Given a number 'n', check whether it is a hoax number or not.
A Hoax Number is defined as a composite number, whose sum of digits is equal to the sum of digits of its distinct prime factors. It may be noted here that, 1 is not considered a prime number, hence it is not included in the sum of digits of distinct prime factors.
Examples :
Input : 22
Output : A Hoax Number
Explanation : The distinct prime factors of 22
are 2 and 11. The sum of their digits are 4,
i.e 2 + 1 + 1 and sum of digits of 22 is also 4.
Input : 84
Output : A Hoax Number
Explanation : The distinct prime factors of 84
are 2, 3 and 7. The sum of their digits are 12,
i.e 2 + 3 + 4 and sum of digits of 84 is also 12.
Input : 19
Output : Not a Hoax Number
Explanation : By definition, a hoax number is
a composite number.
The definition of Hoax numbers bears close resemblance with the definition of Smith Numbers. In fact, some of the Hoax numbers are also Smith numbers. It is apparent that those hoax numbers that do not have repeated factors in their prime decomposition, i.e square free number are also eligible Smith numbers.
Implementation
1) First generate all distinct prime factors of the number 'n'.
2) If the 'n' is not a prime number, find the sum of digits of the factors obtained in step 1.
3) Find the sum of digits of 'n'.
4) Check if the sums obtained in steps 2 and 3 are equal or not.
5) If the sums are equal, 'n' is a hoax number.
C++
// CPP code to check if a number is a hoax
// number or not.
#include <bits/stdc++.h>
using namespace std;
// Function to find distinct prime factors
// of given number n
vector<int> primeFactors(int n)
{
vector<int> res;
if (n % 2 == 0) {
while (n % 2 == 0)
n = n / 2;
res.push_back(2);
}
// n is odd at this point, since it is no
// longer divisible by 2. So we can test
// only for the odd numbers, whether they
// are factors of n
for (int i = 3; i <= sqrt(n); i = i + 2) {
// Check if i is prime factor
if (n % i == 0) {
while (n % i == 0)
n = n / i;
res.push_back(i);
}
}
// This condition is to handle the case
// when n is a prime number greater than 2
if (n > 2)
res.push_back(n);
return res;
}
// Function to calculate sum of digits of
// distinct prime factors of given number n
// and sum of digits of number n and compare
// the sums obtained
bool isHoax(int n)
{
// Distinct prime factors of n are being
// stored in vector pf
vector<int> pf = primeFactors(n);
// If n is a prime number, it cannot be a
// hoax number
if (pf[0] == n)
return false;
// Finding sum of digits of distinct prime
// factors of the number n
int all_pf_sum = 0;
for (int i = 0; i < pf.size(); i++) {
// Finding sum of digits in current
// prime factor pf[i].
int pf_sum;
for (pf_sum = 0; pf[i] > 0;
pf_sum += pf[i] % 10, pf[i] /= 10)
;
all_pf_sum += pf_sum;
}
// Finding sum of digits of number n
int sum_n;
for (sum_n = 0; n > 0; sum_n += n % 10,
n /= 10)
;
// Comparing the two calculated sums
return sum_n == all_pf_sum;
}
// Driver Method
int main()
{
int n = 84;
if (isHoax(n))
cout << "A Hoax Number\n";
else
cout << "Not a Hoax Number\n";
return 0;
}
// Java code to check if a number is
// a hoax number or not.
import java.io.*;
import java.util.*;
public class GFG {
// Function to find distinct
// prime factors of given
// number n
static List<Integer> primeFactors(int n)
{
List<Integer> res = new ArrayList<Integer>();
if (n % 2 == 0)
{
while (n % 2 == 0)
n = n / 2;
res.add(2);
}
// n is odd at this point,
// since it is no longer
// divisible by 2. So we
// can test only for the
// odd numbers, whether they
// are factors of n
for (int i = 3; i <= Math.sqrt(n);
i = i + 2)
{
// Check if i is prime factor
if (n % i == 0)
{
while (n % i == 0)
n = n / i;
res.add(i);
}
}
// This condition is to
// handle the case when
// n is a prime number
// greater than 2
if (n > 2)
res.add(n);
return res;
}
// Function to calculate
// sum of digits of distinct
// prime factors of given
// number n and sum of
// digits of number n and
// compare the sums obtained
static boolean isHoax(int n)
{
// Distinct prime factors
// of n are being
// stored in vector pf
List<Integer> pf = primeFactors(n);
// If n is a prime number,
// it cannot be a hoax number
if (pf.get(0) == n)
return false;
// Finding sum of digits of distinct
// prime factors of the number n
int all_pf_sum = 0;
for (int i = 0; i < pf.size(); i++)
{
// Finding sum of digits in current
// prime factor pf[i].
int pf_sum;
for (pf_sum = 0; pf.get(i) > 0;
pf_sum += pf.get(i) % 10,
pf.set(i,pf.get(i) / 10));
all_pf_sum += pf_sum;
}
// Finding sum of digits of number n
int sum_n;
for (sum_n = 0; n > 0; sum_n += n % 10,
n /= 10)
;
// Comparing the two calculated sums
return sum_n == all_pf_sum;
}
// Driver Code
public static void main(String args[])
{
int n = 84;
if (isHoax(n))
System.out.print( "A Hoax Number\n");
else
System.out.print("Not a Hoax Number\n");
}
}
// This code is contributed by
// Manish Shaw (manishshaw1)
# Python3 code to check if a number is a hoax
# number or not.
import math
# Function to find distinct prime factors
# of given number n
def primeFactors(n) :
res = []
if (n % 2 == 0) :
while (n % 2 == 0) :
n = int(n / 2)
res.append(2)
# n is odd at this point, since it is no
# longer divisible by 2. So we can test
# only for the odd numbers, whether they
# are factors of n
for i in range(3,int(math.sqrt(n)),2):
# Check if i is prime factor
if (n % i == 0) :
while (n % i == 0) :
n = int(n / i)
res.append(i)
# This condition is to handle the case
# when n is a prime number greater than 2
if (n > 2) :
res.append(n)
return res
# Function to calculate sum of digits of
# distinct prime factors of given number n
# and sum of digits of number n and compare
# the sums obtained
def isHoax(n) :
# Distinct prime factors of n are being
# stored in vector pf
pf = primeFactors(n)
# If n is a prime number, it cannot be a
# hoax number
if (pf[0] == n) :
return False
# Finding sum of digits of distinct prime
# factors of the number n
all_pf_sum = 0
for i in range(0,len(pf)):
# Finding sum of digits in current
# prime factor pf[i].
pf_sum = 0
while (pf[i] > 0):
pf_sum += pf[i] % 10
pf[i] = int(pf[i] / 10)
all_pf_sum += pf_sum
# Finding sum of digits of number n
sum_n = 0;
while (n > 0):
sum_n += n % 10
n = int(n / 10)
# Comparing the two calculated sums
return sum_n == all_pf_sum
# Driver Method
n = 84;
if (isHoax(n)):
print ("A Hoax Number\n")
else:
print ("Not a Hoax Number\n")
# This code is contributed by Manish Shaw
# (manishshaw1)
// C# code to check if a number is
// a hoax number or not.
using System;
using System.Collections.Generic;
class GFG {
// Function to find distinct
// prime factors of given
// number n
static List<int> primeFactors(int n)
{
List<int> res = new List<int>();
if (n % 2 == 0)
{
while (n % 2 == 0)
n = n / 2;
res.Add(2);
}
// n is odd at this point,
// since it is no longer
// divisible by 2. So we
// can test only for the
// odd numbers, whether they
// are factors of n
for (int i = 3; i <= Math.Sqrt(n);
i = i + 2)
{
// Check if i is prime factor
if (n % i == 0)
{
while (n % i == 0)
n = n / i;
res.Add(i);
}
}
// This condition is to
// handle the case when
// n is a prime number
// greater than 2
if (n > 2)
res.Add(n);
return res;
}
// Function to calculate
// sum of digits of distinct
// prime factors of given
// number n and sum of
// digits of number n and
// compare the sums obtained
static bool isHoax(int n)
{
// Distinct prime factors
// of n are being
// stored in vector pf
List<int> pf = primeFactors(n);
// If n is a prime number,
// it cannot be a hoax number
if (pf[0] == n)
return false;
// Finding sum of digits of distinct
// prime factors of the number n
int all_pf_sum = 0;
for (int i = 0; i < pf.Count; i++)
{
// Finding sum of digits in current
// prime factor pf[i].
int pf_sum;
for (pf_sum = 0; pf[i] > 0;
pf_sum += pf[i] % 10, pf[i] /= 10);
all_pf_sum += pf_sum;
}
// Finding sum of digits of number n
int sum_n;
for (sum_n = 0; n > 0; sum_n += n % 10,
n /= 10)
;
// Comparing the two calculated sums
return sum_n == all_pf_sum;
}
// Driver Code
public static void Main()
{
int n = 84;
if (isHoax(n))
Console.Write( "A Hoax Number\n");
else
Console.Write("Not a Hoax Number\n");
}
}
// This code is contributed by
// Manish Shaw (manishshaw1)
<script>
// Javascript code to check if a number is a hoax
// number or not.
// Function to find distinct prime factors
// of given number n
function primeFactors(n)
{
var res =[];
if (n % 2 == 0) {
while (n % 2 == 0)
n = parseInt(n / 2);
res.push(2);
}
// n is odd at this point, since it is no
// longer divisible by 2. So we can test
// only for the odd numbers, whether they
// are factors of n
for (var i = 3; i <= Math.sqrt(n); i = i + 2) {
// Check if i is prime factor
if (n % i == 0) {
while (n % i == 0)
n = parseInt(n / i);
res.push(i);
}
}
// This condition is to handle the case
// when n is a prime number greater than 2
if (n > 2)
res.push(n);
return res;
}
// Function to calculate sum of digits of
// distinct prime factors of given number n
// and sum of digits of number n and compare
// the sums obtained
function isHoax(n)
{
// Distinct prime factors of n are being
// stored in vector pf
var pf = primeFactors(n);
// If n is a prime number, it cannot be a
// hoax number
if (pf[0] == n)
return false;
// Finding sum of digits of distinct prime
// factors of the number n
var all_pf_sum = 0;
for (var i = 0; i < pf.length; i++) {
// Finding sum of digits in current
// prime factor pf[i].
var pf_sum;
for (pf_sum = 0; pf[i] > 0;
pf_sum += pf[i] % 10, pf[i] = parseInt(pf[i]/10))
;
all_pf_sum += pf_sum;
}
// Finding sum of digits of number n
var sum_n;
for (sum_n = 0; n > 0; sum_n += n % 10,
n = parseInt(n/10))
;
// Comparing the two calculated sums
return sum_n == all_pf_sum;
}
// Driver Method
var n = 84;
if (isHoax(n))
document.write( "A Hoax Number");
else
document.write( "Not a Hoax Number");
// This code is contributed by rrrtnx.
</script>
<?php
// PHP code to check if a number
// is a hoax number or not.
// Function to find distinct prime
// factors of given number n
function primeFactors($n)
{
$res = array();
if ($n % 2 == 0)
{
while ($n % 2 == 0)
$n = (int)$n / 2;
array_push($res, 2);
}
// n is odd at this point, since
// it is no longer divisible by 2.
// So we can test only for the odd
// numbers, whether they are factors of n
for ($i = 3; $i <= sqrt($n); $i = $i + 2)
{
// Check if i is prime factor
if ($n % $i == 0)
{
while ($n % $i == 0)
$n = (int)$n / $i;
array_push($res, $i);
}
}
// This condition is to handle
// the case when n is a prime
// number greater than 2
if ($n > 2)
array_push($res, $n);
return $res;
}
// Function to calculate sum
// of digits of distinct prime
// factors of given number n
// and sum of digits of number
// n and compare the sums obtained
function isHoax($n)
{
// Distinct prime factors
// of n are being stored
// in vector pf
$pf = primeFactors($n);
// If n is a prime number,
// it cannot be a hoax number
if ($pf[0] == $n)
return false;
// Finding sum of digits of distinct
// prime factors of the number n
$all_pf_sum = 0;
for ($i = 0; $i < count($pf); $i++)
{
// Finding sum of digits in
// current prime factor pf[i].
$pf_sum;
for ($pf_sum = 0; $pf[$i] > 0;
$pf_sum += $pf[$i] % 10,
$pf[$i] /= 10)
;
$all_pf_sum += $pf_sum;
}
// Finding sum of digits of number n
for ($sum_n = 0; $n > 0;
$sum_n += $n % 10, $n /= 10)
;
// Comparing the two calculated sums
return $sum_n == $all_pf_sum;
}
// Driver Code
$n = 84;
if (isHoax($n))
echo ("A Hoax Number\n");
else
echo ("Not a Hoax Number\n");
// This code is contributed by
// Manish Shaw(manishshaw1)
?>
Output :
A Hoax Number
Time Complexity: O(?n log n)
Auxiliary Space: O(n)
Please suggest if someone has a better solution which is more efficient in terms of space and time.
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