Lucky Numbers - GeeksforGeeks

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
1,2,3,4,5,6,7,8,9,10,11,12,14,15,16,17,18,19,……

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)

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

Algorithm:
Before every iteration, if we calculate position of the given no, then in a given iteration, we can determine if the no will be deleted. Suppose calculated position for the given no. is P before some iteration, and each Ith no. is going to be removed in this iteration, if P < I then input no is lucky, if P is such that P%I == 0 (I is a divisor of P), then input no is not lucky.

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;  
      
    /*variable next_position is just for  
    readability of the program we can 
    remove it and use n only */
    int next_position = n;  
    if(counter > n)  
        return 1;  
    if(n % counter == 0)  
        return 0;  
      
    /*calculate next position of input no*/
    next_position -= next_position / 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. ohterwise returns 0*/
bool isLucky(int n) 
{ 
    static int counter = 2; 
      
    /*variable next_position is just for readability of 
        the program we can remove it and use n only */
    int next_position = n; 
    if(counter > n) 
        return 1; 
    if(n%counter == 0) 
        return 0;      
      
    /*calculate next position of input no*/
    next_position -= next_position/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. ohterwise returns 0 
    static boolean isLucky(int n) 
    { 
        // variable next_position is just for readability of 
        // the program we can remove it and use n only  
        int next_position = n; 
        if(counter > n) 
            return true; 
        if(n%counter == 0) 
            return false;       
   
        // calculate next position of input no 
        next_position -= next_position/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 
      
    # just for readability, can be removed and used n instead 
    next_position = n 
      
    if isLucky.counter > n: 
        return 1
    if n % isLucky.counter == 0: 
        return 0
      
    # Calculate next position of input number 
    next_position = next_position - next_position / 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 

Output:

5 is not a lucky no.

C#

// C# program to check Lucky Number 
using System; 
  
class GFG { 
      
    public static int counter = 2;  
  
    // Returns 1 if n is a lucky no. 
    // ohterwise returns 0 
    static bool isLucky(int n) 
    { 
          
        // variable next_position is  
        // just for readability of 
        // the program we can remove 
        // it and use n only  
        int next_position = n; 
          
        if(counter > n) 
            return true; 
        if(n % counter == 0) 
            return false;      
  
        // calculate next position of 
        // input no 
        next_position -= next_position  
                            / 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. ohterwise returns 0 */
function isLucky($n) 
{ 
    $counter = 2; 
      
    /* variable next_position is  
       just for readability of 
       the program we can remove  
       it and use n only */
    $next_position = $n; 
      
    if($counter > $n) 
        return 1; 
    if($n % $counter == 0) 
        return 0;  
      
    /* calculate next position 
       of input no */
    $next_position -= $next_position / $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. 
?> 

Output:

5 is not a lucky no.

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.

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My Personal Notes

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

C++

C

Java

Python

C#

PHP

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