Numbers with sum of digits equal to the sum of digits of its all prime factor

Given a range, the task is to find the count of the numbers in the given range such that the sum of its digit is equal to the sum of all its prime factors digits sum.

Examples:

Input: l = 2, r = 10
Output: 5
2, 3, 4, 5 and 7 are such numbers

Input: l = 15, r = 22
Output: 3
17, 19 and 22 are such numbers
As, 17 and 19 are already prime.
Prime Factors of 22 = 2 * 11 i.e 
For 22, Sum of digits is 2+2 = 4
For 2 * 11, Sum of digits is 2 + 1 + 1 = 4

Approach: An efficient solution is to modify Sieve of Eratosthenes such that for each non-prime number it stores smallest prime factor(prefactor).

  1. Preprocess to find the smallest prime factorfor all the numbers between 2 and MAXN. This can be done by breaking up the number into its prime factors in constant time because for each number if it is a prime, it has no prefactor.
  2. Otherwise, we can break it up to into a prime factor and the other part of the number which may or may not be prime.
  3. And repeat this process of extracting factors till it becomes a prime.
  4. Then check if the digits of that number is equal to the digits of prime factors by adding th digits of smallest prime factor i.e.

    Digits_Sum of SPF[n] + Digits_Sum of (n / SPF[n])

  5. Now make prefix sum array that counts how many valid numbers are there up to a number N. For each query, print:

    ans[R] – ans[L-1]

Below is the implementation of above approach:

C++

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// C++ program to Find the count of the numbers
// in the given range such that the sum of its
// digit is equal to the sum of all its prime
// factors digits sum.
#include <bits/stdc++.h>
using namespace std;
  
// maximum size of number
#define MAXN 100005
  
// array to store smallest prime factor of number
int spf[MAXN] = { 0 };
  
// array to store sum of digits of a number
int sum_digits[MAXN] = { 0 };
  
// boolean array to check given number is countable
// for required answer or not.
bool isValid[MAXN] = { 0 };
  
// prefix array to store answer
int ans[MAXN] = { 0 };
  
// Calculating SPF (Smallest Prime Factor) for every
// number till MAXN.
void Smallest_prime_factor()
{
    // marking smallest prime factor for every
    // number to be itself.
    for (int i = 1; i < MAXN; i++)
        spf[i] = i;
  
    // separately marking spf for every even
    // number as 2
    for (int i = 4; i < MAXN; i += 2)
        spf[i] = 2;
  
    for (int i = 3; i * i <= MAXN; i += 2)
  
        // checking if i is prime
        if (spf[i] == i)
  
            // marking SPF for all numbers divisible by i
            for (int j = i * i; j < MAXN; j += i)
  
                // marking spf[j] if it is not
                // previously marked
                if (spf[j] == j)
                    spf[j] = i;
}
  
// Function to find sum of digits in a number
int Digit_Sum(int copy)
{
    int d = 0;
    while (copy) {
        d += copy % 10;
        copy /= 10;
    }
  
    return d;
}
  
// find sum of digits of all numbers up to MAXN
void Sum_Of_All_Digits()
{
    for (int n = 2; n < MAXN; n++) {
        // add sum of digits of least 
        // prime factor and n/spf[n]
        sum_digits[n] = sum_digits[n / spf[n]] 
                           + Digit_Sum(spf[n]);
  
        // if it is valid make isValid true
        if (Digit_Sum(n) == sum_digits[n])
            isValid[n] = true;
    }
  
    // prefix sum to compute answer
    for (int n = 2; n < MAXN; n++) {
        if (isValid[n])
            ans[n] = 1;
        ans[n] += ans[n - 1];
    }
}
  
// Driver code
int main()
{
    Smallest_prime_factor();
    Sum_Of_All_Digits();
  
    // decleartion
    int l, r;
  
    // print answer for required range
    l = 2, r = 3;
    cout << "Valid numbers in the range " << l << " " 
         << r << " are " << ans[r] - ans[l - 1] << endl;
  
    // print answer for required range
    l = 2, r = 10;
    cout << "Valid numbers in the range " << l << " " 
         << r << " are " << ans[r] - ans[l - 1] << endl;
  
  return 0;
}

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Java

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// Java program to Find the count 
// of the numbers in the given 
// range such that the sum of its
// digit is equal to the sum of 
// all its prime factors digits sum.
import java.io.*;
  
class GFG 
{
  
// maximum size of number
static int MAXN = 100005;
  
// array to store smallest 
// prime factor of number
static int spf[] = new int[MAXN];
  
// array to store sum 
// of digits of a number
static int sum_digits[] = new int[MAXN];
  
// boolean array to check
// given number is countable
// for required answer or not.
static boolean isValid[] = new boolean[MAXN];
  
// prefix array to store answer
static int ans[] = new int[MAXN];
  
// Calculating SPF (Smallest
// Prime Factor) for every
// number till MAXN.
static void Smallest_prime_factor()
{
    // marking smallest prime factor 
    // for every number to be itself.
    for (int i = 1; i < MAXN; i++)
        spf[i] = i;
  
    // separately marking spf 
    // for every even number as 2
    for (int i = 4; i < MAXN; i += 2)
        spf[i] = 2;
  
    for (int i = 3
             i * i <= MAXN; i += 2)
  
        // checking if i is prime
        if (spf[i] == i)
  
            // marking SPF for all
            // numbers divisible by i
            for (int j = i * i; 
                     j < MAXN; j += i)
  
                // marking spf[j] if it
                // is not previously marked
                if (spf[j] == j)
                    spf[j] = i;
}
  
// Function to find sum 
// of digits in a number
static int Digit_Sum(int copy)
{
    int d = 0;
    while (copy > 0
    {
        d += copy % 10;
        copy /= 10;
    }
  
    return d;
}
  
// find sum of digits of 
// all numbers up to MAXN
static void Sum_Of_All_Digits()
{
    for (int n = 2; n < MAXN; n++)
    {
        // add sum of digits of least 
        // prime factor and n/spf[n]
        sum_digits[n] = sum_digits[n / spf[n]] 
                          + Digit_Sum(spf[n]);
  
        // if it is valid make isValid true
        if (Digit_Sum(n) == sum_digits[n])
            isValid[n] = true;
    }
  
    // prefix sum to compute answer
    for (int n = 2; n < MAXN; n++) 
    {
        if (isValid[n])
            ans[n] = 1;
        ans[n] += ans[n - 1];
    }
}
  
// Driver code
public static void main (String[] args) 
{
    Smallest_prime_factor();
    Sum_Of_All_Digits();
      
    // declaration
    int l, r;
      
    // print answer for required range
    l = 2; r = 3;
    System.out.println("Valid numbers in the range "
                               l + " " + r + " are "
                              (ans[r] - ans[l - 1] ));
      
    // print answer for required range
    l = 2; r = 10;
    System.out.println("Valid numbers in the range "
                               l + " " + r + " are "
                               (ans[r] - ans[l - 1]));
}
}
  
// This code is contributed
// by Inder

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Python 3

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# Python 3 program to Find the count of 
# the numbers in the given range such
# that the sum of its digit is equal to
# the sum of all its prime factors digits sum.
  
# maximum size of number
MAXN = 100005
  
# array to store smallest prime
# factor of number
spf = [0] * MAXN
  
# array to store sum of digits of a number
sum_digits = [0] * MAXN
  
# boolean array to check given number 
# is countable for required answer or not.
isValid = [0] * MAXN
  
# prefix array to store answer
ans = [0]*MAXN
  
# Calculating SPF (Smallest Prime Factor) 
# for every number till MAXN.
def Smallest_prime_factor():
  
    # marking smallest prime factor
    # for every number to be itself.
    for i in range(1, MAXN):
        spf[i] = i
  
    # separately marking spf for 
    # every even number as 2
    for i in range(4, MAXN, 2):
        spf[i] = 2
  
    i = 3
    while i * i <= MAXN: 
  
        # checking if i is prime
        if (spf[i] == i):
  
            # marking SPF for all numbers
            # divisible by i
            for j in range(i * i, MAXN, i):
  
                # marking spf[j] if it is not
                # previously marked
                if (spf[j] == j):
                    spf[j] = i
                      
        i += 2
  
# Function to find sum of digits 
# in a number
def Digit_Sum(copy):
      
    d = 0
    while (copy) :
        d += copy % 10
        copy //= 10
  
    return d
  
# find sum of digits of all
# numbers up to MAXN
def Sum_Of_All_Digits():
  
    for n in range(2, MAXN) :
          
        # add sum of digits of least 
        # prime factor and n/spf[n]
        sum_digits[n] = (sum_digits[n // spf[n]] +
                         Digit_Sum(spf[n]))
  
        # if it is valid make isValid true
        if (Digit_Sum(n) == sum_digits[n]):
            isValid[n] = True
  
    # prefix sum to compute answer
    for n in range(2, MAXN) :
        if (isValid[n]):
            ans[n] = 1
        ans[n] += ans[n - 1]
  
# Driver code
if __name__ == "__main__":
      
    Smallest_prime_factor()
    Sum_Of_All_Digits()
  
    # print answer for required range
    l = 2
    r = 3
    print("Valid numbers in the range", l, r,
                  "are", ans[r] - ans[l - 1])
  
    # print answer for required range
    l = 2
    r = 10
    print("Valid numbers in the range", l, r, 
                  "are", ans[r] - ans[l - 1])
  
# This code is contributed by ita_c

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C#

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// C# program to Find the count 
// of the numbers in the given 
// range such that the sum of its
// digit is equal to the sum of 
// all its prime factors digits sum.
using System;
  
class GFG 
{
  
// maximum size of number
static int MAXN = 100005;
  
// array to store smallest 
// prime factor of number
static int []spf = new int[MAXN];
  
// array to store sum 
// of digits of a number
static int []sum_digits = new int[MAXN];
  
// boolean array to check
// given number is countable
// for required answer or not.
static bool []isValid = new bool[MAXN];
  
// prefix array to store answer
static int []ans = new int[MAXN];
  
// Calculating SPF (Smallest
// Prime Factor) for every
// number till MAXN.
static void Smallest_prime_factor()
{
    // marking smallest prime factor 
    // for every number to be itself.
    for (int i = 1; i < MAXN; i++)
        spf[i] = i;
  
    // separately marking spf 
    // for every even number as 2
    for (int i = 4; i < MAXN; i += 2)
        spf[i] = 2;
  
    for (int i = 3; 
             i * i <= MAXN; i += 2)
  
        // checking if i is prime
        if (spf[i] == i)
  
            // marking SPF for all
            // numbers divisible by i
            for (int j = i * i; 
                     j < MAXN; j += i)
  
                // marking spf[j] if it
                // is not previously marked
                if (spf[j] == j)
                    spf[j] = i;
}
  
// Function to find sum 
// of digits in a number
static int Digit_Sum(int copy)
{
    int d = 0;
    while (copy > 0) 
    {
        d += copy % 10;
        copy /= 10;
    }
  
    return d;
}
  
// find sum of digits of 
// all numbers up to MAXN
static void Sum_Of_All_Digits()
{
    for (int n = 2; n < MAXN; n++)
    {
        // add sum of digits of least 
        // prime factor and n/spf[n]
        sum_digits[n] = sum_digits[n / spf[n]] +
                              Digit_Sum(spf[n]);
  
        // if it is valid make 
        // isValid true
        if (Digit_Sum(n) == sum_digits[n])
            isValid[n] = true;
    }
  
    // prefix sum to compute answer
    for (int n = 2; n < MAXN; n++) 
    {
        if (isValid[n])
            ans[n] = 1;
        ans[n] += ans[n - 1];
    }
}
  
// Driver code
public static void Main () 
{
    Smallest_prime_factor();
    Sum_Of_All_Digits();
      
    // declaration
    int l, r;
      
    // print answer for required range
    l = 2; r = 3;
    Console.WriteLine("Valid numbers in the range "
                              l + " " + r + " are "
                             (ans[r] - ans[l - 1] ));
      
    // print answer for required range
    l = 2; r = 10;
    Console.WriteLine("Valid numbers in the range "
                              l + " " + r + " are "
                              (ans[r] - ans[l - 1]));
}
}
  
// This code is contributed
// by Subhadeep

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Output:

Valid numbers in the range 2 3 are 2
Valid numbers in the range 2 10 are 5


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