Valid Palindrome - Solution
Solutions and explanations
Video Explanation
class Solution:
def isPalindrome(self, s: str) -> bool:
filtered = [char.lower() for char in s if char.isalnum()]
return filtered == filtered[::-1]
Complexity Analysis
- Time Complexity:
O(n) - Space Complexity:
O(n)
Here, n is the length of the input string.
class Solution:
def isPalindrome(self, s: str) -> bool:
left, right = 0, len(s) - 1
while left < right:
if not s[left].isalnum():
left += 1
elif not s[right].isalnum():
right -= 1
elif s[left].lower() != s[right].lower():
return False
else:
left += 1
right -= 1
return True
Complexity Analysis
- Time Complexity:
O(n) - Space Complexity:
O(1)
Here, n is the length of the input string.