Given a string s1 and a string s2, write a function to check whether s2 is a rotation of s1.
Examples:
Input: S1 = ABCD, S2 = CDAB
Output: Strings are rotations of each other
Input: S1 = ABCD, S2 = ACBD
Output: Strings are not rotations of each other
Naive Approach: Follow the given steps to solve the problem
- Find all the positions of the first character of the original string in the string to be checked.
- For every position found, consider it to be the starting index of the string to be checked.
- Beginning from the new starting index, compare both strings and check whether they are equal or not.
- (Suppose the original string to is s1, string to be checked to be s2,n is the length of strings and j is the position of the first character of s1 in s2, then for i < (length of original string) , check if s1[i]==s2[(j+1)%n). Return false if any character mismatch is found, else return true.
- Repeat 3rd step for all positions found.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool checkString(string& s1, string& s2, int indexFound,
int Size)
{
for ( int i = 0; i < Size; i++) {
if (s1[i] != s2[(indexFound + i) % Size])
return false ;
}
return true ;
}
int main()
{
string s1 = "abcd" ;
string s2 = "cdab" ;
if (s1.length() != s2.length()) {
cout << "s2 is not a rotation on s1" << endl;
}
else {
vector< int > indexes;
int Size = s1.length();
char firstChar = s1[0];
for ( int i = 0; i < Size; i++) {
if (s2[i] == firstChar) {
indexes.push_back(i);
}
}
bool isRotation = false ;
for ( int idx : indexes) {
isRotation = checkString(s1, s2, idx, Size);
if (isRotation)
break ;
}
if (isRotation)
cout << "Strings are rotations of each other"
<< endl;
else
cout
<< "Strings are not rotations of each other"
<< endl;
}
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static boolean checkString(String s1, String s2,
int indexFound, int Size)
{
for ( int i = 0 ; i < Size; i++) {
if (s1.charAt(i)
!= s2.charAt((indexFound + i) % Size))
return false ;
}
return true ;
}
public static void main(String args[])
{
String s1 = "abcd" ;
String s2 = "cdab" ;
if (s1.length() != s2.length()) {
System.out.println(
"s2 is not a rotation on s1" );
}
else {
ArrayList<Integer> indexes = new ArrayList<
Integer>();
int Size = s1.length();
char firstChar = s1.charAt( 0 );
for ( int i = 0 ; i < Size; i++) {
if (s2.charAt(i) == firstChar) {
indexes.add(i);
}
}
boolean isRotation = false ;
for ( int idx : indexes) {
isRotation = checkString(s1, s2, idx, Size);
if (isRotation)
break ;
}
if (isRotation)
System.out.println(
"Strings are rotations of each other" );
else
System.out.println(
"Strings are not rotations of each other" );
}
}
}
|
Python3
def checkString(s1, s2, indexFound, Size):
for i in range (Size):
if (s1[i] ! = s2[(indexFound + i) % Size]):
return False
return True
s1 = "abcd"
s2 = "cdab"
if ( len (s1) ! = len (s2)):
print ( "s2 is not a rotation on s1" )
else :
indexes = []
Size = len (s1)
firstChar = s1[ 0 ]
for i in range (Size):
if (s2[i] = = firstChar):
indexes.append(i)
isRotation = False
for idx in indexes:
isRotation = checkString(s1, s2, idx, Size)
if (isRotation):
break
if (isRotation):
print ( "Strings are rotations of each other" )
else :
print ( "Strings are not rotations of each other" )
|
C#
using System;
public class GFG {
public static bool checkString( string s1, string s2,
int indexFound, int Size)
{
for ( int i = 0; i < Size; i++) {
if (s1[i] != s2[(indexFound + i) % Size])
return false ;
}
return true ;
}
static public void Main()
{
string s1 = "abcd" ;
string s2 = "cdab" ;
if (s1.Length != s2.Length) {
Console.WriteLine( "s2 is not a rotation on s1" );
}
else {
int [] indexes = new int [1000];
int j = 0;
int Size = s1.Length;
char firstChar = s1[0];
for ( int i = 0; i < Size; i++) {
if (s2[i] == firstChar) {
indexes[j] = i;
j++;
}
}
bool isRotation = false ;
for ( int idx = 0; idx < indexes.Length; idx++) {
isRotation = checkString(s1, s2, idx, Size);
if (isRotation)
break ;
}
if (isRotation)
Console.WriteLine(
"Strings are rotations of each other" );
else
Console.WriteLine(
"Strings are not rotations of each other" );
}
}
}
|
Javascript
<script>
function checkString(s1, s2, indexFound, Size)
{
for (let i = 0; i < Size; i++)
{
if (s1[i] != s2[(indexFound + i) % Size]) return false ;
}
return true ;
}
let s1 = "abcd" ;
let s2 = "cdab" ;
if (s1.length != s2.length)
{
document.write( "s2 is not a rotation on s1" );
}
else
{
let indexes = [];
let Size = s1.length;
let firstChar = s1[0];
for (let i = 0; i < Size; i++)
{
if (s2[i] == firstChar)
{
indexes.push(i);
}
}
let isRotation = false ;
for (let idx of indexes)
{
isRotation = checkString(s1, s2, idx, Size);
if (isRotation)
break ;
}
if (isRotation)document.write( "s2 is rotation of s1" )
else document.write( "s2 is not a rotation of s1" )
}
</script>
|
Output
Strings are rotations of each other
Time Complexity: O(n*n) in the worst case, where n is the length of the string.
Auxiliary Space: O(1)
Program to check if strings are rotations of each other or not using queue:
Follow the given steps to solve the problem
- If the size of both strings is not equal, then it can never be possible.
- Push the original string into a queue q1.
- Push the string to be checked inside another queue q2.
- Keep popping q2‘s and pushing it back into it till the number of such operations is less than the size of the string.
- If q2 becomes equal to q1 at any point during these operations, it is possible. Else not.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool check_rotation(string s, string goal)
{
if (s.size() != goal.size())
return false ;
queue< char > q1;
for ( int i = 0; i < s.size(); i++) {
q1.push(s[i]);
}
queue< char > q2;
for ( int i = 0; i < goal.size(); i++) {
q2.push(goal[i]);
}
int k = goal.size();
while (k--) {
char ch = q2.front();
q2.pop();
q2.push(ch);
if (q2 == q1)
return true ;
}
return false ;
}
int main()
{
string str1 = "AACD" , str2 = "ACDA" ;
if (check_rotation(str1, str2))
printf ( "Strings are rotations of each other" );
else
printf ( "Strings are not rotations of each other" );
return 0;
}
|
Java
import java.util.*;
class GFG {
static boolean check_rotation(String s, String goal)
{
if (s.length() != goal.length())
return false ;
Queue<Character> q1 = new LinkedList<>();
for ( int i = 0 ; i < s.length(); i++) {
q1.add(s.charAt(i));
}
Queue<Character> q2 = new LinkedList<>();
for ( int i = 0 ; i < goal.length(); i++) {
q2.add(goal.charAt(i));
}
int k = goal.length();
while (k > 0 ) {
k--;
char ch = q2.peek();
q2.remove();
q2.add(ch);
if (q2.equals(q1))
return true ;
}
return false ;
}
public static void main(String[] args)
{
String str1 = "AACD" ;
String str2 = "ACDA" ;
if (check_rotation(str1, str2))
System.out.println(
"Strings are rotations of each other" );
else
System.out.printf(
"Strings are not rotations of each other" );
}
}
|
Python3
def check_rotation(s, goal):
if ( len (s) ! = len (goal)):
skip
q1 = []
for i in range ( len (s)):
q1.insert( 0 , s[i])
q2 = []
for i in range ( len (goal)):
q2.insert( 0 , goal[i])
k = len (goal)
while (k > 0 ):
ch = q2[ 0 ]
q2.pop( 0 )
q2.append(ch)
if (q2 = = q1):
return True
k - = 1
return False
if __name__ = = "__main__" :
string1 = "AACD"
string2 = "ACDA"
if check_rotation(string1, string2):
print ( "Strings are rotations of each other" )
else :
print ( "Strings are not rotations of each other" )
|
C#
using System;
using System.Threading;
using System.Collections.Generic;
public class GFG
{
public static bool check_rotation(String s, String goal)
{
if (s.Length != goal.Length)
{
return false ;
}
var q1 = new LinkedList< char >();
for ( int i = 0; i < s.Length; i++)
{
q1.AddLast(s[i]);
}
var q2 = new LinkedList< char >();
for ( int i = 0; i < goal.Length; i++)
{
q2.AddLast(goal[i]);
}
var k = goal.Length;
while (k > 0)
{
k--;
var ch = q2.First.Value;
q2.RemoveFirst();
q2.AddLast(ch);
if (!q2.Equals(q1))
{
return true ;
}
}
return false ;
}
public static void Main(String[] args)
{
var str1 = "AACD" ;
var str2 = "ACDA" ;
if (GFG.check_rotation(str1, str2))
{
Console.WriteLine( "Strings are rotations of each other" );
}
else
{
Console.Write( "Strings are not rotations of each other" );
}
}
}
|
Javascript
<script>
function check_rotation(s, goal){
if (s.length != goal.length){
return false ;
}
let q1 = []
for (let i=0;i<s.length;i++)
q1.push(s[i])
let q2 = []
for (let i=0;i<goal.length;i++)
q2.push(goal[i])
let k = goal.length
while (k--){
let ch = q2[0]
q2.shift()
q2.push(ch)
if (JSON.stringify(q2) == JSON.stringify(q1))
return true
}
return false
}
let s1 = "ABCD"
let s2 = "CDAB"
if (check_rotation(s1, s2))
document.write(s2, " is a rotated form of " , s1, "</br>" )
else
document.write(s2, " is not a rotated form of " , s1, "</br>" )
let s3 = "ACBD"
if (check_rotation(s1, s3))
document.write(s3, " is a rotated form of " , s1, "</br>" )
else
document.write(s3, " is not a rotated form of " , s1, "</br>" )
</script>
|
Output
Strings are rotations of each other
Time Complexity: O(N1 * N2), where N1 and N2 are the lengths of the strings.
Auxiliary Space: O(N)
Efficient Approach: Follow the given steps to solve the problem
- Create a temp string and store concatenation of str1 to str1 in temp, i.e temp = str1.str1
- If str2 is a substring of temp then str1 and str2 are rotations of each other.
Example:
str1 = “ABACD”, str2 = “CDABA”
temp = str1.str1 = “ABACDABACD”
Since str2 is a substring of temp, str1 and str2 are rotations of each other.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool areRotations(string str1, string str2)
{
if (str1.length() != str2.length())
return false ;
string temp = str1 + str1;
return (temp.find(str2) != string::npos);
}
int main()
{
string str1 = "AACD" , str2 = "ACDA" ;
if (areRotations(str1, str2))
printf ( "Strings are rotations of each other" );
else
printf ( "Strings are not rotations of each other" );
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int areRotations( char * str1, char * str2)
{
int size1 = strlen (str1);
int size2 = strlen (str2);
char * temp;
void * ptr;
if (size1 != size2)
return 0;
temp = ( char *) malloc ( sizeof ( char ) * (size1 * 2 + 1));
temp[0] = ' ' ;
strcat (temp, str1);
strcat (temp, str1);
ptr = strstr (temp, str2);
free (temp);
if (ptr != NULL)
return 1;
else
return 0;
}
int main()
{
char * str1 = "AACD" ;
char * str2 = "ACDA" ;
if (areRotations(str1, str2))
printf ( "Strings are rotations of each other" );
else
printf ( "Strings are not rotations of each other" );
getchar ();
return 0;
}
|
Java
class StringRotation {
static boolean areRotations(String str1, String str2)
{
return (str1.length() == str2.length())
&& ((str1 + str1).indexOf(str2) != - 1 );
}
public static void main(String[] args)
{
String str1 = "AACD" ;
String str2 = "ACDA" ;
if (areRotations(str1, str2))
System.out.println(
"Strings are rotations of each other" );
else
System.out.printf(
"Strings are not rotations of each other" );
}
}
|
Python3
def areRotations(string1, string2):
size1 = len (string1)
size2 = len (string2)
temp = ''
if size1 ! = size2:
return 0
temp = string1 + string1
if (temp.count(string2) > 0 ):
return 1
else :
return 0
if __name__ = = "__main__" :
string1 = "AACD"
string2 = "ACDA"
if areRotations(string1, string2):
print ( "Strings are rotations of each other" )
else :
print ( "Strings are not rotations of each other" )
|
C#
using System;
class GFG {
static bool areRotations(String str1, String str2)
{
return (str1.Length == str2.Length)
&& ((str1 + str1).IndexOf(str2) != -1);
}
public static void Main()
{
String str1 = "FGABCDE" ;
String str2 = "ABCDEFG" ;
if (areRotations(str1, str2))
Console.Write( "Strings are"
+ " rotation s of each other" );
else
Console.Write( "Strings are "
+ "not rotations of each other" );
}
}
|
PHP
<?php
function areRotations( $str1 , $str2 )
{
if ( strlen ( $str1 ) != strlen ( $str2 ))
{
return false;
}
$temp = $str1 . $str1 ;
if ( strpos ( $temp , $str2 ) != false)
{
return true;
}
else
{
return false;
}
}
$str1 = "AACD" ;
$str2 = "ACDA" ;
if (areRotations( $str1 , $str2 ))
{
echo "Strings are rotations " .
"of each other" ;
}
else
{
echo "Strings are not " .
"rotations of each other" ;
}
?>
|
Javascript
<script>
function areRotations( str1, str2)
{
return (str1.length == str2.length) &&
((str1 + str1).indexOf(str2) != -1);
}
var str1 = "AACD" ;
var str2 = "ACDA" ;
if (areRotations(str1, str2))
document.write( "Strings are rotations of each other" );
else
document.write( "Strings are not rotations of each other" );
</script>
|
Output
Strings are rotations of each other
Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(N)
No extra space approach:
Instead of creating a new string as str1+str1, we use two pointers.
One pointer per string, also a count of the length of string that has been found the same since the last mismatch.
Every time there is a mismatch, we reset one of the pointer to 0 and start comparing again, once one pointer has run over any one string twice, we know we have checked for all possible rotations. Therefore we can return 0.
We keep track of a reset variable that helps us implement this, since i should only be incremented only when current character at i has been compared with current at j, when j is reset, the value of j changes to 0, now if we increment i without comparing s1[i]==s2[0] then we miss this comparison (we skip it). Therefore a reset variable keeps track of if there has been a reset given there was a mismatch, if yes then it increments i else i is only incremented when a match is found.
C++
#include <bits/stdc++.h>
using namespace std;
int isCyclicRotation(string& p, string& q)
{
int i = 0, j = 0, k = 0, n = p.length();
bool reset = false ;
while (i < 2 * n) {
if (k == n)
return 1;
if (p[i % n] == q[j % n]) {
i++;
j++;
k++;
}
else if (reset) {
reset = false ;
i++;
}
else {
reset = true ;
j = 0;
k = 0;
}
}
return 0;
}
int main()
{
string str1 = "AACD" , str2 = "ACDA" ;
if (isCyclicRotation(str1, str2)) {
cout << "YES" ;
}
else {
cout << "NO" ;
}
return 0;
}
|
Java
import java.util.Scanner;
public class GFG {
public static boolean isCyclicRotation(String p,
String q)
{
int i = 0 , j = 0 , k = 0 , n = p.length();
boolean reset = false ;
while (i < 2 * n) {
if (k == n) {
return true ;
}
if (p.charAt(i % n) == q.charAt(j % n)) {
i++;
j++;
k++;
}
else if (reset) {
reset = false ;
i++;
}
else {
reset = true ;
j = 0 ;
k = 0 ;
}
}
return false ;
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
String str1 = "AACD" , str2 = "ACDA" ;
if (isCyclicRotation(str1, str2)) {
System.out.println( "YES" );
}
else {
System.out.println( "NO" );
}
}
}
|
Python3
def isCyclicRotation(p, q):
i, j, k, n = 0 , 0 , 0 , len (p)
reset = False
while i < 2 * n:
if k = = n:
return 1
if p[i % n] = = q[j % n]:
i + = 1
j + = 1
k + = 1
elif reset:
reset = False
i + = 1
else :
reset = True
j = 0
k = 0
return 0
str1 = "AACD"
str2 = "ACDA"
if isCyclicRotation(str1, str2):
print ( "YES" )
else :
print ( "NO" )
|
C#
using System;
public class GFG {
static bool IsCyclicRotation( string p, string q)
{
int i = 0, j = 0, k = 0, n = p.Length;
bool reset = false ;
while (i < 2 * n) {
if (k == n) {
return true ;
}
if (p[i % n] == q[j % n]) {
i++;
j++;
k++;
}
else if (reset) {
reset = false ;
i++;
}
else {
reset = true ;
j = 0;
k = 0;
}
}
return false ;
}
static void Main( string [] args)
{
string str1 = "AACD" , str2 = "ACDA" ;
if (IsCyclicRotation(str1, str2)) {
Console.WriteLine( "YES" );
}
else {
Console.WriteLine( "NO" );
}
}
}
|
Javascript
function isCyclicRotation(p, q) {
let i = 0, j = 0, k = 0, n = p.length;
let reset = false ;
while (i < 2 * n) {
if (k === n)
return true ;
if (p[i % n] === q[j % n]) {
i++;
j++;
k++;
}
else if (reset) {
reset = false ;
i++;
}
else {
reset = true ;
j = 0;
k = 0;
}
}
return false ;
}
let str1 = "AACD" , str2 = "ACDA" ;
if (isCyclicRotation(str1, str2)) {
console.log( "YES" );
}
else {
console.log( "NO" );
}
|
Time Complexity: O(2*n) where n is the length of the string, this can be approximated to O(n)
Auxiliary Space: O(1) i.e. constant extra space.
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