Solidify the DP Formulation! | Day 2 Doubt Session | Dynamic Programming | Vivek Gupta
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- Опубліковано 28 гру 2024
- In this DP workshop, we are going to learn many DP formulations that are going to make solving DP problems easy for you. Register today if you haven’t.
🔴 Practice Problem Link: www.learning.a...
🔴 Registration link for the Workshop: algozenith.com...
🔴 Discord for Discussion: / discord
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give the link for cleaning window problem please
Solved all
Though I did cses DP
But these problems were refreshing
Thanks Vivek
22:40
we can reduce the space to O(N*W)
In the transition part
for(int take=0;taken
Great Video man
Very nice
These session helps develop thinking. Great bro. Waiting for more and more content from you
Solved Brick Colouring on my own🤤 Felt Amazing
Thank you for this series.
Though that long long thing took me some time to figure out😂
Great explanation 👍
thank you vivek bhai for the sereies you are clearing all the concept and the quiz section which you take in begining is unique and good brush up for session
Great Session💥
at 17:09 , i dont get the use of min(Wsum , W) ,especifically the use of Wsum
Great discussion 😊
the series is amazing bhaiya. so many previous doubts being cleared up in each video and so many new things to learn!
bhaiya its a great effort by you. thank you very much🙏🙂
great 🔥🔥
Thanks for sharing . It helps a lot.
*hey bro, excellent session* 🤗
i was able to think 3rd prblm : )
I have a doubt. While calculating complexity we multiply the max limit of the parameters passed and multiply this with (1+transition states). But what if some parameters passed are co-dependent? Then the time complexity would depend on only one of the co-dependent pair of elements right?
nice
At 22:34 if I don't take the next item why the sum doesn't become zero since we are starting all over again to take the items?
nice session
hey vivek , I tried a lot figure out the dp state for below question but couldn't able to do that can u help me to find the dp state.
Alexa to color all the walls on the occasion of New Year. Alexa can color each wall with one of the K colors. The cost associated with coloring each wall with a particular color is represented by a 2D costs array of size N * K. For example, costs[0][0] is the cost of coloring wall 0 with color 0; costs[1][2] is the cost of coloring wall 1 with color 2, and so on... Find the minimum cost to color all the walls such that no two adjacent walls have the same color.
TC-----> (N*k)
dp(i,last colour used) … transitions can be made O(1) using optimizations
@@vivekgupta3484 can u explain how transitions can be made O(1)?
You will have to maintain previous i's 2 best shops. If the best one is of the same colour, take the second best.@@rahulyela
@@vivekgupta3484 Got it vivek 👍
Thanks for replying ❤️
"Something in Input that changes through test cases then we can't cache it" can someone explain it more please?
can someone provide recursive code for the third problem
Bricks colouring problem?
👍
Going to Practice Problem Link says 404 error. Can you please fix it.
Cfbr
+1
:)