Valid Palindrome - Leetcode 125 - Python

0:00 - Read the problem
1:40 - Explaining Solution #1
3:45 - Coding Solution #1
5:40 - Explaining Solution #2
9:35 - Coding Solution #2

leetcode 125

#coding #interview #python
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[davidfranke]

It's so damn satisfying having done a solution and seeing that Neetcode uses the absolute same method.

[callmebiz]

I'm currently prepping to interview with Google in a couple months. Just wanted to let you know you've been an extremely helpful resource in getting ready for this! Thank you so much and hoping all is well :)

[johns]

Yes!!! Thank you for continuing with the LC 75 series!! We all really appreciate it.

[HowToAiNow]

The reason why the first solution is faster than the second one is that the method isalnum() is written in C behind the scenes while the solution you achieved is written in pure Python. C is a much more performant language than python, so thats why. Interestingly, a large part of Python methods are written in C.

[Extremesarova]

Actually, we can write function for checking alphanumeric characters like this and it will work
def isalnum(c: str) -> bool:
return (("a" <= c.lower() <= "z") or ("0" <= c <= "9"))
We don't need to use ord()

[wasbashing]

I picked up blind-75 aswell! Keep up the grind as always 💪

[turtlenekk]

Heres an optimization ive noticed....you can avoid having to write while l < r in the checking parts again by changing it to an if statement and using continue.. so it would be something like:
if not alphanum(s[l]) :
l+=1
continue
same for right pointer
if not alphanum(s[r]) :
r-=1
continue

[youngmoneymahini]

Could someone elaborate on why he includes "l < r" and "r > l" in the internal while loops? My assumption is that the outermost while loop already accounts for that piece of logic (while l < r:...)

[nero]

I'd let out a big sigh if the interviewer asked for another approach after I show him the first solution

[adventurer]

The nested while loops is a bit hard to read. We could easily do with if else statements:
while l < r:
if not self.alphaNum(s[l]):
l += 1
elif not self.alphaNum(s[r]):
r -= 1
elif s[l].lower() == s[r].lower():
l += 1
r -= 1
else:
return False
return True

[ah-rdk]

7:03 Yeah, but ord() and chr() are also built-in functions like isallnum().

[AustinCS]

I'm really glad that you use a different solution here that is actually O(n) rather than the advanced algorithms course. I see in the advanced algorithms course that you clean the string using appending method to a new string. In fact, if this is a paid course it is IMPERATIVE that you clarify that the method used there is polynomial time to clean the string. Because under the hood you are doing 1 + 2 + 3 ... N operations since it instantiates a new string each time you append a char. That's the impression I'm under, someone correct me if i'm wrong, and if that's not clarified in the course then it's really going to be a shame when someone fails an interview because of it.

[DonJulio-hruj]

The first solution is actually O(n^2) time where n is number of alphanumeric characters.
This is because python strings are immutable. "Appending" a character creates a new string each iteration.
It's better to build the new string with a list, and then using join later.

[SteeleJackson]

i ended up using this is_alnum function instead of the one used in the solution. it made more sense to me than using the <= approach:

Def is_alnum(self, c):
return ord(c) in range(48, 58) or ord(c) in range(65, 91) or ord(c) in range(97, 123)

the ranges correspond to the ascii vales for digits 0-9, uppercase letters, and lowercase letters respectively.

[Xushkkkk]

Helpful video. One can also perform a regex check to make sure the character values are in the alphanumeric range

[ShadyAlwidyan]

Thank you for making these resources available

[johns]

After LC 75 is done, can you do the SeanPrashad 170 questions? If you have both playlists, that would be HUGE

[kapilpatel]

I solved this already, but coming here I see instead of remembering ASCII code it is fine to use the function and this is was a saviour.

[m_jdm]

Ok, I'm going to be positive and say that, this is a great question. What it thought me stayed and made me a better programmer.

[crikxouba]

Took me minutes to a variant of the first solution (using comprehension list), whereas other medium challenges in LeetCode take me hours and I often need to check the solution to get it, the difficulty level is all over the place on LeetCode, I don't get how they rank it.