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(n)
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" );
}
}
|
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>
|
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" ;
}
?>
|
Output
Strings are rotations of each other
Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(N)
Another Approach: Comparing the prefix and suffix
- Check if sizes of two strings are not same, then return false.
- Check if ith character is equal to the first character of str2
- Check prefix of str1 with suffix of str2
- Check suffix of str2 with prefix of str1
- If none of the above cases satisfy then answer must be false so, return false
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 ;
for ( int i = 0; i < str1.length(); i++) {
if (str1[i] == str2[0]) {
if (str1.substr(i)
== str2.substr(0, str1.length() - i)) {
if (str1.substr(0, i)
== str2.substr(str1.length() - i))
return true ;
}
}
}
return false ;
}
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;
}
|
Java
class StringRotation {
static boolean areRotations(String s1, String s2)
{
if (s1.length() != s2.length()) {
return false ;
}
else {
for ( int i = 0 ; i < s1.length(); i++) {
if (s1.charAt(i) == s2.charAt( 0 )) {
if (s1.substring(i).equals(s2.substring(
0 , s1.length() - i))) {
if (s1.substring( 0 , i).equals(
s2.substring(s1.length()
- i)))
return true ;
}
}
}
}
return false ;
}
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" );
}
}
|
Python
def areRotations(s1, s2):
if len (s1) ! = len (s2):
return False
else :
for i in range ( len (s1)):
if s1[i] = = s2[ 0 ]:
if s1[i:] = = s2[: len (s1) - i]:
if s1[:i] = = s2[ len (s1) - i:]:
return True
return False
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)
{
if (str1.Length != str2.Length)
return false ;
else {
for ( int i = 0; i < str1.Length; i++) {
if (str1[i] == str2[0]) {
if (str1.Substring(i)
== str2.Substring(0, str1.Length
- i)) {
if (str1.Substring(0, i)
== str2.Substring(str1.Length
- i))
return true ;
}
}
}
}
return false ;
}
public static void Main()
{
String str1 = "FGABCDE" ;
String str2 = "ABCDEFG" ;
if (areRotations(str1, str2))
Console.Write( "Strings are"
+ " rotations of each other" );
else
Console.Write( "Strings are "
+ "not rotations of each other" );
}
}
|
Javascript
<script>
function areRotations( str1, str2)
{
if (str1.length !== str2.length) {
return false ;
} else {
for (let i = 0; i < str1.length; i++) {
if (str1[i] === str2[0]) {
if (str1.substr(i) === str2.substr(0, str1.length - i)) {
if (str1.substr(0, i) === str2.substr(str1.length - i)) {
return true ;
}
}
}
}
}
return false ;
}
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*N), where N is the length of the string and another N because we compare two string.
Auxiliary Space: O(N) , Because we are creating substring
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