it may have been easier to understand by keeping the/one example

Great explanation. Thank you so much. Keep this great work up !! 🥰

Keep up the good work!

good explanation = thank you so much + please continue❤❤

Thanks a lot this was helpful👍👍

给我讲懂了，太强了

What is your intution when you come across this problem?

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very easy and intuitive solution

Thanks!

Nice reading comprehension

Thanks for the solution, this would be very hard without sortedlist.

Hey buddy I have no idea about this but Im still watching..!

This was helpful thanks!

why do you draw that grid's border as a semi-circle? that should be a straight line god damn it!

this is a 4 color problem

class Solution: def findTheDistanceValue(self, arr1: List[int], arr2: List[int], d: int) -> int: myset = set(arr2) count = 0 def process(i,d): ans = [i] while d>=0: ans.extend([i-d,i+d]) d-=1 return ans for i in arr1: acceptable_values=process(i,d) ll=list(filter(lambda x:x in myset,acceptable_values)) if len(ll)>=1: continue else: count+=1 return count

Elegant

we can use return list(A.intersection(P)) to make it mush easier to write

If someone is good at codeforces problems can they easily become good at leetcode problems

Thanks for the video, helped me out. keep up the good work! 🌸

I have a doubt why we cannot make prefix sum of size same as nums size, why we initialize prefix sum 0 at 0 index

Thank you bro, this solution really helped me...

Nice explanation!

thank you!

good idea bro thanks

thanks broo

Nyc Explaination Bro

good explanation = thank you so much❤❤❤ please keep doing this kind of videos , because are very helpful for us

Thanks a lot!!!!! <3

code does work when hamsters =".HH.H.H.H.."

promo sm

Thank you so much for the video! It really helped me learn sliding window, such a smart way to solve it, the logic behind it is quite tricky though.

Nice Explanation

Great solution and easy to understand thanks for the whiteboard explanation.

So far. easiest explanation found on youtube ..Thanks !

great solution and explanation!

Can we use vanilla dijkstra instead?

Thanks, good explanation

it is not binary search solution((

Explained it in precise and concise manner ..amazing

your channel is a goldmine got my weekends covered now🙏

Thanks

The time complexity is O(n^4), sorry for that I didn't solve it online, after carefully checking, it should be this: we have s1, s2 with cache, it's n^2, we have for loop, it's now n^3, but we also have s1 == s2, it's another O(n), so the time complexity is O(n^4), if you have any questions, please make a comment.

awesome bro.....thankx

time is O(m + n)

Time is O(n^2), there is for loop inside.

Cool video, thanks!

I think they have added new test cases to this problem, this solution gives TLE now. But my intuition is that you can actually abstract the inner for-loop into its own method (example: def parseSubstring(...){}. If the for-loop in this new method is somehow transformed into a recursive call, then you can add @cache decorator above the new parseSubstring(...) method signature to memoize the results and prevent TLE. I'm too lazy to spend time implementing this though since I'd rather use my time to review other problems for my Amazon on-site lol.....

This really helped me understand this problem better, thanks