Lucky numbers are a 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,………
Given an integer n, write a function to say whether this number is lucky or not.
Examples:
Input: n = 7
Output: 7 is a lucky numberInput: n = 9
Output: 9 is not a lucky number
To solve the problem follow the below idea:
Before every iteration, if we calculate the 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
if 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 the 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
Below is the recursive implementation of the above approach:
C++
// C++ program for Lucky Numbers#include <bits/stdc++.h>using namespace std;/* 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; // Function call if (isLucky(x)) cout << x << " is a lucky no."; else cout << x << " is not a lucky no.";}// This code is contributed// by rathbhupendra |
C
// C program for Lucky Numbers#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 codeint main(){ int x = 5; // Function call 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. // 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 code public static void main(String[] args) { int x = 5; // Function call 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
# Python3 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 = 5# Function callif 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. // 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 code public static void Main() { int x = 5; // Function call 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 $x = 5; // Function call if(isLucky($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.
Time Complexity: O(n)
Auxiliary Space: O(1)
Dry run of the above approach:
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 bugs in the given programs or other ways to solve the same problem.
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