This is a porbleminha that although simple, has no magic to solve:
is a simple algorithm, that we can think and understand how to solve "manually" - and the way is to understand the algorithm, and then transcribe to C.
You have a set of tools that are the functions for pairing string of the standard C library, and checking the characters directly. You can also write your functions if there are none.
But then - realize that it's an algorithm in "steps" - it's not a 3 or 4 line thing - why you'll have to at the very least find the starting point of the search, and compare the characters from there.
So, the pseudo-algorithm in Portuguese could start like this:
- prepare an empty S3 response string.
- find if there is, and the position N of the first character of S2 within S1;
- If there is no N, return saying that the suffix is the empty string
- create a counter i = 0;
- checkpoint
- copy the character S1 [N] to the response string S3
- Increase the value of i
- If the character in S1 [N + i] is the end of the string (character '\ x00') - the answer is what we have in S3: add a string terminator in S3 and return S3 even if we reached the end of S2)
- compare the character N + i in S2 with the character in position (0 + i) of S2;
- If they are different, clear the string S3, and return an empty suffix (** but see note below)
- return to "checkpoint"
Poronto - now you translate this to "C" - and notice that in the description in Portuguese comes naturally what will be a loop "for" or "while" in a programming language.
** - there is a catch in this algorithm there - if the first letter of the second word occurs more than once in the first word, the above description does not account. (for example: "cheating", and "rate" -
he will find the first "t" of "cheat", but will stop the search on "o" - when the suffix is after the second "t") - or you repeat the search for "t", or - to start all , look for the first letter of S2 in S1, but from right to left - this second alternative is a lot easier, but I think you will not have a ready-made function in C's stdlib for this - then you'll do your own little function for that Believe me, it's worth it):
- search for the letter X in S1 from the end:
- make i equal the total length of S1
- heckpoint check if the character S1 [i] is X - if yes, return i
- decrement i
- If i is less than zero, return i (and take advantage of
return of -1 as an indication that the letter does not exist in S1)
- go to checkpoint
- Ready now you already have where to start. It is not the purpose of the S.O. to give ready answers - so try to translate this to C.
