Permutation in String - Leetcode 567 - Fixed Sliding Window (Python) 

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This is a more clear explanation than Neetcode. Solidified the concept right away. Thanks!

OMG! This is the clearest explanation I have seen! Really hope you keep making videos!

way simpler than whole Leet code discussion forum and editorial, you have my respect sir !

Better than Neetcode's explanation on the same problem, and more simple/ straightforward too! Came here cuz the neetcode one was too confusing and long

The best explanation I have come across for this problem! thank you so much

After watching 3 other videos, this is the clearest of them all, finally understood this problem. Fantastic explanations, pls don't stop making leetcode videos.

Very clear and detailed line by line explanation which helped me to understand the algorithm better. Thank you for your efforts !!!!

Another video you explained better than anyone else! Thanks greg!

The count array gets rid of the permutation problem and the sliding window trick gives a nicely scalable iteration. Edit: watched to the end... Nice

This is an epic explanation! Thank you!

Thank you so much for making this video! It is crystal clear!

Simple and elegant, thank you for the great video

Really helpful and well explained

for i in range(abs(len(s2)-len(s1))+1): if sorted(s1)==sorted(s2[i:i+len(s1)]): return True return False take alook mr.greg just tried

It's easy to understand, thank you!

Comparing two arrays and checking if they are equal should ideally be O(n1) right? What am I understanding wrong?

Could someone please explain to me why this is faster than the solution I wrote? def check_inclusion(s1: str, s2: str) -> bool: n1, n2 = len(s1), len(s2) if n1 > n2: return False permutations = Counter(s1) for i in range(n2): if permutations == Counter(s2[i:i + n1]): return True return False

just do this class Solution: def checkInclusion(self, s1: str, s2: str) -> bool: check = Counter(s1) temp = {} k=len(s1) for i in range(len(s2)): if i

Why we build up s2 count array in the first for loop? Why can't we just build it from start in the second loop?

elegant.

Good video

drawing tool name please

confusing solution

again, why don't you stick to dictionaries lol

you just got a new sub, thank you