CareerCup

Concatante the original string with itself and search for second string.
using strstr()

algo

mid=strlen(a)/2;
for(i=0;i<=n/2;i++)
temp=a[mid+i+1];
a[mid+i+1]=a[i];
a[i]=temp;

algo

mid=strlen(a)/2;
for(i=0;i<=n/2;i++)
temp=a[mid+i+1];
a[mid+i+1]=a[i];
a[i]=temp;

use strstr() or its algorithm thereof
then use a match chars, wrapping around if end of string is reached

it's basically asking for a strstr algo

suppose the length is odd or even
then we can get it solved

int compare(char *s,char *t)
{
char *ptr,ch=*s;
ptr=strchr(t,ch);
if(!strncmp(s,ptr,strlen(ptr)))
{
s=s+strlen(ptr);
if(!strncmp(t,s,strlen(s)))
return(1);
else
return(0);
}
else return(0);
}

first find the position POS firstCharOf s2 in s1 in O(n)
then
.................
count=0;
if(len1!=len2) return false;
while(count!=len2)
{
if(s1[POS]!=s2[count])return false;

count++;POS++;
if(s1[POS]=='\0') POS=0;
}
return true;

Double S2 to S2S2 like "cdabcdab", use KMP to search if S1 "abcd" in S2S2.

Come back to check again whether there is a false match like S2="cdaaaaaaaab";

#include <stdio.h>
#include <string.h>
#include <unistd.h>

int main()
{
char *len1,len2,rot=0;
char str1[]="ceaser";
char str2[] = "erceas";

char str3[20];

strcpy(str3,str1);

strcat(str3,str1);
len1 = strstr(str3,str2);

if(len1 == NULL)
printf("false");
else
printf("true");
}

Using python

str1 = 'abcd'
str2 = 'bcda'
l1 = list(str1)
l2 = list(str2)
is_found = False
for each in range(1, len(str1)):
l1 = [l1.pop()] + l1[:len(str1)-1]
if "".join(l1) == str2:
is_found = True
break
if is_found == True:
print ("{} and {} are same strings that can be rotated".format(str1, str2))
else:
print("not found")