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