Brilliant Numbers

Brilliant Number is a number N which is the product of two primes with the same number of digits.
Few Brilliant numbers are: 

4, 6, 9, 10, 14, 15, 21, 25, 35, 49…. 

Check if N is a Brilliant number

Given a number N, the task is to check if N is a Brilliant Number or not. If N is a Brilliant Number then print “Yes” else print “No”.
Examples:

Input: N = 1711 
Output: Yes 
Explanation:
1711 = 29*59 and both 29 and 59 have two digits.
Input: N = 16 
Output: No 

Approach: The idea is to find all the primes less than or equal to the given number N using Sieve of Eratosthenes. Once we have an array that tells all primes, we can traverse through this array to find a pair with a given product. We will find Two Prime Numbers with given product using the sieve of Eratosthenes and check if the pair has the same number of digits or not.
Below is the implementation of the above approach:

C++

// C++ implementation for the
// above approach

#include <bits/stdc++.h>
using namespace std;

// Function to generate all prime
// numbers less than n
bool SieveOfEratosthenes(int n,
                bool isPrime[])
{
    // Initialize all entries of
    // boolean array as true.
    // A value in isPrime[i]
    // will finally be false
    // if i is Not a prime
    isPrime[0] = isPrime[1] = false;
    for (int i = 2; i <= n; i++)
        isPrime[i] = true;

    for (int p = 2; p * p <= n; p++) {

        // If isPrime[p] is not changed,
        // then it is a prime
        if (isPrime[p] == true) {

            // Update all multiples of p
            for (int i = p * 2; i <= n; i += p)
                isPrime[i] = false;
        }
    }
}

// Function to return the number
// of digits in a number
int countDigit(long long n)
{
    return floor(log10(n) + 1);
}

// Function to check if N is a
// Brilliant number
bool isBrilliant(int n)
{
    int flag = 0;

    // Generating primes using Sieve
    bool isPrime[n + 1];
    SieveOfEratosthenes(n, isPrime);

    // Traversing all numbers
    // to find first pair
    for (int i = 2; i < n; i++) {
        int x = n / i;

        if (isPrime[i] &&
          isPrime[x] and x * i == n) {
            if (countDigit(i) == countDigit(x))
                return true;
        }
    }

    return false;
}

// Driver Code
int main()
{
    // Given Number
    int n = 1711;

    // Function Call
    if (isBrilliant(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java implementation for the
// above approach
import java.util.*;
class GFG{

// Function to generate all prime
// numbers less than n
static void SieveOfEratosthenes(int n,
                    boolean isPrime[])
{
    // Initialize all entries of
    // boolean array as true.
    // A value in isPrime[i]
    // will finally be false
    // if i is Not a prime
    isPrime[0] = isPrime[1] = false;
    for (int i = 2; i <= n; i++)
        isPrime[i] = true;

    for (int p = 2; p * p <= n; p++)
    {

        // If isPrime[p] is not changed,
        // then it is a prime
        if (isPrime[p] == true)
        {

            // Update all multiples of p
            for (int i = p * 2; i <= n; i += p)
                isPrime[i] = false;
        }
    }
}

// Function to return the number
// of digits in a number
static int countDigit(int n)
{
    int count = 0;
        while (n != 0)
        {
            n = n / 10;
            ++count;
        }
        return count;
}

// Function to check if N is a
// Brilliant number
static boolean isBrilliant(int n)
{
    int flag = 0;

    // Generating primes using Sieve
    boolean isPrime[] = new boolean[n + 1];
    SieveOfEratosthenes(n, isPrime);

    // Traversing all numbers
    // to find first pair
    for (int i = 2; i < n; i++)
    {
        int x = n / i;

        if (isPrime[i] &&
        isPrime[x] && (x * i) == n)
        {
            if (countDigit(i) == countDigit(x))
                return true;
        }
    }
    return false;
}

// Driver Code
public static void main (String[] args)
{
    // Given Number
    int n = 1711;

    // Function Call
    if (isBrilliant(n))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}

// This code is contributed by Ritik Bansal

Python3

# Python3 program for the
# above approach
import math

# Function to generate all prime
# numbers less than n
def SieveOfEratosthenes(n, isPrime):

    # Initialize all entries of
    # boolean array as true.
    # A value in isPrime[i]
    # will finally be false
    # if i is Not a prime
    isPrime[0] = isPrime[1] = False

    for i in range(2, n + 1, 1):
        isPrime[i] = True

    p = 2
    while(p * p <= n ):

        # If isPrime[p] is not changed,
        # then it is a prime
        if (isPrime[p] == True):

            # Update all multiples of p
            for i in range(p * 2, n + 1, p):
                isPrime[i] = False

        p += 1

# Function to return the number
# of digits in a number
def countDigit(n):

    return math.floor(math.log10(n) + 1)

# Function to check if N is a
# Brilliant number
def isBrilliant(n):

    flag = 0

    # Generating primes using Sieve
    isPrime = [0] * (n + 1)
    SieveOfEratosthenes(n, isPrime)

    # Traversing all numbers
    # to find first pair
    for i in range(2, n, 1):
        x = n // i

        if (isPrime[i] and
            isPrime[x] and x * i == n):
            if (countDigit(i) == countDigit(x)):
                return True

    return False

# Driver Code

# Given Number
n = 1711

# Function Call
if (isBrilliant(n)):
    print("Yes")
else:
    print("No")

# This code is contributed by sanjoy_62

C#

// C# implementation for the
// above approach
using System;
class GFG{

// Function to generate all prime
// numbers less than n
static void SieveOfEratosthenes(int n,
                       bool []isPrime)
{

    // Initialize all entries of
    // boolean array as true.
    // A value in isPrime[i]
    // will finally be false
    // if i is Not a prime
    isPrime[0] = isPrime[1] = false;
    for(int i = 2; i <= n; i++)
       isPrime[i] = true;

    for(int p = 2; p * p <= n; p++)
    {

       // If isPrime[p] is not changed,
       // then it is a prime
       if (isPrime[p] == true)
       {

           // Update all multiples of p
           for(int i = p * 2; i <= n; i += p)
              isPrime[i] = false;
       }
    }
}

// Function to return the number
// of digits in a number
static int countDigit(int n)
{
    int count = 0;
    while (n != 0)
    {
        n = n / 10;
        ++count;
    }
    return count;
}

// Function to check if N is a
// Brilliant number
static bool isBrilliant(int n)
{
    //int flag = 0;

    // Generating primes using Sieve
    bool []isPrime = new bool[n + 1];
    SieveOfEratosthenes(n, isPrime);

    // Traversing all numbers
    // to find first pair
    for(int i = 2; i < n; i++)
    {
       int x = n / i;

       if (isPrime[i] &&
           isPrime[x] && (x * i) == n)
       {
           if (countDigit(i) == countDigit(x))
               return true;
       }
    }
    return false;
}

// Driver Code
public static void Main()
{
    // Given Number
    int n = 1711;

    // Function Call
    if (isBrilliant(n))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by Code_Mech

Javascript

<script>
// Javascript implementation for the
// above approach


    // Function to generate all prime
    // numbers less than n
    function SieveOfEratosthenes( n, isPrime) {
        // Initialize all entries of
        // let array as true.
        // A value in isPrime[i]
        // will finally be false
        // if i is Not a prime
        isPrime[0] = isPrime[1] = false;
        for ( let i = 2; i <= n; i++)
            isPrime[i] = true;

        for (let  p = 2; p * p <= n; p++) {

            // If isPrime[p] is not changed,
            // then it is a prime
            if (isPrime[p] == true) {

                // Update all multiples of p
                for (let  i = p * 2; i <= n; i += p)
                    isPrime[i] = false;
            }
        }
    }

    // Function to return the number
    // of digits in a number
    function countDigit( n) {
        let count = 0;
        while (n != 0) {
            n = parseInt(n / 10);
            ++count;
        }
        return count;
    }

    // Function to check if N is a
    // Brilliant number
    function isBrilliant( n) {
        let flag = 0;

        // Generating primes using Sieve
        let isPrime = Array(n + 1).fill(true);
        SieveOfEratosthenes(n, isPrime);

        // Traversing all numbers
        // to find first pair
        for ( let i = 2; i < n; i++) {
            let x = n / i;

            if (isPrime[i] && isPrime[x] && (x * i) == n) {
                if (countDigit(i) == countDigit(x))
                    return true;
            }
        }
        return false;
    }

    // Driver Code

        // Given Number
        let n = 1711;

        // Function Call
        if (isBrilliant(n))
            document.write("Yes");
        else
            document.write("No");

// This code contributed by Rajput-Ji

</script>

Output:

Yes

Time Complexity: O(n)

Auxiliary Space: O(n)

Reference: http://oeis.org/A078972