Find all numbers between range L to R such that sum of digit and sum of square of digit is prime

Given a range L and R count all numbers between L to R such that sum of digit of each number and sum of square of digit of each number is Prime.

Note: 10 <= [L, R] <= 108

Examples:



Input: L = 10, R = 20
Output: 4
Such types of numbers are: 11 12 14 16

Input: L = 100, R = 130
Output: 9

Such types of numbers are : 101 102 104 106 110 111 113 119 120

Naive Approach:
Just get the sum of digit of each number and sum of the square of digit of each number and to check whether they both are prime or not.

Efficient Approach:

    In this approach, there is an observation that can do the optimization:

  • Now if look closely into range the number is 108 ie., and largest number less than this will be 99999999 and maximum number can be formed is 8 * ( 9 * 9 ) = 648 (as the sum of digit square is 92 + 92 + … 8times) so we need only primes upto 648 only which can be done using Sieve of Eratosthenes.
  • Now iterate for each number in the range and check whether it satisfies above condition or not.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Sieve of prime numbers
void primesieve(vector<bool>& prime)
{
    // Sieve to store whether a
    // number is prime or not in
    // O(nlog(log(n)))
    prime[1] = false;
  
    for (int p = 2; p * p <= 650; p++) {
        if (prime[p] == true) {
            for (int i = p * 2; i <= 650; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to return sum of digit
// and sum of square of digit
pair<int, int> sum_sqsum(int n)
{
  
    int sum = 0;
    int sqsum = 0;
    int x;
  
    // Until number is not
    // zero
    while (n) {
        x = n % 10;
        sum += x;
        sqsum += x * x;
        n /= 10;
    }
  
    return (make_pair(sum, sqsum));
}
  
// Function to return the count
// of number form L to R
// whose sum of digits and
// sum of square of digits
// are prime
int countnumber(int L, int R)
{
  
    vector<bool> prime(651, true);
  
    primesieve(prime);
  
    int cnt = 0;
  
    // Iterate for each value
    // in the range of L to R
    for (int i = L; i <= R; i++) {
  
        // digit.first stores sum of digits
        // digit.second stores sum of
        // square of digit
        pair<int, int> digit = sum_sqsum(i);
  
        // If sum of digits and sum of
        // square of digit both are
        // prime then increment the count
        if (prime[digit.first]
            && prime[digit.second]) {
            cnt += 1;
        }
    }
  
    return cnt;
}
  
// Driver Code
int main()
{
  
    int L = 10;
    int R = 20;
  
    cout << countnumber(L, R);
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG 
{
static class pair
    int first, second; 
    public pair(int first, int second) 
    
        this.first = first; 
        this.second = second; 
    
}
  
// Sieve of prime numbers
static void primesieve(boolean []prime)
{
    // Sieve to store whether a
    // number is prime or not in
    // O(nlog(log(n)))
    prime[1] = false;
  
    for (int p = 2; p * p <= 650; p++)
    {
        if (prime[p] == true)
        {
            for (int i = p * 2; i <= 650; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to return sum of digit
// and sum of square of digit
static pair sum_sqsum(int n)
{
    int sum = 0;
    int sqsum = 0;
    int x;
  
    // Until number is not
    // zero
    while (n > 0)
    {
        x = n % 10;
        sum += x;
        sqsum += x * x;
        n /= 10;
    }
    return (new pair(sum, sqsum));
}
  
// Function to return the count
// of number form L to R
// whose sum of digits and
// sum of square of digits
// are prime
static int countnumber(int L, int R)
{
    boolean []prime = new boolean[651];
  
    Arrays.fill(prime, true);
    primesieve(prime);
  
    int cnt = 0;
  
    // Iterate for each value
    // in the range of L to R
    for (int i = L; i <= R; i++) 
    {
  
        // digit.first stores sum of digits
        // digit.second stores sum of
        // square of digit
        pair digit = sum_sqsum(i);
  
        // If sum of digits and sum of
        // square of digit both are
        // prime then increment the count
        if (prime[digit.first] && 
            prime[digit.second]) 
        {
            cnt += 1;
        }
    }
    return cnt;
}
  
// Driver Code
public static void main(String[] args) 
{
    int L = 10;
    int R = 20;
  
    System.out.println(countnumber(L, R));
}
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 implementation of the approach 
from math import sqrt
  
# Sieve of prime numbers 
def primesieve(prime) :
  
    # Sieve to store whether a 
    # number is prime or not in 
    # O(nlog(log(n))) 
    prime[1] = False
  
    for p in range(2, int(sqrt(650)) + 1) :
        if (prime[p] == True) :
            for i in range(p * 2, 651, p) : 
                prime[i] = False
  
# Function to return sum of digit 
# and sum of square of digit 
def sum_sqsum(n) :
  
    sum = 0
    sqsum = 0
  
    # Until number is not 
    # zero 
    while (n) :
        x = n % 10
        sum += x; 
        sqsum += x * x; 
        n //= 10
  
    return (sum, sqsum); 
  
# Function to return the count 
# of number form L to R 
# whose sum of digits and 
# sum of square of digits 
# are prime 
def countnumber(L, R): 
  
    prime = [True] * 651
  
    primesieve(prime); 
  
    cnt = 0
  
    # Iterate for each value 
    # in the range of L to R 
    for i in range(L, R + 1) :
          
        # digit.first stores sum of digits 
        # digit.second stores sum of 
        # square of digit 
        digit = sum_sqsum(i); 
  
        # If sum of digits and sum of 
        # square of digit both are 
        # prime then increment the count 
        if (prime[digit[0]] and prime[digit[1]]) :
            cnt += 1
  
    return cnt; 
  
# Driver Code 
if __name__ == "__main__"
  
    L = 10
    R = 20
  
    print(countnumber(L, R)); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
      
class GFG 
{
public class pair
    public int first, second; 
    public pair(int first, int second) 
    
        this.first = first; 
        this.second = second; 
    
}
  
// Sieve of prime numbers
static void primesieve(bool []prime)
{
    // Sieve to store whether a
    // number is prime or not in
    // O(nlog(log(n)))
    prime[1] = false;
  
    for (int p = 2; p * p <= 650; p++)
    {
        if (prime[p] == true)
        {
            for (int i = p * 2;
                     i <= 650; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to return sum of digit
// and sum of square of digit
static pair sum_sqsum(int n)
{
    int sum = 0;
    int sqsum = 0;
    int x;
  
    // Until number is not
    // zero
    while (n > 0)
    {
        x = n % 10;
        sum += x;
        sqsum += x * x;
        n /= 10;
    }
    return (new pair(sum, sqsum));
}
  
// Function to return the count
// of number form L to R
// whose sum of digits and
// sum of square of digits
// are prime
static int countnumber(int L, int R)
{
    bool []prime = new bool[651];
    for (int i = 0; i < 651; i++) 
        prime[i] = true;
    primesieve(prime);
  
    int cnt = 0;
  
    // Iterate for each value
    // in the range of L to R
    for (int i = L; i <= R; i++) 
    {
  
        // digit.first stores sum of digits
        // digit.second stores sum of
        // square of digit
        pair digit = sum_sqsum(i);
  
        // If sum of digits and sum of
        // square of digit both are
        // prime then increment the count
        if (prime[digit.first] && 
            prime[digit.second]) 
        {
            cnt += 1;
        }
    }
    return cnt;
}
  
// Driver Code
public static void Main(String[] args) 
{
    int L = 10;
    int R = 20;
  
    Console.WriteLine(countnumber(L, R));
}
}
  
// This code is contributed by 29AjayKumar

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Output:

4

Note:

    If there are multiple queries asked to find out numbers between range from L and R there will be 2 approaches:

  1. Store all number which satisfies above condition in another array and use binary search to find out how many elements in array such that it less than R , say cnt1 , and how many elements in array such that it less than L , say cnt2 . Return cnt1 – cnt2
    Time Complexity: O(log(N)) per query.
  2. We can use prefix array or DP approach such that it already stores how many no. are good of above type from index 0 to i, and return total count by giving DP[R] – DP[L-1]
    Time Complexity: O(1) per query.


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