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.
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;
}
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Output:
Strings are rotations of each other
Time Complexity: O(n*n), where n is the length of the string.
Auxiliary Space: O(N) where n is the length of the new string.
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 implementation of strstr is done using KMP matcher then complexity of the above program is (-)(n1 + n2) where n1 and n2 are lengths of strings. KMP matcher takes (-)(n) time to find a substring in a string of length n where length of substring is assumed to be smaller than the string. Please refer complete article on A Program to check if strings are rotations of each other or not for more details!