Lucky Numbers

Lucky numbers are subset of integers. Rather than going into much theory, let us see the process of arriving at lucky numbers, 
Take the set of integers 
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,……
First, delete every second number, we get following reduced set. 
1,3,5,7,9,11,13,15,17,19,…………
Now, delete every third number, we get 
1, 3, 7, 9, 13, 15, 19,….….
Continue this process indefinitely…… 
Any number that does NOT get deleted due to above process is called “lucky”.
Therefore, set of lucky numbers is 1, 3, 7, 13,………
Now, given an integer ‘n’, write a function to say whether this number is lucky or not. 

    bool isLucky(int n)

Algorithm:
Before every iteration, if we calculate position of the given number, then in a given iteration, we can determine if the number will be deleted. Suppose calculated position for the given number is P before some iteration, and each Ith number is going to be removed in this iteration, i

f P < I then input number is lucky, 

if P is such that P%I == 0 (I is a divisor of P), then input no is not lucky. 

How to calculate Next position of  the number:

We know that initially the position of the number is nth itself. Now any next position will be equal to the previous position minus the number of elements (or say items) removed.

That is,  next_position = current_position – count of numbers removed

For example, take the case of n=13.

We have: Initial position: n, ie. 13 itself.  

1,2,3,4,5,6,7,8,9,10,11,12,13

Now after removing every second elements , we actually removed n/2 elements. So  now the position of 13 will be : n-n/2=13-6=7 (n=13),  i=2

1,3,5,7,9,11,13.

After that, we remove n/3 items. Note that n now is n=7. So position of 13 : n-n/3 = 7-7/3 = 7-2 = 5 (n=7),  i=3

1,3,7,9,13

So next it will be : n-n/4 = 5-5/4 = 4 (n=5),  i=4

1,3,7,13

So now i=5, but since position of 13 is 4 only, so it will be saved. Hence a lucky number!  n=4,  i=5

Recursive Way:

C++

// C++ program for Lucky Numbers
#include <bits/stdc++.h>
using namespace std;
#define bool int

/* Returns 1 if n is a lucky no.
otherwise returns 0*/
bool isLucky(int n)
{
    static int counter = 2;

    if(counter > n)
        return 1;
    if(n % counter == 0)
        return 0;

    /*calculate next position of input no.
      Variable "next_position" is just for
    readability of the program we can
    remove it and update in "n" only */
    int next_position = n - (n/counter);

    counter++;
    return isLucky(next_position);
}

// Driver Code
int main()
{
    int x = 5;
    if( isLucky(x) )
        cout << x << " is a lucky no.";
    else
        cout << x << " is not a lucky no.";
}

// This code is contributed
// by rathbhupendra

C

#include <stdio.h>
#define bool int

/* Returns 1 if n is a lucky no. otherwise returns 0*/
bool isLucky(int n)
{
    static int counter = 2;

    if(counter > n)
        return 1;
    if(n%counter == 0)
        return 0;

    /*calculate next position of input no.
      Variable "next_position" is just for
    readability of the program we can
    remove it and update in "n" only */
    int next_position = n - (n/counter);

    counter++;
    return isLucky(next_position);
}

/*Driver function to test above function*/
int main()
{
    int x = 5;
    if( isLucky(x) )
        printf("%d is a lucky no.", x);
    else
        printf("%d is not a lucky no.", x);
    getchar();
}

Java

// Java program to check Lucky Number
import java.io.*;

class GFG
{
    public static int counter = 2;

    // Returns 1 if n is a lucky no.
    // otherwise returns 0
    static boolean isLucky(int n)
    {
        if(counter > n)
            return true;
        if(n%counter == 0)
            return false;

        /*calculate next position of input no.
        Variable "next_position" is just for
        readability of the program we can
        remove it and update in "n" only */
        int next_position = n - (n/counter);

        counter++;
        return isLucky(next_position);
    }

    // driver program
    public static void main (String[] args)
    {
        int x = 5;
        if( isLucky(x) )
            System.out.println(x+" is a lucky no.");
        else
            System.out.println(x+" is not a lucky no.");
    }
}

// Contributed by Pramod Kumar

Python

# Python program to check for lucky number

# Returns 1 if n is a lucky number
# otherwise returns 0
def isLucky(n):

    # Function attribute will act
    # as static variable

    if isLucky.counter > n:
        return 1
    if n % isLucky.counter == 0:
        return 0

    #calculate next position of input no.
      #Variable "next_position" is just for
    #readability of the program we can
    #remove it and update in "n" only
    next_position = n - (n/isLucky.counter)

    isLucky.counter = isLucky.counter + 1

    return isLucky(next_position)


# Driver Code

isLucky.counter = 2 # Acts as static variable
x = 5
if isLucky(x):
    print x,"is a Lucky number"
else:
    print x,"is not a Lucky number"

# Contributed by Harshit Agrawal

C#

// C# program to check Lucky Number
using System;

class GFG {

    public static int counter = 2;

    // Returns 1 if n is a lucky no.
    // otherwise returns 0
    static bool isLucky(int n)
    {
        if(counter > n)
            return true;
        if(n % counter == 0)
            return false;

       /*calculate next position of input no.
        Variable "next_position" is just for
        readability of the program we can
        remove it and update in "n" only */
        int next_position = n - (n/counter);

        counter++;

        return isLucky(next_position);
    }

    // driver program
    public static void Main ()
    {
        int x = 5;

        if( isLucky(x) )
            Console.Write(x + " is a "
                         + "lucky no.");
        else
            Console.Write(x + " is not"
                      + " a lucky no.");
    }
}

// This code is contributed by
// nitin mittal.

PHP

<?php
// PHP program for Lucky Numbers

/* Returns 1 if n is a lucky
   no. otherwise returns 0 */
function isLucky($n)
{
    $counter = 2;

    if($counter > $n)
        return 1;
    if($n % $counter == 0)
        return 0;

    /*calculate next position of input no.
      Variable "next_position" is just for
    readability of the program we can
    remove it and update in "n" only */
    $next_position = $n - ($n / $counter);

    $counter++;
    return isLucky($next_position);
}

    // Driver Code
    $x = 5;
    if(isLucky($x) )
        echo $x ," is a lucky no.";
    else
        echo $x ," is not a lucky no.";

// This code is contributed by anuj_67.
?>

Javascript

<script>

// C++ program for Lucky Numbers

/* Returns 1 if n is a lucky no.
otherwise returns 0*/
function isLucky(n)
{
    let counter = 2;

    if(counter > n)
        return 1;
    if(n % counter == 0)
        return 0;

   /*calculate next position of input no.
      Variable "next_position" is just for
    readability of the program we can
    remove it and update in "n" only */
    let next_position = n - Math.floor(n/counter);

    counter++;
    return isLucky(next_position);
}

// Driver Code

    let x = 5;
    if( isLucky(x) )
        document.write(x + " is a lucky no.");
    else
        document.write(x + " is not a lucky no.");


// This code is contributed by Mayank Tyagi

</script>

Output

5 is not a lucky no.

Time Complexity: O(n)

Auxiliary Space: O(1)

Example:
Let’s us take an example of 19
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,15,17,18,19,20,21,…… 
1,3,5,7,9,11,13,15,17,19,….. 
1,3,7,9,13,15,19,………. 
1,3,7,13,15,19,……… 
1,3,7,13,19,………
In next step every 6th no .in sequence will be deleted. 19 will not be deleted after this step because position of 19 is 5th after this step. Therefore, 19 is lucky. Let’s see how above C code finds out: 

Current function call Position after this call Counter for next call Next Call isLucky(19 ) 10 3 isLucky(10) isLucky(10) 7 4 isLucky(7) isLucky(7) 6 5 isLucky(6) isLucky(6) 5 6 isLucky(5) 

When isLucky(6) is called, it returns 1 (because counter > n).
Please write comments if you find any bug in the given programs or other ways to solve the same problem.