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 Codeint 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; 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 Numberimport 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) { 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 0def 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 CodeisLucky.counter = 2 # Acts as static variablex = 5if 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 Numberusing System;class GFG { public static int counter = 2; // Returns 1 if n is a lucky no. // ohterwise 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. ohterwise 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> |
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:
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.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
In case you wish to attend live classes with industry experts, please refer DSA Live Classes
