers // in the range [L, R] static void countNumbersInRange(int l, int r) { // Function Call to find primes findPrimes(); int result = prefix[r] - prefix[l - 1]; // Print the result System.out.print(result + "\n"); } // Driver Code public static void main(String[] args) { // Input range int l, r; l = 5; r = 20; // Function Call countNumbersInRange(l, r); } } // This code is contributed by sapnasingh4991 |
chevron_right
filter_none
C#
// C# program for the above approach using System; class GFG{ static int maxN = 1000000; // Create an array for storing primes static int []arr = new int[1000001]; // Create a prefix array that will // contain whether sum is prime or not static int []prefix = new int[1000001]; // Function to find primes in the range // and check whether the sum of digits // of a prime number is prime or not static void findPrimes() { // Initialise Prime array arr[] for (int i = 1; i <= maxN; i++) arr[i] = 1; // Since 0 and 1 are not prime // numbers we mark them as '0' arr[0] = 0; arr[1] = 0; // Using Sieve Of Eratosthenes for (int i = 2; i * i <= maxN; i++) { // if the number is prime if (arr[i] == 1) { // Mark all the multiples // of i starting from sqaure // of i with '0' ie. composite for (int j = i * i; j <= maxN; j += i) { //'0' represents not prime arr[j] = 0; } } } // Initialise a sum variable as 0 int sum = 0; prefix[0] = 0; for (int i = 1; i <= maxN; i++) { // Check if the number is prime if (arr[i] == 1) { // A temporary variable to // store the number int temp = i; sum = 0; // Loop to calculate the // sum of digits while (temp > 0) { int x = temp % 10; sum += x; temp = temp / 10; // Check if the sum of prime // number is prime if (arr[sum] == 1) { // if prime mark 1 prefix[i] = 1; } else { // If not prime mark 0 prefix[i] = 0; } } } } // computing prefix array for (int i = 1; i <= maxN; i++) { prefix[i] += prefix[i - 1]; } } // Function to count the prime numbers // in the range [L, R] static void countNumbersInRange(int l, int r) { // Function Call to find primes findPrimes(); int result = prefix[r] - prefix[l - 1]; // Print the result Console.Write(result + "\n"); } // Driver Code public static void Main() { // Input range int l, r; l = 5; r = 20; // Function Call countNumbersInRange(l, r); } } // This code is contributed by Code_Mech |
chevron_right
filter_none
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(N)
Recommended Posts:
- Count Numbers in Range with difference between Sum of digits at even and odd positions as Prime
- Sum of all the prime numbers with the count of digits ≤ D
- Count occurrences of a prime number in the prime factorization of every element from the given range
- Print prime numbers with prime sum of digits in an array
- Count of Double Prime numbers in a given range L to R
- Sum of numbers in a range [L, R] whose count of divisors is prime
- Count of natural numbers in range [L, R] which are relatively prime with N
- Numbers in range [L, R] such that the count of their divisors is both even and prime
- Count numbers from range whose prime factors are only 2 and 3
- Count numbers from range whose prime factors are only 2 and 3 using Arrays | Set 2
- Queries for the difference between the count of composite and prime numbers in a given range
- Count numbers in a range having GCD of powers of prime factors equal to 1
- Sum of prime numbers without odd prime digits
- Numbers with sum of digits equal to the sum of digits of its all prime factor
- Number of Co-prime pairs obtained from the sum of digits of elements in the given range
- Count of Prime digits in a Number
- Prime numbers in a given range using STL | Set 2
- Sum of all the prime numbers in a given range
- Find all the prime numbers of given number of digits
- Absolute difference between the Product of Non-Prime numbers and Prime numbers of an Array
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
