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 * 105
s
consists only of printable ASCII characters.
Valid Palindrome Solutions
✅Time: O(n)
✅Space: O(1)
C++
class Solution {
public:
bool isPalindrome(string s) {
int l = 0;
int r = s.length() - 1;
while (l < r) {
while (l < r && !isalnum(s[l]))
++l;
while (l < r && !isalnum(s[r]))
--r;
if (tolower(s[l]) != tolower(s[r]))
return false;
++l;
--r;
}
return true;
}
};
Java
class Solution {
public boolean isPalindrome(String s) {
int l = 0;
int r = s.length() - 1;
while (l < r) {
while (l < r && !Character.isLetterOrDigit(s.charAt(l)))
++l;
while (l < r && !Character.isLetterOrDigit(s.charAt(r)))
--r;
if (Character.toLowerCase(s.charAt(l)) != Character.toLowerCase(s.charAt(r)))
return false;
++l;
--r;
}
return true;
}
}
Python
class Solution:
def isPalindrome(self, s: str) -> bool:
l = 0
r = len(s) - 1
while l < r:
while l < r and not s[l].isalnum():
l += 1
while l < r and not s[r].isalnum():
r -= 1
if s[l].lower() != s[r].lower():
return False
l += 1
r -= 1
return True
Watch Tutorial
Checkout more Solutions here