It would be great if you could explain some of the Q4’s on the weekly/biweekly contests. I can never usually solve them but I can (usually) solve the average hard question.
Hi, sir, this is Purna from India, I always watch your videos for the problem-solving, "Wow, this video is an absolute gem! The way it breaks down the complex problem into easy-to-understand segments is truly impressive. The visuals are captivating, and the narration is clear and engaging. I learned so much from watching this - it's evident that a lot of effort and research went into creating such a valuable solution. This is definitely going into my 'favorites' playlist, and I can't wait to share it with my friends. Keep up the fantastic work!"
Hey NeetCode! How are you doing? In the top-down implementation, we can make an optimisation: if the first dfs returns true, there is no need to run the second one. So in line 13, I would add this condition: if !res and i < len(nums) - 2: Also, we can change line 16 to res = dfs(i + 3), as we already know res has been set to false, if we are going to execute this expression. Thanks for your amazing video! Massimo.
Can someone tell me why are we not returning true when we find one valid partition. Like in [4, ,4, ,4, 5, 6] we have [4, ,4] as valid partition but still we go and check for the partition [4, ,5, 6]. Why is this needed?
![Lp. Сердце Вселенной #1 НАЧАЛО ПУТЕШЕСТВИЯ [Новый сезон] • Майнкрафт](http://i.ytimg.com/vi/380Q7gLWGvk/mqdefault.jpg)
Excellent explanation!!!! Keep uploading daily leetcode problems and if possible weekly and biweekly contests too. It'll be very much helpful!!!!
It would be great if you could explain some of the Q4’s on the weekly/biweekly contests. I can never usually solve them but I can (usually) solve the average hard question.
Hi, sir, this is Purna from India, I always watch your videos for the problem-solving,
"Wow, this video is an absolute gem! The way it breaks down the complex problem into easy-to-understand segments is truly impressive. The visuals are captivating, and the narration is clear and engaging. I learned so much from watching this - it's evident that a lot of effort and research went into creating such a valuable solution. This is definitely going into my 'favorites' playlist, and I can't wait to share it with my friends. Keep up the fantastic work!"
That's a really nice explanation, I checked 3 different explanation for bottom up but yours is very simple and intuitive!
Awesome to see both bottom-up and top-down approaches. Keep it up
Hey NeetCode! How are you doing?
In the top-down implementation, we can make an optimisation: if the first dfs returns true, there is no need to run the second one. So in line 13, I would add this condition:
if !res and i < len(nums) - 2:
Also, we can change line 16 to res = dfs(i + 3), as we already know res has been set to false, if we are going to execute this expression.
Thanks for your amazing video!
Massimo.
Today completed by myself haha, no need to watch your video😁
didnt really understand the dp caching thing part... do you have any other video that explains caching better ?
Can someone tell me why are we not returning true when we find one valid partition. Like in [4, ,4, ,4, 5, 6] we have [4, ,4] as valid partition but still we go and check for the partition [4, ,5, 6]. Why is this needed?
thanks for the daily
Thanks for sharing such a excellent video!!
Ever considered using manim for animations? Might look prettier and like 3b1b
SUPER CLEAR!!
I wonder how the solution would change is we have to check the string in a circular way. e.g. 74456
Do you use your Microsoft surface to draw on during interviews?
Wow ... You are the king dude !!!!!
The bottom up dp part of the video, I could not follow
I could solve it the brute force but it fails with time limit. Can someone explain how caching works here?
Where did you go?
Awesome video
let's gooooo
First!