Count all prime numbers in a given range whose sum of digits is also prime
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; } |
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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 |
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Python3
# Python3 program for the above approach maxN = 1000000 # Create an array for storing primes arr = [0] * (1000001) # Create a prefix array that will # contain whether sum is prime or not prefix = [0] * (1000001) # Function to find primes in the range # and check whether the sum of digits # of a prime number is prime or not def findPrimes(): # Initialise Prime array arr[] for i in range(1, maxN + 1): 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 i = 2 while i * i <= maxN: # 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 j in range(i * i, maxN, i): # '0' represents not prime arr[j] = 0 i += 1 # Initialise a sum variable as 0 sum = 0 prefix[0] = 0 for i in range(1, maxN + 1): # Check if the number is prime if (arr[i] == 1): # A temporary variable to # store the number temp = i sum = 0 # Loop to calculate the # sum of digits while (temp > 0): 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 i in range(1, maxN + 1): prefix[i] += prefix[i - 1] # Function to count the prime numbers # in the range [L, R] def countNumbersInRange(l, r): # Function call to find primes findPrimes() result = (prefix[r] - prefix[l - 1]) # Print the result print(result) # Driver Code if __name__ == "__main__": # Input range l = 5 r = 20 # Function call countNumbersInRange(l, r) # This code is contributed by chitranayal |
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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 |
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Output:
3
Time Complexity: O(N)
Auxiliary Space: O(N)
