Created
March 24, 2012 21:39
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For an arbitrary string, detect the largest palindrome* present within it. | |
* a sequence of characters the same forwards and backwards at least 3 characters long. | |
For instance in "abbabcdedc", the longest palindrome would be "cdedc". | |
Answers? |
Author
joshbuddy
commented
Mar 24, 2012
via email
Up to you, I just wanted to see what other people thought. Was the question too silly?
…On 2012-03-24, at 15:08, Oliver ***@***.*** wrote:
haha, want me to delete that comment so other people can guess without my pre-disposing them?
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Reply to this email directly or view it on GitHub:
https://gist.github.com/2188292
It's pretty Computer Science-y. I thought it might be NP-hard. Looks like there's a linear time solution http://en.wikipedia.org/wiki/Longest_palindromic_substring
not too efficient but short:
def largest_palindrome(s)
(s.split /\s/).map {|x| x.length}.max.downto(1).each do |x|
re = ("(\\w)" * x) + "\\w?"
x.downto(1).each { |y| re += "\\#{y}" }
s.match(re) { |z| return z }
end
end
Here's something in Haskell. I tried to decompose the problem into 1. make substrings, 2. filter out non palindromes, then 3. find the longest string. substrings
time complexity grows like a Trianglular number. I think palindrome
and longest
are both O(n).
Usage: longest (palindromes "abbabcdedc")
substrings [] = []
substrings (x:xs) = decTail (x:xs) ++ substrings xs
where
decTail [] = []
decTail (y:ys) = [y] : [ (y:z) | z <- decTail ys ]
longest [] = []
longest (w:ws) = if length w > length w' then w else w' where w' = longest ws
palindromes w = [ s | s <- substrings w, s == reverse s ]
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