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Tighten up psf/black and flake8 (#2024)

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* Tighten up psf/black and flake8

* Fix some tests

* Fix some E741

* Fix some E741

* updating DIRECTORY.md

Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
This commit is contained in:
Christian Clauss
2020-05-22 08:10:11 +02:00
committed by GitHub
parent 21ed8968c0
commit 1f8a21d727
124 changed files with 583 additions and 495 deletions
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42
strings/manacher.py
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@ -9,9 +9,10 @@ def palindromic_string(input_string):
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)
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
@ -26,7 +27,8 @@ def palindromic_string(input_string):
# append last character
new_input_string += input_string[-1]
# we will store the starting and ending of previous furthest ending palindromic substring
# 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
@ -47,7 +49,7 @@ def palindromic_string(input_string):
# 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
l = i - k + 1 # noqa: E741
r = i + k - 1
# update max_length and start position
@ -72,32 +74,34 @@ if __name__ == "__main__":
"""
...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
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)
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
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 ..
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.
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
"""