Given two integers L and R. The task is to count the prime numbers in the range [L, R], whose single sum is also a prime number.
A single sum is obtained by adding the digits of a number until a single digit is left.
Examples
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}
Their “single sum” of digits is {5, 7, 2, 4, 8, 1}.
Only {5, 7, 2} are prime. Hence the answer is 3.Input: L = 1, R = 10
Output: 4
Explanation: Prime numbers in the range L = 1 to R = 10 are {2, 3, 5, 7}.
Their “single sum” of digits is {2, 3, 5, 7}.
Since all the numbers are prime, hence the answer is 4.
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 single sum of its digits and again check whether the single sum is prime or not. If the single sum is prime, then increment the counter and print the current element in the range [L, R].
Below is the implementation of the above approach.
C++14
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Function to check whether number// is prime or notbool isPrime(int n){ // Corner case if (n <= 1) return false; // Check from 2 to square root of n for (int i = 2; i <= sqrt(n); i++) if (n % i == 0) return false; return true;}// Function to find single digit sumint SingleDigitSum(int& n){ if (n <= 9) return n; return (n % 9 == 0) ? 9 : n % 9;}// Function to find single digit primesint countSingleDigitPrimes(int l, int r){ int count = 0, i; for (i = l; i <= r; i++) { if (isPrime(i) && isPrime(SingleDigitSum(i))) { count++; } } return count;}// Driver Codeint main(){ // Input range int L = 1, R = 10; // Function Call cout << countSingleDigitPrimes(L, R); return 0;} |
Java
// Java program to implement// the above approachclass GFG{ // Function to check whether number // is prime or not static boolean isPrime(int n) { // Corner case if (n <= 1) return false; // Check from 2 to square root of n for (int i = 2; i <= Math.sqrt(n); i++) if (n % i == 0) return false; return true; } // Function to find single digit sum static int SingleDigitSum(int n) { if (n <= 9) return n; return (n % 9 == 0) ? 9 : n % 9; } // Function to find single digit primes static int countSingleDigitPrimes(int l, int r) { int count = 0, i; for (i = l; i <= r; i++) { if (isPrime(i) && isPrime(SingleDigitSum(i))) { count++; } } return count; } // Driver Code public static void main(String args[]) { // Input range int L = 1, R = 10; // Function Call System.out.println(countSingleDigitPrimes(L, R)); }}// This code is contributed by gfgking |
Python3
# Python program for above approachimport math# Function to check whether number# is prime or notdef isPrime(n): # Corner case if n <= 1: return False # Check from 2 to square root of n for i in range(2, math.floor(math.sqrt(n)) + 1): if n % i == 0: return False return True# Function to find single digit sumdef SingleDigitSum(n): if n <= 9: return n return 9 if (n % 9 == 0) else n % 9# Function to find single digit primesdef countSingleDigitPrimes(l, r): count = 0 i = None for i in range(l, r + 1): if isPrime(i) and isPrime(SingleDigitSum(i)): count += 1 return count# Driver Code# Input rangeL = 1R = 10# Function Callprint(countSingleDigitPrimes(L, R))# This code is contributed by gfgking |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to check whether number // is prime or not static bool isPrime(int n) { // Corner case if (n <= 1) return false; // Check from 2 to square root of n for (int i = 2; i <= Math.Sqrt(n); i++) if (n % i == 0) return false; return true; } // Function to find single digit sum static int SingleDigitSum(int n) { if (n <= 9) return n; return (n % 9 == 0) ? 9 : n % 9; } // Function to find single digit primes static int countSingleDigitPrimes(int l, int r) { int count = 0, i; for (i = l; i <= r; i++) { if (isPrime(i) && isPrime(SingleDigitSum(i))) { count++; } } return count; } // Driver Code public static void Main() { // Input range int L = 1, R = 10; // Function Call Console.Write(countSingleDigitPrimes(L, R)); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript program for above approach // Function to check whether number // is prime or not const isPrime = (n) => { // Corner case if (n <= 1) return false; // Check from 2 to square root of n for (let i = 2; i <= Math.sqrt(n); i++) if (n % i == 0) return false; return true; } // Function to find single digit sum const SingleDigitSum = (n) => { if (n <= 9) return n; return (n % 9 == 0) ? 9 : n % 9; } // Function to find single digit primes const countSingleDigitPrimes = (l, r) => { let count = 0, i; for (i = l; i <= r; i++) { if (isPrime(i) && isPrime(SingleDigitSum(i))) { count++; } } return count; } // Driver Code // Input range let L = 1, R = 10; // Function Call document.write(countSingleDigitPrimes(L, R));// This code is contributed by rakeshsahni</script> |
Output
4
Time Complexity: O((R – L)*N^(1/2)) where N is the prime number in the range [L, R].
Auxiliary Space: O(1)
Another method:
The approach for finding single-digit primes within a given range is to use the following observations:
- The only single-digit prime numbers are 2, 3, 5, and 7.
- The sum of the digits of a number is congruent to the number modulo 9.
- If a number is not divisible by 3, then its sum of digits is not divisible by 3.
- If a number is not divisible by 2 or 5, then its sum of digits is not divisible by 2 or 5.
Based on these observations, we can take the following approach:
- Check if the range includes any of the single-digit prime numbers (2, 3, 5, or 7) and count them.
- For each number in the range that is not a single-digit prime number, check if it is divisible by 2, 3, 5, or 7. If it is, skip it. Otherwise, calculate the sum of its digits and check if it is a single-digit prime number. If it is, increment the count.
- Return the count as the result.
Algorithm:
- Declare the ‘countSingleDigitPrimes‘ function that will find the count of single-digit prime numbers in the given range.
- Initialize a variable ‘count‘ to 0, which will keep track of the count of single-digit prime numbers in the range.
- In the first for loop, iterate over numbers 2 to 7 (inclusive), and check if they are prime and if they lie in the given range. If they are both prime and in the range, increment the ‘count’ variable.
- In the second for loop, iterate over numbers in the range that are greater than or equal to 8. For each number, check if it is divisible by 2, 3, 5, or 7, and continue to the next number if any of these conditions are true.
- Calculate the sum of digits of the current number, and if it is 0, 1, 4, 7, or 9, continue to the next number.
- Check if both the current number and its sum of digits are prime, and increment the ‘count’ variable if this condition is true.
- Return the ‘count’ variable as the output of the ‘countSingleDigitPrimes’ function.
Below is the implementation of the above code:
C++
#include <bits/stdc++.h>using namespace std;// Function to check whether number// is prime or notbool isPrime(int n){ // Corner case if (n <= 1) return false; // Check from 2 to square root of n for (int i = 2; i <= sqrt(n); i++) if (n % i == 0) return false; return true;} // Function to find single digit primesint countSingleDigitPrimes(int l, int r){ int count = 0; // Check if range includes any single digit primes for (int i = 2; i <= 7 && i <= r; i++) { if (l <= i && isPrime(i)) { count++; } } // Check remaining numbers for (int i = max(l, 8); i <= r; i++) { if (i % 2 == 0 || i % 5 == 0) { continue; } if (i % 3 == 0 || i % 7 == 0) { continue; } int sum = i % 9; if (sum == 0 || sum == 1 || sum == 4 || sum == 7 || sum == 9) { continue; } if (isPrime(i) && isPrime(sum)) { count++; } } return count;} // Driver Codeint main(){ // Input range int L = 1, R = 10; // Function Call cout << countSingleDigitPrimes(L, R); return 0;} |
Java
import java.util.*;public class Main { // Function to check whether number is prime or not public static boolean isPrime(int n) { // Corner case if (n <= 1) return false; // Check from 2 to square root of n for (int i = 2; i <= Math.sqrt(n); i++) if (n % i == 0) return false; return true; } // Function to find single digit primes public static int countSingleDigitPrimes(int l, int r) { int count = 0; // Check if range includes any single digit primes for (int i = 2; i <= 7 && i <= r; i++) { if (l <= i && isPrime(i)) { count++; } } // Check remaining numbers for (int i = Math.max(l, 8); i <= r; i++) { if (i % 2 == 0 || i % 5 == 0) { continue; } if (i % 3 == 0 || i % 7 == 0) { continue; } int sum = i % 9; if (sum == 0 || sum == 1 || sum == 4 || sum == 7 || sum == 9) { continue; } if (isPrime(i) && isPrime(sum)) { count++; } } return count; } // Driver Code public static void main(String[] args) { // Input range int L = 1, R = 10; // Function Call System.out.println(countSingleDigitPrimes(L, R)); }}// This code is contributed by Prajwal Kandekar |
Python3
import math# Function to check whether number is prime or notdef isPrime(n): # Corner case if n <= 1: return False # Check from 2 to square root of n for i in range(2, int(math.sqrt(n)) + 1): if n % i == 0: return False return True# Function to find single digit primesdef countSingleDigitPrimes(l, r): count = 0 # Check if range includes any single digit primes for i in range(2, 8): if l <= i <= r and isPrime(i): count += 1 # Check remaining numbers for i in range(max(l, 8), r + 1): if i % 2 == 0 or i % 5 == 0: continue if i % 3 == 0 or i % 7 == 0: continue sum = i % 9 if sum == 0 or sum == 1 or sum == 4 or sum == 7 or sum == 9: continue if isPrime(i) and isPrime(sum): count += 1 return count# Driver CodeL, R = 1, 10# Function Callprint(countSingleDigitPrimes(L, R)) |
C#
using System;public class GFG { // Function to check whether number is prime or not public static Boolean isPrime(int n) { // Corner case if (n <= 1) return false; // Check from 2 to square root of n for (int i = 2; i <= Math.Sqrt(n); i++) if (n % i == 0) return false; return true; } // Function to find single digit primes public static int countSingleDigitPrimes(int l, int r) { int count = 0; // Check if range includes any single digit primes for (int i = 2; i <= 7 && i <= r; i++) { if (l <= i && isPrime(i)) { count++; } } // Check remaining numbers for (int i = Math.Max(l, 8); i <= r; i++) { if (i % 2 == 0 || i % 5 == 0) { continue; } if (i % 3 == 0 || i % 7 == 0) { continue; } int sum = i % 9; if (sum == 0 || sum == 1 || sum == 4 || sum == 7 || sum == 9) { continue; } if (isPrime(i) && isPrime(sum)) { count++; } } return count; } // Driver Code public static void Main() { // Input range int L = 1, R = 10; // Function Call Console.Write(countSingleDigitPrimes(L, R)); }}// This code is contributed by shivanisinghss2110 |
Javascript
// Function to check whether number// is prime or notfunction isPrime(n){// Corner caseif (n <= 1) {return false;}// Check from 2 to square root of nfor (let i = 2; i <= Math.sqrt(n); i++) {if (n % i == 0) {return false;}}return true;}// Function to find single digit primesfunction countSingleDigitPrimes(l, r) {let count = 0;// Check if range includes any single digit primesfor (let i = 2; i <= 7 && i <= r; i++) {if (l <= i && isPrime(i)) {count++;}}// Check remaining numbersfor (let i = Math.max(l, 8); i <= r; i++) {if (i % 2 == 0 || i % 5 == 0) {continue;}if (i % 3 == 0 || i % 7 == 0) {continue;}let sum = i % 9;if (sum == 0 || sum == 1 || sum == 4 || sum == 7 || sum == 9) {continue;}if (isPrime(i) && isPrime(sum)) {count++;}}return count;}// Driver Code// Input rangelet L = 1,R = 10;// Function Callconsole.log(countSingleDigitPrimes(L, R));// This code is contributed by Vaibhav |
Output:
4
Time Complexity: O((R – L + 1) * sqrt(R))
Space Complexity: O(1)
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