Given a string s1 and a string s2, write a snippet to say whether s2 is a rotation of s1? (eg given s1 = ABCD and s2 = CDAB, return true, given s1 = ABCD, and s2 = ACBD , return false)
Algorithm: areRotations(str1, str2)
1. Create a temp string and store concatenation of str1 to
str1 in temp.
temp = str1.str1
2. 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. Implementation:
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 <string.h>
# include <stdlib.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
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
(
$temp
.
count
(
$str2
)> 0)
{
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 will be O(n*n), where n is the length of the string.Auxiliary Space: O(n)
Method 2(Using STL):
Algorithm :
If the size of both the 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 are less than the size of the string. If q2 becomes equal to q1 at any point during these operations, it is possible. Else not. Implementation:
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 s1 =
"ABCD"
;
string s2 =
"CDAB"
;
if
(check_rotation(s1, s2))
cout << s2 <<
" is a rotated form of "
<< s1
<< endl;
else
cout << s2 <<
" is not a rotated form of "
<< s1
<< endl;
string s3 =
"ACBD"
;
if
(check_rotation(s1, s3))
cout << s3 <<
" is a rotated form of "
<< s1
<< endl;
else
cout << s3 <<
" is not a rotated form of "
<< s1
<< endl;
return
0;
}
Java import
java.util.*;
class
GFG{
static
boolean
check_rotation(String s, String goal)
{
if
(s.length() != goal.length())
;
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 s1 =
"ABCD"
;
String s2 =
"CDAB"
;
if
(check_rotation(s1, s2))
System.out.print(s2+
" is a rotated form of "
+ s1
+
"\n"
);
else
System.out.print(s2+
" is not a rotated form of "
+ s1
+
"\n"
);
String s3 =
"ACBD"
;
if
(check_rotation(s1, s3))
System.out.print(s3+
" is a rotated form of "
+ s1
+
"\n"
);
else
System.out.print(s3+
" is not a rotated form of "
+ s1
+
"\n"
);
}
}
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__"
:
s1
=
"ABCD"
s2
=
"CDAB"
if
(check_rotation(s1, s2)):
print
(s2,
" is a rotated form of "
, s1)
else
:
print
(s2,
" is not a rotated form of "
, s1)
s3
=
"ACBD"
if
(check_rotation(s1, s3)):
print
(s3,
" is a rotated form of "
, s1)
else
:
print
(s3,
" is not a rotated form of "
, s1)
C# using
System;
using
System.Collections;
public
class
GFG{
public
static
bool
check_rotation(
string
s,
string
goal)
{
if
(s.Length != goal.Length)
return
false
;
Queue q1 =
new
Queue();
for
(
int
i = 0; i < s.Length; i++) {
q1.Enqueue(s[i]);
}
Queue q2 =
new
Queue();
for
(
int
i = 0; i < goal.Length; i++) {
q2.Enqueue(goal[i]);
}
int
k = goal.Length;
for
(
int
i=k-1;i>=0;i--) {
char
ch = (
char
)(q2.Peek());
q2.Dequeue();
q2.Enqueue(ch);
if
(q2.Equals(q1))
return
true
;
}
return
false
;
}
static
public
void
Main (){
string
s1 =
"ABCD"
;
string
s2 =
"CDAB"
;
if
(!check_rotation(s1, s2))
{ Console.Write(s2);
Console.Write(
" is a rotated form of "
);
Console.WriteLine(s1);
}
else
{
Console.Write(s2);
Console.Write(
" is not a rotated form of "
);
Console.WriteLine(s1);
}
string
s3 =
"ACBD"
;
if
(check_rotation(s1, s3))
{ Console.Write(s3);
Console.Write(
" is a rotated form of "
);
Console.WriteLine(s1);
}
else
{
Console.Write(s3);
Console.Write(
" is not a rotated form of "
);
Console.WriteLine(s1);
}
}
}
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 CDAB is a rotated form of ABCD
ACBD is not a rotated form of ABCD Time Complexity will be O(n1 + n2) , where n is the length of the strings.Auxiliary Space: O(n)
Library Functions Used: strstr: strstr finds a sub-string within a string. Prototype: char * strstr(const char *s1, const char *s2); See http://www.lix.polytechnique.fr/Labo/Leo.Liberti/public/computing/prog/c/C/MAN/strstr.htm for more details strcat: strncat concatenate two strings Prototype: char *strcat(char *dest, const char *src); See http://www.lix.polytechnique.fr/Labo/Leo.Liberti/public/computing/prog/c/C/MAN/strcat.htm for more details
Time Complexity: Time complexity of this problem depends on the implementation of strstr function. If the implementation of strstr is done using KMP matcher then complexity of the above program is O(n1 + n2) where n1 and n2 are lengths of strings. KMP matcher takes O(n) time to find a substring in a string of length n where length of substring is assumed to be smaller than the string.
Method 3:
Algorithm:
Find all the positions of first character of 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 original string to be s1 ,string to be checked be s2,n is length of strings and j is the position of 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. Implementation:
C++ #include <iostream>
#include <vector>
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<<
"s2 is rotation of s1"
<<endl;
else
cout<<
"s2 is not a rotation of s1"
<<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(
"s2 is rotation of s1"
);
else
System.out.println(
"s2 is not a rotation of s1"
);
}
}
}
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
(
"s2 is rotation of s1"
)
else
:
print
(
"s2 is not a rotation of s1"
)
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(
"s2 is rotation of s1"
);
else
Console.WriteLine(
"s2 is not a rotation of s1"
);
}
}
}
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 s2 is rotation of s1 Time Complexity will be O(n*n) in the worst case, where n is the length of the string.Auxiliary Space: O(1)
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