My solution for Greplin_problem_1 as a reply to this comment at HN - http://news.ycombinator.com/item?id=2276473

Greplin_problem_1.py

#Two algos for solving problem-1 at http://challenge.greplin.com/
def getPalindrome1(text):
    '''
    A more effecient approach for finding palindromes : 

    Iterate through the text and,

    Case1:
    For each character, check if it forms MIDDLE character of a (ODD-LENGTH) palindrome.
    For ex. Palindrome like rac[e]car, abc[d]cba
    
    Case2:
    For each character, check if it forms LEFT character of a (EVEN-LENGTH) palindrome.
    For ex. Palindrome like ra[c]car, ab[c]cba

    Worst case - O(n2), Average case - O(n).
    '''
    longestPalindrome=''
    
    for idx in range(len(text)):
        #Case1 consider idx'th chr as the middle character
        cur=1
        max_word1,max_word2='',''
        while cur <= min(idx,len(text)-idx-1) and text[idx-cur] == text[cur+idx]:
            max_word1 = text[idx-cur:idx+cur+1]
            cur+=1

        #Case2 consider idx'th chr as the left character
        cur=0
        while cur <= min(idx,len(text)-idx-2) and text[idx-cur] == text[cur+idx+1]:
            max_word2 = text[idx-cur:idx+cur+2]
            cur+=1

        longestPalindrome = max([max_word2,max_word1,longestPalindrome], key=lambda x:len(x))
    return longestPalindrome


def getPalindrome2(text):
    '''
    Good'ol brute force solution to generate all possible strings and see if it's a palindrome (word and it's reverse form are 
    identical).
    *The O(n2) solution will work for this case, as the text input is pretty small.
    '''
    longestPalindrome = ''
    for i in range(len(text)):
        for j in range(i, len(text)):
            word = text[i:j]
            if word == word[::-1]:
                longestPalindrome = max([longestPalindrome, word], key=lambda x:len(x))
    return longestPalindrome