# Finding the largest palindrome in string implementation

I'm trying to solve a problem that asks to find the largest palindrome in a string up to 20,000 characters. I've tried to check every sub string whether it's a palindrome, that worked, but obviously was too slow. After a little googling I found this nice algorithm http://stevekrenzel.com/articles/longest-palnidrome. I've tried to implement it, however I can't get it to work. Also the given string contains illegal characters, so I have to convert it to only legal characters and output the longest palindrome with all characters.

Here's my attempt:

`int len = original.length(); int longest = 0; string answer; for (int i = 0; i < len-1; i++){ int lower(0), upper(0); if (len % 2 == 0){ lower = i; upper = i+1; } else { lower = i; upper = i; } while (lower >= 0 && upper <= len){ string s2 = original.substr(lower,upper-lower+1); string s = convert(s2); if (s == s[s.length()-1]){ lower -= 1; upper += 1; } else { if (s.length() > longest){ longest = s.length(); answer = s2; } break; } } } `

I can't get it to work, I've tried using this exact algorithm on paper and it worked, please help. Here's full code if you need it : http://pastebin.com/sSskr3GY

EDIT:

`int longest = 0; string answer; string converted = convert(original); int len = converted.length(); if (len % 2 == 0){ for (int i = 0; i < len - 1; i++){ int lower(i),upper(i+1); while (lower >= 0 && upper <= len && converted[lower] == converted[upper]){ lower -= 1; upper += 1; } string s = converted.substr(lower+1,upper-lower-1); if (s.length() > longest){ longest = s.length(); answer = s; } } } else { for (int i = 0; i < len; i++){ int lower(i), upper(i); while (lower >= 0 && upper <= len && converted[lower] == converted[upper]){ lower -= 1; upper += 1; } string s = converted.substr(lower+1,upper-lower-1); if (s.length() > longest){ longest = s.length(); answer = s; } } } `

Okay so I fixed the problems, it works perfectly fine but only if the length of converted string is odd. Please help.

I can see two major errors:

1. Whether you initialise your upper/lower pointers to i,i or i,i+1 depends on the parity of the palindrome's length you want to find, not the original string. So (without any further optimisations) you'll need two separate loops with i going from 0 to len (len-1), one for odd palindrome lengths and another one for even.
2. The algorithms should be executed on the converted string only. You have to convert the original string first for it to work.

Consider this string: `abc^ba` (where `^` is an illegal character), the longest palindrome excluding illegal characters is clearly `abcba`, but when you get to `i==2`, and move your lower/upper bounds out by one, they will define the `bc^` substring, after conversion it becomes `bc`, and `b != c` so you concede this palindrome can't be extended.

```#include <iostream> using namespace std;```

``` int main() { string s; cin >> s; signed int i=1; signed int k=0; int ml=0; int mi=0; bool f=0; while(i<s.length()) { if(s[i]!=s[i+1]) { for(k=1;;k++) { if(!(s[i-k]==s[i+k] && (i-k)>=0 && (i+k)<s.length())) { break; } else if(ml < k) { ml=k; mi=i; f=1; } } } i++; } i=0; while(i<s.length()) { if(s[i]==s[i+1]) { for(k=1;;k++) { if(!(s[i-k]==s[k+1+i] && (i-k)>=0 && (k+i)<s.length())) { break; } else if(ml < k) { ml=k; mi=i; } } } i++; } if(ml < 1) { cout << "No Planidrom found"; return 0; } if(f==0) { cout << s.substr(mi-ml,2*ml+2); } else { cout << s.substr(mi-ml,2*ml+1); } return 0; ```

```} ```

@biziclop : As you said.. i used 2 while loops. one for even and one for old palindrom string. finally i was able to fix it. thanks for your suggestion.

Category:c# Views:1 Time:2011-03-14

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