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Java Program to Count of rotations required to generate a sorted array
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Java Program to Count rotations which are divisible by 10

Last Updated : 29 Jul, 2022
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Given a number N, the task is to count all the rotations of the given number which are divisible by 10.
Examples: 
 

Input: N = 10203 
Output: 2 
Explanation: 
There are 5 rotations possible for the given number. They are: 02031, 20310, 03102, 31020, 10203 
Out of these rotations, only 20310 and 31020 are divisible by 10. So 2 is the output. 
Input: N = 135 
Output: 0 
 


 


Naive Approach: The naive approach for this problem is to form all the possible rotations. It is known that for a number of size K, the number of possible rotations for this number N is K. Therefore, find all the rotations and for every rotation, check if the number is divisible by 10 or not. The time complexity for this approach is quadratic. 
Efficient Approach: The efficient approach lies behind the concept that in order to check whether a number is divisible by 10 or not, we simply check if the last digit is 0. So, the idea is to simply iterate over the given number and find the count of 0's. If the count of 0's is F, then clearly, F out of K rotations will have 0 at the end of the given number N. 
Below is the implementation of the above approach:
 

Java 
// Java implementation to find the // count of rotations which are // divisible by 10  class GFG {      // Function to return the count     // of all rotations which are     // divisible by 10.     static int countRotation(int n)     {         int count = 0;          // Loop to iterate through the         // number         do {             int digit = n % 10;              // If the last digit is 0,             // then increment the count             if (digit == 0)                 count++;             n = n / 10;         } while (n != 0);          return count;     }      // Driver code     public static void main(String[] args)     {         int n = 10203;          System.out.println(countRotation(n));     } }

Output: 
2

 

Time Complexity: O(log10N), where N is the length of the number.
Auxiliary Space: O(1)

Please refer complete article on Count rotations which are divisible by 10 for more details!


Next Article
Java Program to Count of rotations required to generate a sorted array
author
kartik
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Article Tags :
  • Java
  • Mathematical
  • Combinatorial
  • Java Programs
  • DSA
  • rotation
  • Numbers
  • Number Divisibility
Practice Tags :
  • Combinatorial
  • Java
  • Mathematical
  • Numbers

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