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# 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.

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().