Fascinating Number.md

Fascinating Number

Fasinating Number

• Difficulty Level : Medium
• Last Updated : 14 May, 2021

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. 
 

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution. 
 

Approach: 
 

1. Check if the given number has three digits or more. If not, print No.
2. Else, Multiply the given number with 2 and 3.
3. Concatenate these products with the given number to form a string.
4. Traverse this string, keep the frequency count of the digits.
5. Print No if any digit is missing or has appeared multiple times.
6. 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

Output: 

Yes