Therefore, it provides a linear time solution to the longest palindromic substring problem. However, as detected e.g., by Apostolico, Breslauer & Galil (1995), the same algorithm can also be used to find all maximal palindromic substrings anywhere within the input string, again in linear time.
def palindromic_string(input_string): """ >>> palindromic_string('abbbaba') 'abbba' >>> palindromic_string('ababa') 'ababa' Manacher’s algorithm which finds Longest palindromic Substring in linear time. 1. first this convert input_string("xyx") into new_string("x|y|x") where odd positions are actual input characters. 2. for each character in new_string it find corresponding length and store the length and l,r to store previously calculated info.(please look the explanation for details) 3. return corresponding output_string by removing all "|" """ max_length = 0 # if input_string is "aba" than new_input_string become "a|b|a" new_input_string = "" output_string = "" # append each character + "|" in new_string for range(0, length-1) for i in input_string[: len(input_string) - 1]: new_input_string += i + "|" # append last character new_input_string += input_string[-1] # we will store the starting and ending of previous furthest ending palindromic # substring l, r = 0, 0 # length[i] shows the length of palindromic substring with center i length = [1 for i in range(len(new_input_string))] # for each character in new_string find corresponding palindromic string for i in range(len(new_input_string)): k = 1 if i > r else min(length[l + r - i] // 2, r - i + 1) while ( i - k >= 0 and i + k < len(new_input_string) and new_input_string[k + i] == new_input_string[i - k] ): k += 1 length[i] = 2 * k - 1 # does this string is ending after the previously explored end (that is r) ? # if yes the update the new r to the last index of this if i + k - 1 > r: l = i - k + 1 # noqa: E741 r = i + k - 1 # update max_length and start position if max_length < length[i]: max_length = length[i] start = i # create that string s = new_input_string[start - max_length // 2 : start + max_length // 2 + 1] for i in s: if i != "|": output_string += i return output_string if __name__ == "__main__": import doctest doctest.testmod() """ ...a0...a1...a2.....a3......a4...a5...a6.... consider the string for which we are calculating the longest palindromic substring is shown above where ... are some characters in between and right now we are calculating the length of palindromic substring with center at a5 with following conditions : i) we have stored the length of palindromic substring which has center at a3 (starts at l ends at r) and it is the furthest ending till now, and it has ending after a6 ii) a2 and a4 are equally distant from a3 so char(a2) == char(a4) iii) a0 and a6 are equally distant from a3 so char(a0) == char(a6) iv) a1 is corresponding equal character of a5 in palindrome with center a3 (remember that in below derivation of a4==a6) now for a5 we will calculate the length of palindromic substring with center as a5 but can we use previously calculated information in some way? Yes, look the above string we know that a5 is inside the palindrome with center a3 and previously we have have calculated that a0==a2 (palindrome of center a1) a2==a4 (palindrome of center a3) a0==a6 (palindrome of center a3) so a4==a6 so we can say that palindrome at center a5 is at least as long as palindrome at center a1 but this only holds if a0 and a6 are inside the limits of palindrome centered at a3 so finally .. len_of_palindrome__at(a5) = min(len_of_palindrome_at(a1), r-a5) where a3 lies from l to r and we have to keep updating that and if the a5 lies outside of l,r boundary we calculate length of palindrome with bruteforce and update l,r. it gives the linear time complexity just like z-function """