Count rotations of N which are Odd and Even
Last Updated :
16 Nov, 2023
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);
}
}
|
Python3
def count_odd_rotation(n):
s = str (n)
len_s = len (s)
count_odd = 0
count_even = 0
for i in range (len_s):
temp = s[i:] + s[:i]
x = int (temp)
if x % 2 = = 0 :
count_even + = 1
else :
count_odd + = 1
print ( "Odd =" , count_odd)
print ( "Even =" , count_even)
if __name__ = = "__main__" :
n = 1234
count_odd_rotation(n)
|
C#
using System;
public class GFG {
static void CountOddRotation( int n)
{
string s = n.ToString();
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 = int .Parse(temp);
if (x % 2 == 0) {
countEven++;
}
else {
countOdd++;
}
}
Console.WriteLine( "Odd = " + countOdd);
Console.WriteLine( "Even = " + countEven);
}
public static void Main( string [] args)
{
int n = 1234;
CountOddRotation(n);
}
}
|
Javascript
function countOddRotation(n) {
const s = n.toString();
const len = s.length;
let countOdd = 0;
let countEven = 0;
for (let i = 0; i < len; i++) {
const temp = s.substring(i) + s.substring(0, i);
const x = parseInt(temp);
if (x % 2 === 0) {
countEven++;
} else {
countOdd++;
}
}
console.log( "Odd = " + countOdd);
console.log( "Even = " + countEven);
}
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++
#include <bits/stdc++.h>
using namespace std;
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;
}
int main()
{
int n = 1234;
countOddRotations(n);
}
|
Java
class Solution {
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
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)
n = 1234
countOddRotations(n)
|
C#
using System;
class Solution {
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>
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>" );
}
var n = 1234;
countOddRotations(n);
</script>
|
PHP
<?php
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" ;
}
$n = 1234;
countOddRotations( $n );
?>
|
Time Complexity: O(log10n)
Auxiliary Space: O(1)
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