0125 Valid Palindrome
Solved at: 2022-07-14 and 2022-07-26
Question
A phrase is a palindrome if, after converting all uppercase letters into lowercase letters and removing all non-alphanumeric characters, it reads the same forward and backward. Alphanumeric characters include letters and numbers.
Given a string s, return true if it is a palindrome, or false otherwise.
Solution
This is a simple recursion problem.
python
class Solution:
def prune(self, s: str) -> str: answer = "" for c in s: lc = c.lower() if ord(lc) >= ord("a") and ord(lc) <= ord("z"): answer += lc if ord(lc) >= ord("0") and ord(lc) <= ord("9"): answer += lc return answer
def palindromeCore(self, s: str) -> bool: if len(s) <= 1: return True else: return (s[0] == s[-1]) and self.palindromeCore(s[1:-1])
def isPalindrome(self, s: str) -> bool: s = self.prune(s) print(s) if len(s) <= 1: return True return self.palindromeCore(s)But I got Memory Limit Exceeded.
Improved
So I fixed it with a loop-based solution.
python
class Solution: def prune(self, s: str) -> str: answer = "" for c in s: lc = c.lower() if ord(lc) >= ord("a") and ord(lc) <= ord("z"): answer += lc if ord(lc) >= ord("0") and ord(lc) <= ord("9"): answer += lc return answer
def isPalindrome(self, s: str) -> bool: s = self.prune(s) if len(s) <= 1: return True for idx, c in enumerate(s): if s[idx] != s[len(s) - idx - 1]: return False if idx >= len(s)//2: return True return TrueResults
Runtime
- 120 ms, faster than 10.41% of Python3 online submissions for Valid Palindrome.
Memory Usage
- 14.7 MB, less than 42.81% of Python3 online submissions for Valid Palindrome.
Other Answers Online
python
class Solution: def isPalindrome(self, s: str) -> bool: n = s.replace(' ','').lower() s = ''.join(c for c in n if c.isalnum()) p = 0 q = len(s) - 1 while p < q: if s[p] != s[q]: return False else: p += 1 q -= 1 return True- Didn't know there was
c.isalnum().