Given a number N, the task is to check whether it is fascinating or not.
Fascinating Number: When a number( 3 digits or more ) is multiplied by 2 and 3, and when both these products are concatenated with the original number, then it results in all digits from 1 to 9 present exactly once. There could be any number of zeros and are ignored.
Examples:
Input: 192
Output: Yes
After multiplication with 2 and 3, and concatenating with original number, resultant number is 192384576 which contains all digits from 1 to 9.
Input: 853
Output: No
After multiplication with 2 and 3, and concatenating with original number, the resultant number is 85317062559. In this, number 4 is missing and the number 5 has appeared multiple times.
Try It Yourself
Approach:
- Check if the given number has three digits or more. If not, print No.
- Else, Multiply the given number with 2 and 3.
- Concatenate these products with the given number to form a string.
- Traverse this string, keep the frequency count of the digits.
- Print No if any digit is missing or has appeared multiple times.
- Else, print Yes.
Below is the implementation of above approach:
// C++ program to implement
// fascinating number
#include <bits/stdc++.h>
using namespace std;
// function to check if number
// is fascinating or not
bool isFascinating(int num)
{
// frequency count array
// using 1 indexing
int freq[10] = {0};
// obtaining the resultant number
// using string concatenation
string val = "" + to_string(num) +
to_string(num * 2) +
to_string(num * 3);
// Traversing the string
// character by character
for (int i = 0; i < val.length(); i++)
{
// gives integer value of
// a character digit
int digit = val[i] - '0';
// To check if any digit has
// appeared multiple times
if (freq[digit] and digit != 0 > 0)
return false;
else
freq[digit]++;
}
// Traversing through freq array to
// check if any digit was missing
for (int i = 1; i < 10; i++)
{
if (freq[i] == 0)
return false;
}
return true;
}
// Driver code
int main()
{
// Input number
int num = 192;
// Not a valid number
if (num < 100)
cout << "No" << endl;
else
{
// Calling the function to
// check if input number
// is fascinating or not
bool ans = isFascinating(num);
if (ans)
cout << "Yes";
else
cout << "No";
}
}
// This code is contributed
// by Subhadeep
// Java program to implement
// fascinating number
import java.io.*;
import java.util.*;
public class GFG {
// function to check if number
// is fascinating or not
public static boolean isFascinating(
int num)
{
// frequency count array
//using 1 indexing
int[] freq = new int[10];
// obtaining the resultant number
// using string concatenation
String val = "" + num + num * 2 +
num * 3;
// Traversing the string character //by character
for (int i = 0; i < val.length(); i++)
{
// gives integer value of //a character digit
int digit = val.charAt(i) - '0';
// To check if any digit has
// appeared multiple times
if (freq[digit]>0 && digit != 0)
return false;
else
freq[digit]++;
}
// Traversing through freq array to
// check if any digit was missing
for (int i = 1; i < freq.length; i++)
{
if (freq[i] == 0)
return false;
}
return true;
}
// Driver code
public static void main(String args[])
{
// Input number
int num = 192;
// Not a valid number
if (num < 100)
System.out.println("No");
else
{
// Calling the function to check
// if input number is fascinating or not
boolean ans = isFascinating(num);
if (ans)
System.out.println("Yes");
else
System.out.println("No");
}
}
}
# Python 3 program to implement
# fascinating number
# function to check if number
# is fascinating or not
def isFascinating(num) :
# frequency count array
# using 1 indexing
freq = [0] * 10
# obtaining the resultant number
# using string concatenation
val = (str(num) + str(num * 2) +
str(num * 3))
# Traversing the string
# character by character
for i in range(len(val)) :
# gives integer value of
# a character digit
digit = int(val[i])
# To check if any digit has
# appeared multiple times
if freq[digit] and digit != 0 > 0 :
return False
else :
freq[digit] += 1
# Traversing through freq array to
# check if any digit was missing
for i in range(1, 10) :
if freq[i] == 0 :
return False
return True
# Driver Code
if __name__ == "__main__" :
# Input number
num = 192
# Not a valid number
if num < 100 :
print("No")
else :
# Calling the function to
# check if input number
# is fascinating or not
ans = isFascinating(num)
if ans :
print("Yes")
else :
print("No")
# This code is contributed by ANKITRAI1
// C# program to implement
// fascinating number
using System;
class GFG
{
// function to check if number
// is fascinating or not
public static bool isFascinating(int num)
{
// frequency count array
// using 1 indexing
int[] freq = new int[10];
// obtaining the resultant number
// using string concatenation
String val = "" + num.ToString() +
(num * 2).ToString() +
(num * 3).ToString();
// Traversing the string
// character by character
for (int i = 0; i < val.Length; i++)
{
// gives integer value of
// a character digit
int digit = val[i] - '0';
// To check if any digit has
// appeared multiple times
if (freq[digit] && digit != 0 > 0 )
return false;
else
freq[digit]++;
}
// Traversing through freq array to
// check if any digit was missing
for (int i = 1; i < freq.Length; i++)
{
if (freq[i] == 0)
return false;
}
return true;
}
// Driver code
static void Main()
{
// Input number
int num = 192;
// Not a valid number
if (num < 100)
Console.WriteLine("No");
else
{
// Calling the function to check
// if input number is fascinating or not
bool ans = isFascinating(num);
if (ans)
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
}
// This code is contributed by mits
<?php
// PHP program to implement
// fascinating number
// function to check if number
// is fascinating or not
function isFascinating($num)
{
// frequency count array
// using 1 indexing
$freq = array_fill(0, 10, NULL);
// obtaining the resultant number
// using string concatenation
$val = "" . $num . ($num * 2).
($num * 3);
// Traversing the string
// character by character
for ($i = 0; $i < strlen($val); $i++)
{
// gives integer value of
// a character digit
$digit = $val[$i] - '0';
// To check if any digit has
// appeared multiple times
if ($freq[$digit] > 0 && $digit != 0)
return false;
else
$freq[$digit]++;
}
// Traversing through freq array to
// check if any digit was missing
for ($i = 1; $i < 10; $i++)
{
if ($freq[$i] == 0)
return false;
}
return true;
}
// Driver code
// Input number
$num = 192;
// Not a valid number
if ($num < 100)
echo "No" ;
else
{
// Calling the function to
// check if input number
// is fascinating or not
$ans = isFascinating($num);
if ($ans)
echo "Yes";
else
echo "No";
}
// This code is contributed
// by ChitraNayal
?>
<script>
// Javascript program to implement
// fascinating number
// function to check if number
// is fascinating or not
function isFascinating(num)
{
// frequency count array
//using 1 indexing
let freq = new Array(10);
for(let i=0;i<freq.length;i++)
{
freq[i]=0;
}
// obtaining the resultant number
// using string concatenation
let val = "" + num + num * 2 +
num * 3;
// Traversing the string character //by character
for (let i = 0; i < val.length; i++)
{
// gives integer value of //a character digit
let digit = val[i].charCodeAt(0) -
'0'.charCodeAt(0);
// To check if any digit has
// appeared multiple times
if (freq[digit]>0 && digit != 0)
return false;
else
freq[digit]++;
}
// Traversing through freq array to
// check if any digit was missing
for (let i = 1; i < freq.length; i++)
{
if (freq[i] == 0)
return false;
}
return true;
}
// Driver code
// Input number
let num = 192;
// Not a valid number
if (num < 100)
document.write("No");
else
{
// Calling the function to check
// if input number is fascinating or not
let ans = isFascinating(num);
if (ans)
document.write("Yes");
else
document.write("No");
}
// This code is contributed by rag2127
</script>
Output:
Yes
Time Complexity: O(|num|)
Auxiliary Space: O(10)