_125. Valid Palindrome.md

125. Valid Palindrome

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First completed : June 07, 2024

Last updated : July 01, 2024

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Related Topics : Two Pointers, String

Acceptance Rate : 49.09 %

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Solutions

• e125.c
• e125.py

C

e125.c

bool isPalindrome(char* s) {
    int left = 0;
    int right = 0;

    while (s[right + 1]) {
        right++;
    }

    while (left < right) {
        if (s[left] <= 90 && s[left] >= 65) {
            s[left] += 32;
        } else if (!((s[left] <= 57 && s[left] >= 48) || s[left] <= 122 && s[left] >= 97)) {
            left++;
            continue;
        }

        if (s[right] <= 90 && s[right] >= 65) {
            s[right] += 32;
        } else if (!((s[right] <= 57 && s[right] >= 48) || (s[right] <= 122 && s[right] >= 97))) {
            right--;
            continue;
        }
        if (s[left] != s[right]) {
            return false;
        }
        left++;
        right--;
    }

    return true;
}

Python

e125.py

class Solution:
    def isPalindrome(self, s: str) -> bool:
        singleStr = re.sub('[^A-Za-z0-9]', '', s).lower()

        for i in range(len(singleStr) // 2) :
            if not singleStr[i] == singleStr[len(singleStr) - i - 1] :
                return False 

        return True