Given a number n, find n’th smart number (1<=n<=1000). Smart number is a number which has at least three distinct prime factors. We are given an upper limit on value of result as MAX For example 30 is 1st smart number because it has 2, 3, 5 as it’s distinct prime factors. 42 is 2nd smart number because it has 2, 3, 7 as it’s distinct prime factors. Examples:
Input : n = 1 Output: 30 // three distinct prime factors 2, 3, 5 Input : n = 50 Output: 273 // three distinct prime factors 3, 7, 13 Input : n = 1000 Output: 2664 // three distinct prime factors 2, 3, 37
The idea is based on Sieve of Eratosthenes. We use an array to use an array prime[] to keep track of prime numbers. We also use the same array to keep track of the count of prime factors seen so far. Whenever the count reaches 3, we add the number to result.
- Take an array primes[] and initialize it with 0.
- Now we know that first prime number is i = 2 so mark primes[2] = 1 i.e; primes[i] = 1 indicates that ‘i’ is prime number.
- Now traverse the primes[] array and mark all multiples of ‘i’ by condition primes[j] -= 1 where ‘j’ is multiple of ‘i’, and check the condition primes[j]+3 = 0 because whenever primes[j] become -3 it indicates that previously it had been multiple of three distinct prime factors. If condition primes[j]+3=0 becomes true that means ‘j’ is a Smart Number so store it in a array result[].
- Now sort array result[] and return result[n-1].
Below is the implementation of above idea.
// C++ implementation to find n'th smart number #include<bits/stdc++.h> using namespace std;
// Limit on result const int MAX = 3000;
// Function to calculate n'th smart number int smartNumber( int n)
{ // Initialize all numbers as not prime
int primes[MAX] = {0};
// iterate to mark all primes and smart number
vector< int > result;
// Traverse all numbers till maximum limit
for ( int i=2; i<MAX; i++)
{
// 'i' is maked as prime number because
// it is not multiple of any other prime
if (primes[i] == 0)
{
primes[i] = 1;
// mark all multiples of 'i' as non prime
for ( int j=i*2; j<MAX; j=j+i)
{
primes[j] -= 1;
// If i is the third prime factor of j
// then add it to result as it has at
// least three prime factors.
if ( (primes[j] + 3) == 0)
result.push_back(j);
}
}
}
// Sort all smart numbers
sort(result.begin(), result.end());
// return n'th smart number
return result[n-1];
} // Driver program to run the case int main()
{ int n = 50;
cout << smartNumber(n);
return 0;
} |
// Java implementation to find n'th smart number import java.util.*;
import java.lang.*;
class GFG {
// Limit on result
static int MAX = 3000 ;
// Function to calculate n'th smart number
public static int smartNumber( int n)
{
// Initialize all numbers as not prime
Integer[] primes = new Integer[MAX];
Arrays.fill(primes, new Integer( 0 ));
// iterate to mark all primes and smart
// number
Vector<Integer> result = new Vector<>();
// Traverse all numbers till maximum
// limit
for ( int i = 2 ; i < MAX; i++)
{
// 'i' is maked as prime number
// because it is not multiple of
// any other prime
if (primes[i] == 0 )
{
primes[i] = 1 ;
// mark all multiples of 'i'
// as non prime
for ( int j = i* 2 ; j < MAX; j = j+i)
{
primes[j] -= 1 ;
// If i is the third prime
// factor of j then add it
// to result as it has at
// least three prime factors.
if ( (primes[j] + 3 ) == 0 )
result.add(j);
}
}
}
// Sort all smart numbers
Collections.sort(result);
// return n'th smart number
return result.get(n- 1 );
}
// Driver program to run the case
public static void main(String[] args)
{
int n = 50 ;
System.out.println(smartNumber(n));
}
} // This code is contributed by Prasad Kshirsagar |
# Python3 implementation to find # n'th smart number # Limit on result MAX = 3000 ;
# Function to calculate n'th # smart number def smartNumber(n):
# Initialize all numbers as not prime
primes = [ 0 ] * MAX ;
# iterate to mark all primes
# and smart number
result = [];
# Traverse all numbers till maximum limit
for i in range ( 2 , MAX ):
# 'i' is maked as prime number because
# it is not multiple of any other prime
if (primes[i] = = 0 ):
primes[i] = 1 ;
# mark all multiples of 'i' as non prime
j = i * 2 ;
while (j < MAX ):
primes[j] - = 1 ;
# If i is the third prime factor of j
# then add it to result as it has at
# least three prime factors.
if ( (primes[j] + 3 ) = = 0 ):
result.append(j);
j = j + i;
# Sort all smart numbers
result.sort();
# return n'th smart number
return result[n - 1 ];
# Driver Code n = 50 ;
print (smartNumber(n));
# This code is contributed by mits |
// C# implementation to find n'th smart number using System.Collections.Generic;
class GFG {
// Limit on result
static int MAX = 3000;
// Function to calculate n'th smart number
public static int smartNumber( int n)
{
// Initialize all numbers as not prime
int [] primes = new int [MAX];
// iterate to mark all primes and smart
// number
List< int > result = new List< int >();
// Traverse all numbers till maximum
// limit
for ( int i = 2; i < MAX; i++)
{
// 'i' is maked as prime number
// because it is not multiple of
// any other prime
if (primes[i] == 0)
{
primes[i] = 1;
// mark all multiples of 'i'
// as non prime
for ( int j = i*2; j < MAX; j = j+i)
{
primes[j] -= 1;
// If i is the third prime
// factor of j then add it
// to result as it has at
// least three prime factors.
if ( (primes[j] + 3) == 0)
result.Add(j);
}
}
}
// Sort all smart numbers
result.Sort();
// return n'th smart number
return result[n-1];
}
// Driver program to run the case
public static void Main()
{
int n = 50;
System.Console.WriteLine(smartNumber(n));
}
} // This code is contributed by mits |
<?php // PHP implementation to find n'th smart number // Limit on result $MAX = 3000;
// Function to calculate n'th smart number function smartNumber( $n )
{ global $MAX ;
// Initialize all numbers as not prime
$primes = array_fill (0, $MAX ,0);
// iterate to mark all primes and smart number
$result = array ();
// Traverse all numbers till maximum limit
for ( $i =2; $i < $MAX ; $i ++)
{
// 'i' is maked as prime number because
// it is not multiple of any other prime
if ( $primes [ $i ] == 0)
{
$primes [ $i ] = 1;
// mark all multiples of 'i' as non prime
for ( $j = $i *2; $j < $MAX ; $j = $j + $i )
{
$primes [ $j ] -= 1;
// If i is the third prime factor of j
// then add it to result as it has at
// least three prime factors.
if ( ( $primes [ $j ] + 3) == 0)
array_push ( $result , $j );
}
}
}
// Sort all smart numbers
sort( $result );
// return n'th smart number
return $result [ $n -1];
} // Driver program to run the case $n = 50;
echo smartNumber( $n );
// This code is contributed by mits ?> |
<script> // JavaScript implementation to find n'th smart number // Limit on result const MAX = 3000; // Function to calculate n'th smart number function smartNumber(n)
{ // Initialize all numbers as not prime
let primes = new Array(MAX).fill(0);
// iterate to mark all primes and smart number
let result = [];
// Traverse all numbers till maximum limit
for (let i=2; i<MAX; i++)
{
// 'i' is maked as prime number because
// it is not multiple of any other prime
if (primes[i] == 0)
{
primes[i] = 1;
// mark all multiples of 'i' as non prime
for (let j=i*2; j<MAX; j=j+i)
{
primes[j] -= 1;
// If i is the third prime factor of j
// then add it to result as it has at
// least three prime factors.
if ( (primes[j] + 3) == 0)
result.push(j);
}
}
}
// Sort all smart numbers
result.sort((a,b)=>a-b);
// return n'th smart number
return result[n-1];
} // Driver program to run the case let n = 50; document.write(smartNumber(n)); // This code is contributed by shinjanpatra </script> |
Output:
273
Time Complexity: O(MAX)
Auxiliary Space: O(MAX)