Length of the Longest Palindromic Substring

Question

A palindrome is a string which is the same when read forwards and backwards. For example "racecar" is a palindrome, but "Racecar" is not. A substring is a contiguous set of characters in a larger string. Your task is to write a program or method which takes a string or character array as input and outputs the length of the longest substring of that string which is a palindrome.

Examples

Input

banana

Ouput

5

This is because "anana" is a 5 character substring of "banana"

Input

abracadabra

Output

3

This is because both "aca" and "ada" are 3 characters.

Input

True

Output

1

This string has no palindromes that are more than 1 character.

If the input is the empty string the output must be 0.

This is code-golf so the shortest submission in any language wins.

Note: There have been slight variations of this question in the past, which have all had wording issues or been asking a variation of this question. You can see these:

Find the Longest Palindrome in a String by Removing Characters

Longest reverse palindromic DNA substring

How do I find the longest palindrome in a string?

Finding "sub-palindromes". - The purpose of this challenge is to print all palindromic substring, rather than the length of the longest. As such many of the answers to this question do not have close analogues.

Answer 1

Brachylog, 9 bytes

{s.↔}ᶠlᵐ⌉

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Explanation

{   }ᶠ       Find all…
 s.            …substrings of the input…
  .↔           …which are their own reverse
      lᵐ     Map length
        ⌉    Max

Answer 2

Jelly, 6 bytes

Saved 1 byte thanks to Dennis (pointing out that Þ is stable and thus works instead of Ðf).

ẆŒḂÞṪL

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How?

ẆŒḂÞṪL – Full program.

Ẇ      – All contiguous non-empty subsequences.
   Þ   – Stable sort by...
 ŒḂ    – 1 if the element is palindrome, 0 otherwise.
    Ṫ  – Get the last element.
     L – And take its length.

Note that the substrings are ordered by length, in Jelly.

Answer 3

Pyth, 7 bytes

le_I#.:

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How?

le_I#.: – Full program.

     .: – All contiguous non-empty subsequences.
    #   – Keep those...
  _I    – That are invariant under reversal.
 e      – Get the last element.
l       – And take its length.

Note that the substrings are ordered by length, in Pyth.

Answer 4

05AB1E, 9 8 bytes

ŒʒÂQ}€gM

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Answer 5

Java 8, 163 bytes

s->{int r=0,l=s.length(),i=0,j,T;for(String t;i<l;i++)for(j=i;++j<=l;r=t.contains(new StringBuffer(t).reverse())&(T=t.length())>r?T:r)t=s.substring(i,j);return r;}

Explanation:

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s->{                 // Method with String parameter and integer return-type
  int r=0,           //  Result-String, starting empty
      l=s.length(),  //  The length of the input-String
      i=0,j,         //  Index-integers
      T;             //  Temp integer
  for(String t;      //  Temp String
      i<l;i++)       //  Loop `i` over the String
    for(j=i;++j<=l;  //   Inner loop `j` over the String
        r=           //     After every iteration, set the result to:
          t.contains(new StringBuffer(t).reverse())
                     //      If the temp-String is a palindrome,
          &(T=t.length())>r?
                     //      and the length of the substring is larger than the result
           T         //       Set this length as the new result
          :          //      Else:
           r)        //       Keep the result the same
      t=s.substring(i,j);
                     //    Set the temp String to the substring from index `i` to `j`
  return r;}         //  Return the result

Answer 6

Husk, 8 bytes

L►LfS=↔Q

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How?

L►LfS=↔Q – Full program.

       Q – All contiguous non-empty subsequences.
   f     – Keep those...
    S=   – That are equal to themselves when...
      ↔  – Reversed.
 ►L      – Get the maximum element, sorted by length.
L        – Take its length.

Alternative: ▲mLfS=↔Q. Note that the substrings are not ordered by length, in Husk.

Answer 7

Retina, 42 bytes

Lv`(.)*.?(?<-1>\1)*(?(1)(?!))
N$^`
$.&
\G.

Try it online! Link includes test cases. Explanation:

Lv`(.)*.?(?<-1>\1)*(?(1)(?!))

For each character in the string, find the longest palindrome that starts at that point.

N$^`
$.&

Sort the found palindromes in reverse order of length.

\G.

Count the length of the first (i.e. longest) palindrome.

Answer 8

Python 3, 58 bytes

f=lambda s:len(s)if s==s[::-1]else max(f(s[1:]),f(s[:-1]))

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Answer 9

J, 20 bytes

[:>./@,#\(#*]-:|.)\]

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