C++ Program to Check whether all the rotations of a given number is greater than or equal to the given number or not
Given an integer x, the task is to find if every k-cycle shift on the element produces a number greater than or equal to the same element.
A k-cyclic shift of an integer x is a function that removes the last k digits of x and inserts them in its beginning.
For example, the k-cyclic shifts of 123 are 312 for k=1 and 231 for k=2. Print Yes if the given condition is satisfied else print No.
Examples:
Input: x = 123
Output : Yes
The k-cyclic shifts of 123 are 312 for k=1 and 231 for k=2.
Both 312 and 231 are greater than 123.
Input: 2214
Output: No
The k-cyclic shift of 2214 when k=2 is 1422 which is smaller than 2214
Approach 1: Simply find all the possible k cyclic shifts of the number and check if all are greater than the given number or not.
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>
using namespace std;
void CheckKCycles( int n, string s)
{
bool ff = true ;
int x = 0;
for ( int i = 1; i < n; i++)
{
x = (s.substr(i) + s.substr(0, i)).length();
if (x >= s.length())
{
continue ;
}
ff = false ;
break ;
}
if (ff)
{
cout << ( "Yes" );
}
else
{
cout << ( "No" );
}
}
int main()
{
int n = 3;
string s = "123" ;
CheckKCycles(n, s);
return 0;
}
|
Java
public class Main {
static void checkKCycles( int n, String s) {
boolean ff = true ;
int x = 0 ;
for ( int i = 1 ; i < n; i++) {
String rotatedString = s.substring(i) + s.substring( 0 , i);
x = rotatedString.length();
if (x >= s.length()) {
continue ;
}
ff = false ;
break ;
}
if (ff) {
System.out.println( "Yes" );
} else {
System.out.println( "No" );
}
}
public static void main(String[] args) {
int n = 3 ;
String s = "123" ;
checkKCycles(n, s);
}
}
|
Python
def check_k_cycles(n, s):
ff = True
x = 0
for i in range ( 1 , n):
x = len (s[i:] + s[:i])
if x > = len (s):
continue
ff = False
break
if ff:
print ( "Yes" )
else :
print ( "No" )
n = 3
s = "123"
check_k_cycles(n, s)
|
C#
using System;
class Program
{
static void CheckKCycles( int n, string s)
{
bool ff = true ;
int x = 0;
for ( int i = 1; i < n; i++)
{
string rotatedString = s.Substring(i) + s.Substring(0, i);
x = rotatedString.Length;
if (x >= s.Length)
{
continue ;
}
ff = false ;
break ;
}
if (ff)
{
Console.WriteLine( "Yes" );
}
else
{
Console.WriteLine( "No" );
}
}
static void Main( string [] args)
{
int n = 3;
string s = "123" ;
CheckKCycles(n, s);
}
}
|
Javascript
function checkKCycles(n, s) {
let ff = true ;
let x = 0;
for (let i = 1; i < n; i++) {
x = (s.substring(i) + s.substring(0, i)).length;
if (x >= s.length) {
continue ;
}
ff = false ;
break ;
}
if (ff) {
console.log( "Yes" );
} else {
console.log( "No" );
}
}
let n = 3;
let s = "123" ;
checkKCycles(n, s);
|
Time Complexity: O(N2), where N represents the length of the given string.
The time complexity of the program is O(N2) because first it runs a loop for traversing the string and inside that substring function is used.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Approach 2:
- Define a function CheckKCycles that takes an integer n and a string s as input.
- Initialize a boolean variable isKCycle as true.
- Iterate over the string starting from index 1 up to n-1.
- Compare characters s[i] and s[i % n].
- If the characters are not equal, set isKCycle as false and break the loop.
- After the loop, check the value of isKCycle.
- If it is true, print Yes to indicate a K-Cycle.
- If it is false, print No to indicate it is not a K-Cycle.
Below is the implementation of the above approach:
C++
#include <iostream>
#include <string>
void CheckKCycles( int n, const std::string& s) {
bool isKCycle = true ;
for ( int i = 1; i < n; i++) {
if (s[i] != s[i % n]) {
isKCycle = false ;
break ;
}
}
if (isKCycle) {
std::cout << "Yes" ;
} else {
std::cout << "No" ;
}
}
int main() {
int n = 3;
std::string s = "123" ;
CheckKCycles(n, s);
return 0;
}
|
Java
public class Main {
static void checkKCycles( int n, String s) {
boolean isKCycle = true ;
for ( int i = 1 ; i < n; i++) {
if (s.charAt(i) != s.charAt(i % n)) {
isKCycle = false ;
break ;
}
}
if (isKCycle) {
System.out.println( "Yes" );
} else {
System.out.println( "No" );
}
}
public static void main(String[] args) {
int n = 3 ;
String s = "123" ;
checkKCycles(n, s);
}
}
|
Python3
def check_k_cycles(n, s):
is_k_cycle = all (s[i] = = s[i % n] for i in range ( 1 , n))
if is_k_cycle:
print ( "Yes" )
else :
print ( "No" )
def main():
n = 3
s = "123"
check_k_cycles(n, s)
if __name__ = = "__main__" :
main()
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Check whether all the rotations of a given number is greater than or equal to the given number or not for more details!
Last Updated :
27 Dec, 2023
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