Shortest palindrome¶
Time: O(N); Space: O(N); hard
Given a string s, you are allowed to convert it to a palindrome by adding characters in front of it.
Find and return the shortest palindrome you can find by performing this transformation.
Example 1:
Input: s = “aacecaaa”
Output: “aaacecaaa”
Example 2:
Input: s = “abcd”
Output: “dcbabcd”
[1]:
class Solution1(object):
"""
KMP Algorithm
Time: O(N)
Space: O(N)
"""
def shortestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
def getPrefix(pattern):
prefix = [-1] * len(pattern)
j = -1
for i in range(1, len(pattern)):
while j > -1 and pattern[j+1] != pattern[i]:
j = prefix[j]
if pattern[j+1] == pattern[i]:
j += 1
prefix[i] = j
return prefix
if not s:
return s
A = s + s[::-1]
prefix = getPrefix(A)
i = prefix[-1]
while i >= len(s):
i = prefix[i]
return s[i+1:][::-1] + s
[2]:
sol = Solution1()
s = "aacecaaa"
assert sol.shortestPalindrome(s) == "aaacecaaa"
s = "abcd"
assert sol.shortestPalindrome(s) == "dcbabcd"
[3]:
class Solution2(object):
"""
Manacher's Algorithm
Time: O(N)
Space: O(N)
"""
def shortestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
def preProcess(s):
if not s:
return ['^', '$']
string = ['^']
for c in s:
string += ['#', c]
string += ['#', '$']
return string
string = preProcess(s)
palindrome = [0] * len(string)
center, right = 0, 0
for i in range(1, len(string) - 1):
i_mirror = 2 * center - i
if right > i:
palindrome[i] = min(right - i, palindrome[i_mirror])
else:
palindrome[i] = 0
while string[i + 1 + palindrome[i]] == string[i - 1 - palindrome[i]]:
palindrome[i] += 1
if i + palindrome[i] > right:
center, right = i, i + palindrome[i]
max_len = 0
for i in range(1, len(string) - 1):
if i - palindrome[i] == 1:
max_len = palindrome[i]
return s[len(s)-1:max_len-1:-1] + s
[4]:
sol = Solution2()
s = "aacecaaa"
assert sol.shortestPalindrome(s) == "aaacecaaa"
s = "abcd"
assert sol.shortestPalindrome(s) == "dcbabcd"