Given a number n, the task is to count all rotations of the given number which are odd and even.
Examples:
Input: n = 1234
Output: Odd = 2, Even = 2
Total rotations: 1234, 2341, 3412, 4123
Odd rotations: 2341 and 4123
Even rotations: 1234 and 3412
Input: n = 246
Output: Odd = 0, Even = 3
Brute force approach:
The brute force approach is very simple .
The steps are as follows:
1.Convert the number to a string.
2.Loop through all possible rotations of the string.
3.Check if the current rotation is odd or even.
4.Increment the count of odd or even rotations accordingly.
5.Print the counts.
C++
#include <bits/stdc++.h>using namespace std;void countoddrotation(int n){ string s = to_string(n); int len = s.length(); int count_odd = 0, count_even = 0; for (int i = 0; i < len; i++) { string temp = s.substr(i) + s.substr(0, i); int x = stoi(temp); if (x % 2 == 0) { count_even++; } else { count_odd++; } } cout << "Odd = "<< count_odd<<endl<<"Even = "<< count_even;}int main() { int n = 1234; countoddrotation(n); return 0;} |
Java
import java.util.*;public class Main { static void countOddRotation(int n) { String s = Integer.toString(n); int len = s.length(); int countOdd = 0, countEven = 0; for (int i = 0; i < len; i++) { String temp = s.substring(i) + s.substring(0, i); int x = Integer.parseInt(temp); if (x % 2 == 0) { countEven++; } else { countOdd++; } } System.out.println("Odd = " + countOdd); System.out.println("Even = " + countEven); } public static void main(String[] args) { int n = 1234; countOddRotation(n); }} |
Output:
Odd = 2
Even = 2
Time Complexity: O(n*n)
Auxiliary Space: O(n)
Efficient Approach: For large numbers, it is difficult to rotate and check whether it is odd or not for every rotation. Hence, in this approach, check the count of odd digits and even digits present in the number. These will be the answer to this problem.
Below is the implementation of the above approach:
Implementation:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to count of all rotations// which are odd and evenvoid countOddRotations(int n){ int odd_count = 0, even_count = 0; do { int digit = n % 10; if (digit % 2 == 1) odd_count++; else even_count++; n = n / 10; } while (n != 0); cout << "Odd = " << odd_count << endl; cout << "Even = " << even_count << endl;}// Driver Codeint main(){ int n = 1234; countOddRotations(n);} |
Java
// Java implementation of the above approachclass Solution { // Function to count of all rotations // which are odd and even static void countOddRotations(int n) { int odd_count = 0, even_count = 0; do { int digit = n % 10; if (digit % 2 == 1) odd_count++; else even_count++; n = n / 10; } while (n != 0); System.out.println("Odd = " + odd_count); System.out.println("Even = " + even_count); } public static void main(String[] args) { int n = 1234; countOddRotations(n); }} |
Python3
# Python implementation of the above approach# Function to count of all rotations# which are odd and evendef countOddRotations(n): odd_count = 0; even_count = 0 while n != 0: digit = n % 10 if digit % 2 == 0: odd_count += 1 else: even_count += 1 n = n//10 print("Odd =", odd_count) print("Even =", even_count)# Driver coden = 1234countOddRotations(n)# This code is contributed by Shrikant13 |
C#
// CSharp implementation of the above approachusing System;class Solution { // Function to count of all rotations // which are odd and even static void countOddRotations(int n) { int odd_count = 0, even_count = 0; do { int digit = n % 10; if (digit % 2 == 1) odd_count++; else even_count++; n = n / 10; } while (n != 0); Console.WriteLine("Odd = " + odd_count); Console.WriteLine("Even = " + even_count); } public static void Main() { int n = 1234; countOddRotations(n); }} |
Javascript
<script>// Javascript implementation of the above approach// Function to count of all rotations// which are odd and evenfunction countOddRotations(n){ var odd_count = 0, even_count = 0; do { var digit = n % 10; if (digit % 2 == 1) odd_count++; else even_count++; n = parseInt(n / 10); } while (n != 0); document.write("Odd = " + odd_count + "<br>"); document.write("Even = " + even_count + "<br>");}// Driver Codevar n = 1234;countOddRotations(n);// This code is contributed by rutvik_56.</script> |
PHP
<?php// PHP implementation of the above approach// Function to count of all rotations// which are odd and evenfunction countOddRotations($n){ $odd_count = 0; $even_count = 0; do { $digit = $n % 10; if ($digit % 2 == 1) $odd_count++; else $even_count++; $n = (int)($n / 10); } while ($n != 0); echo "Odd = ", $odd_count, "\n"; echo "Even = ", $even_count, "\n";}// Driver Code$n = 1234;countOddRotations($n);// This code is contributed by ajit..?> |
Output
Odd = 2 Even = 2
Time Complexity: O(log10n)
Auxiliary Space: O(1)
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