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# Valid Palindrome LeetCode Solution

## Problem – 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 LeetCode Solution in Java

``````public class Solution {
public boolean isPalindrome(String s) {
if (s.isEmpty()) {
return true;
}
int head = 0, tail = s.length() - 1;
cTail = s.charAt(tail);
} else if(!Character.isLetterOrDigit(cTail)) {
tail--;
} else {
return false;
}
tail--;
}
}

return true;
}
}
``````

### Valid Palindrome LeetCode Solution in Python

``````def isPalindrome(self, s):
l, r = 0, 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``````

### Valid Palindrome LeetCode Solution in C++

``````bool isPalindrome(string s) {
for (int i = 0, j = s.size() - 1; i < j; i++, j--) { // Move 2 pointers from each end until they collide
while (isalnum(s[i]) == false && i < j) i++; // Increment left pointer if not alphanumeric
while (isalnum(s[j]) == false && i < j) j--; // Decrement right pointer if no alphanumeric
if (toupper(s[i]) != toupper(s[j])) return false; // Exit and return error if not match
}

return true;
}
``````
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