125. Valid Palindrome

Sat Dec 20

LeetCode problem solving: checking if string is palindrome using two pointers

What does the problem want?

Finding a poalindorme type of question is a classic coding test that has been around for ages. This question in particular asks to check if we only consider the letters, whether or not it is a valid palindrome.

Brainstorm ideas

To solve this, I’d assume the first step would be to get rid of all white spaces, commas, other notations, and to convert all Upper-case letters to lower-case. After that, we can just comparae reversed string nad the string and return the comparison. That in itself is a pretty solid solution to finding a valid palindrome and can be written as the following.

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class Solution:
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    def isPalindrome(self, s: str) -> bool:
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        s = re.sub(r'[^A-Za-z0-9]', '', s)
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        s = s.lower()
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        reverse = s[::-1]
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        return s == reverse

But you’ll notice that this solution has a time and space complexity of O(n).

We can reduce the space complexity to O(1) if we use two-pointers, rather than reversing the array itself. In that case, we can compare the two ends on each side and check for negative results. If everything passes, it must be a palindrome, and we can return true.

Solution

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class Solution:
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    def isPalindrome(self, s: str) -> bool:
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        left, right = 0, len(s)-1
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        while left < right:
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            while left < right and not s[left].isalnum():
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                left += 1
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            while left < right and not s[right].isalnum():
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                right -= 1
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            if s[left].lower() != s[right].lower():
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                return False
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            left += 1
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            right -= 1
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        return True

Time complexity

This has a time complexity of O(n). Although we do have a nested while-loop, keep in mind that these loops do not exceed n and only act within the bounds/limit that we had originally set, which is the left < right. Therefore, even with the nested loops, this solution still retains the time complexity of O(n).

Space complexity

Since we don’t have any external variables that stoe any values, this has a time compelxity of O(1).