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.

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.

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 True

Results

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

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().