Valid Palindrome
Problem
https://leetcode.com/problems/valid-palindrome/
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.
Example 1:
Input: s = "A man, a plan, a canal: Panama"
Output: true
Explanation: "amanaplanacanalpanama" is a palindrome.
Example 2:
Input: s = "race a car"
Output: false
Explanation: "raceacar" is not a palindrome.
Example 3:
Input: s = " "
Output: true
Explanation: s is an empty string "" after removing non-alphanumeric characters.
Since an empty string reads the same forward and backward, it is a palindrome.
Constraints:
1 <= s.length <= 2 * 10:sup:`5`sconsists only of printable ASCII characters.
Pattern
Two Pointers, String
Approaches
Explanation
First, we convert s to all lowercase. Rather than remove all
non-alphanumeric characters, we can ignore them when checking if s is a
palindrome. We do this by using two pointers i and j at the beginning
of s and at the end of s. When both pointers are on non-alphanumeric
characters, move them inwards.
Code
def isPalindrome(s: str) -> bool:
"""Check if ``s`` is a palindrome, ignoring non-alphanumeric characters."""
s = s.lower()
s = "".join(filter(str.isalnum, s))
left = 0
right = len(s) - 1
while left < right:
if s[left] != s[right]:
return False
else:
left += 1
right -= 1
return True
Test
>>> from valid_palindrome__two_pointers import isPalindrome
>>> isPalindrome('A man, a plan, a canal: Panama')
True
>>> isPalindrome('race a car')
False
>>> isPalindrome(' ')
True
>>> isPalindrome('0P')
False
Complexity
\(n\) is the length of s.
Measure |
Complexity |
Notes |
|---|---|---|
Time |
\(O(n)\) |
single pass through |
Auxiliary Space |
\(O(n)\) |
lowercase copy |
- valid_palindrome__two_pointers.isPalindrome(s: str) bool
Check if
sis a palindrome, ignoring non-alphanumeric characters.