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:
C++
// C++ implementation to find the// count of rotations which are// divisible by 10#include <bits/stdc++.h>using namespace std;// Function to return the count of// all the rotations which are// divisible by 10.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 codeint main(){ int n = 10203; cout << countRotation(n);} |
C#
// CSharp implementation to find the// count of rotations which are// divisible by 10using System;class Solution { // 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 % 2 == 0) count++; n = n / 10; } while (n != 0); return count; } // Driver code public static void Main() { int n = 10203; Console.Write(countRotation(n)); }} |
Java
// Java implementation to find the// count of rotations which are// divisible by 10class 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)); }} |
Python
# Python3 implementation to find the # count of rotations which are # divisible by 10# Function to return the count of # all rotations which are divisible # by 10.def countRotation(n): count = 0; # Loop to iterate through the # number while n > 0: digit = n % 10 # If the last digit is 0, # then increment the count if(digit % 2 == 0): count = count + 1 n = int(n / 10) return count; # Driver code if __name__ == "__main__" : n = 10203; print(countRotation(n)); |
Javascript
<script>// Javascript implementation to find the// count of rotations which are// divisible by 10// Function to return the count of// all the rotations which are// divisible by 10.function countRotation(n){ let count = 0; // Loop to iterate through the // number do { let digit = n % 10; // If the last digit is 0, // then increment the count if (digit == 0) count++; n = parseInt(n / 10); } while (n != 0); return count;}// Driver codelet n = 10203;document.write(countRotation(n));</script> |
2
Time Complexity: O(log10N), where N is the length of the number.
Auxiliary Space: O(1)
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