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.
#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;
} |
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);
}
} |
# function to count the number of odd and even numbers in a rotated number def count_odd_rotation(n):
# Convert the input number to a string to manipulate its digits
s = str (n)
len_s = len (s) # Get the length of the string
count_odd = 0 # Initialize a counter for odd numbers
count_even = 0 # Initialize a counter for even numbers
# Iterate through all possible rotations of the number
for i in range (len_s):
# Create a rotated string by slicing and reordering the original string
temp = s[i:] + s[:i]
x = int (temp) # Convert the rotated string back to an integer
# Check if the rotated number is even or odd
if x % 2 = = 0 :
count_even + = 1 # Increment the even count if it's even
else :
count_odd + = 1 # Increment the odd count if it's odd
# Print the counts of odd and even numbers
print ( "Odd =" , count_odd)
print ( "Even =" , count_even)
if __name__ = = "__main__" :
n = 1234
count_odd_rotation(n)
|
// C# implementation for the approach using System;
public class GFG {
// Function to count of all rotations
// which are odd and even
static void CountOddRotation( int n)
{
// Convert the integer to a string to manipulate its
// digits
string s = n.ToString();
int len = s.Length;
int countOdd = 0, countEven = 0;
for ( int i = 0; i < len; i++) {
// Create a rotated string by moving the first
// 'i' digits to the end
string temp
= s.Substring(i) + s.Substring(0, i);
// Convert the rotated string back to an integer
int x = int .Parse(temp);
if (x % 2 == 0) {
countEven++; // Count even rotations
}
else {
countOdd++; // Count odd rotations
}
}
Console.WriteLine( "Odd = " + countOdd);
Console.WriteLine( "Even = " + countEven);
}
public static void Main( string [] args)
{
int n = 1234;
// Call the function
CountOddRotation(n);
}
} |
// Function to count the number of odd and even rotations of an integer function countOddRotation(n) {
// Convert the integer to a string to manipulate its digits
const s = n.toString();
const len = s.length; // Get the length of the number as a string
let countOdd = 0; // Initialize a counter for odd rotations
let countEven = 0; // Initialize a counter for even rotations
// Iterate through all possible rotations of the number
for (let i = 0; i < len; i++) {
// Create a rotated version of the number by shifting digits
const temp = s.substring(i) + s.substring(0, i);
const x = parseInt(temp); // Convert the rotated string back to an integer
// Check if the rotated number is even or odd
if (x % 2 === 0) {
countEven++; // Increment the count for even rotations
} else {
countOdd++; // Increment the count for odd rotations
}
}
// Display the counts of odd and even rotations
console.log( "Odd = " + countOdd);
console.log( "Even = " + countEven);
} // Driver Code const 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++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Function to count of all rotations // which are odd and even 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);
cout << "Odd = " << odd_count << endl;
cout << "Even = " << even_count << endl;
} // Driver Code int main()
{ int n = 1234;
countOddRotations(n);
} |
// Java implementation of the above approach 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 );
System.out.println( "Odd = " + odd_count);
System.out.println( "Even = " + even_count);
}
public static void main(String[] args)
{
int n = 1234 ;
countOddRotations(n);
}
} |
# Python implementation of the above approach # Function to count of all rotations # which are odd and even def 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 code n = 1234
countOddRotations(n) # This code is contributed by Shrikant13 |
// CSharp implementation of the above approach using 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);
}
} |
<script> // Javascript implementation of the above approach // Function to count of all rotations // which are odd and even function 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 Code var n = 1234;
countOddRotations(n); // This code is contributed by rutvik_56. </script> |
<?php // PHP implementation of the above approach // Function to count of all rotations // which are odd and even function 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.. ?> |
Odd = 2 Even = 2
Time Complexity: O(log10n)
Auxiliary Space: O(1)