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This simple LeetCode problem (Problem #141) demonstrates the power of two pointers to reduce runtime.
Problem:
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.
Solution
def isPalindrome(s: str) -> bool:
l = [x for x in list(s.lower()) if x.isalnum()]
left = 0
right = len(l)-1
while left < right:
if l[left] == l[right]:
left += 1
right -= 1
else:
return False
return True
Runtime: Runtime is $O(n)$ and the space complexity is $O(n)$ (as we are creating a list of size $n$).