Efficient way to check if a string is rotated palindrome

Question

I was looking for method to find efficient way to check if a string is rotated palindrome or not and apart from O(n^2) solution I found SO post which talks of O(n) solution which seems quite complex. I have written below code and I think it achieves the solution in O(n) as well in a much simpler hash table format.Can someone have a look at it and see if I am missing something here or my solution is correct ?

/*Code doesnt cover scenarios where character is caps and space*/

int main(void)
{
    char str[] = "123217898";
    char *pstr = str;
    int length = strlen(str);
    int notdouble = 0;


    int arr[256] = { 0 };

    for (int i = 0; i < length; i++)
    {
        arr[*pstr]++;
        pstr++;
    }

    pstr = str;

    for (int i = 0; i < length; i++)
    {
        /*check if its divisible by 2 and greater than 2*/
        if (arr[*pstr] % 2 != 0 && arr[*pstr] > 2)
        {
            notdouble = 5;   // just assign random value greater than 1
        }

        if (arr[*pstr] == 1)
        {
            notdouble++;
        }
        pstr++;
    }

    if (notdouble > 1)
    {
        printf("string is not palindrome \n");
    }
    else
    {
        printf("string is palindrome \n");
    }

    return 0;
}

I'm voting to close this question as off-topic because it's about code-review (try http://codereview.stackexchange.com). — Oliver Charlesworth, Jan 10 '16 at 23:15
Previously asked [question](http://stackoverflow.com/questions/12286091/efficient-way-to-verify-if-a-string-is-a-rotated-palindrome) — Weather Vane, Jan 10 '16 at 23:16
@WeatherVane - I myself have provided that link - question talks about alternative solution which is much simpler. I am looking for answer to check if there is some hole in my logic because no one has suggested it — oneday, Jan 10 '16 at 23:18
Oh - sorry. It was the top entry when I googled "rotated palindrome" — Weather Vane, Jan 10 '16 at 23:19
@OliverCharlesworth - More than code review I am looking for validation of part of logic which provides much simpler alternative than the complex method associated. If people dont agree here I dont mind posting it on code review .. — oneday, Jan 10 '16 at 23:19
This may be a good question for [codereview.se], so long as: **(A)** _the code works_, **and (B)** _it's not hypothetical or incomplete in any way_. Please read the [on-topic guide](http://codereview.stackexchange.com/help/on-topic) before posting, if you choose to go to [Code Review](http://codereview.stackexchange.com/questions/ask). If you have any questions or concerns, join us at our [CR Help Desk](http://chat.stackexchange.com/rooms/34045). — Quill, Jan 10 '16 at 23:23
Your code doesn't seem to solve the problem you are describing, as it just tests whether characters are repeated or not... It shouldn't be hard to provide a string which repeats characters, but are neither a palindrome nor a rotated palindrome. I.e. '123321', rotated: '211233', not legal: '221133'. Not tested, but the last should falsely be reported. — holroy, Jan 10 '16 at 23:26
@holroy - yes thats the logic I am pursuing that as long as all characters are multiple of two with max 1 character not being multiple of 2 its a palindrome. Curious why would string '221133' not be rotated palindrome ? — oneday, Jan 10 '16 at 23:27
There's a reasonably simple expected linear-time solution that uses Rabin fingerprints, but this is not that. — David Eisenstat, Jan 10 '16 at 23:31
@oneday, How would you rotate it to become a palindrome? 322113? 332211?, 133221? 113322? 211332? Niether of these are palindromic in the normal sense of palindromes — holroy, Jan 10 '16 at 23:32
@oneday: Recall that, when not rotated, a palindrome is a string that reads the same from either end. You can't rotate '221133' to form a palindrome. — 500 - Internal Server Error, Jan 10 '16 at 23:32
@holroy - I see ur point - what I did was tested with bunch of palindrome words from google and some random words and it turned out true - but I was skeptical of solution myself :) so the post. thanks for the answer it does show my solution is wrong — oneday, Jan 10 '16 at 23:36
@500-InternalServerError - got the point - solves my confusion. — oneday, Jan 10 '16 at 23:37

score 3 · Accepted Answer · answered Jan 10 '16 at 23:41

3

Your code does not test for a rotated palindrome, it merely checks if there is a even amount of repeated characters. That is not conformant with being a palindrome, nor a rotated palindrome.

In addition, a palindrome or rotated palindrome can contain one character with an odd count.

In other words the following two examples are wrongly classified:

"221133" – This is not a rotated palindrome, even though there is even number of repeated characters.
"114321234" - This a rotated palindrome, but the odd count of 1's is three, not one...

answered Jan 10 '16 at 23:41

holroy

3,047
25
41

thanks for the pointer. it seems with rotation I assumed all permutation will occur once and thats where the algorithm fails. and yes as u pointed out there can be instances where one character count can be 3 as well ( which I think I can hack into code) but that wont solve problem as my code assumes all permutation. thanks once again !!! – oneday Jan 11 '16 at 03:20
The code checks for odd concurrence counts, allowing, if anything, exactly _one_ character occurring once: _permuted_ palindrome (with not all palindrome anagrams accepted). – greybeard Jan 13 '16 at 10:13
@greybeard yes, but when allowing multiple occurrences of characters (and rotation) the total may be an odd number of characters – holroy Jan 13 '16 at 11:45
@greybeard the following is a palindrome: 555123454321555, where the number of 5's is odd and not one. – holroy Jan 13 '16 at 11:49
@greybeard that is rotated. Move the first character to the end, and you'll see the palindrome – holroy Jan 13 '16 at 11:53

Efficient way to check if a string is rotated palindrome

1 Answers1