mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-11-24 21:41:08 +00:00
109 lines
3.9 KiB
Python
109 lines
3.9 KiB
Python
def palindromic_string(input_string: str) -> str:
|
||
"""
|
||
>>> 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
|
||
start = 0
|
||
for j in range(len(new_input_string)):
|
||
k = 1 if j > r else min(length[l + r - j] // 2, r - j + 1)
|
||
while (
|
||
j - k >= 0
|
||
and j + k < len(new_input_string)
|
||
and new_input_string[k + j] == new_input_string[j - k]
|
||
):
|
||
k += 1
|
||
|
||
length[j] = 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 j + k - 1 > r:
|
||
l = j - k + 1 # noqa: E741
|
||
r = j + k - 1
|
||
|
||
# update max_length and start position
|
||
if max_length < length[j]:
|
||
max_length = length[j]
|
||
start = j
|
||
|
||
# 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 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
|
||
"""
|