4.5 0/1 Knapsack - Two Methods - Dynamic Programming 

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bro did you watch the video.

@@jay-rathod-01 He is not wrong

i dont think it is necessary as by the end all of the objects are considered and you would have filled the cells with maximum profit

Just for future readers: the order of the items does not matter; no need to sort them. ``` import random m = 8 p = 1, 2, 5, 6 w = 2, 3, 4, 5 items = list(zip(p, w)) random.shuffle(items) results = [0] * (m + 1) for item_profit, item_weight in items: for target_weight in range(m, item_weight - 1, -1): results[target_weight] = max(results[target_weight], results[target_weight - item_weight] + item_profit) print(results) ```

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agree

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Hey can you tell me that while we eliminate an ordered pair using dominance rule...so for applying that rule our ordered pairs should be arranfed in ascending order of profits right??

yes profits should be in ascending order @@aarushitiwari4787

@@aarushitiwari4787 arranging the weights in ascending is the key here...because then we always check which prev weight to include or not

Pasd hua kya😂

Same bro

My brain got damaged

Thnx

why 0/1 knapsack needs a 2d array to memoize whereas house robber needs 1d array ? both are similar problems

@@jaideepsingh7955 In house robber we calculate Profit/weight where as in 0/1 Knapsack we can't divide the objects thus we store the profit value..

Brain left

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Hey, we do that formula just at last row right..??

@@samhita2917 Yes.

haha.. can't hold his Indian back for long

Love the explanation! Can you elaborate what is meant by TC, TLE and SC?

@@alexwhitewood6480 TC : Time complexity TLE: Time limit Exceeded SC: Space complexity

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but its 2^n bro

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It came back at 19:24 for the explanation using the Set method

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It came back at 1:55

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Peddinti Kartik cmt Ĺugj

Even mine too

Yup

First of all sort both the array wrt to wi arr in increasing order. Then try to make the table

wait u dont think he is ugly tho that was slightly off-putting lol 😅

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-VJbnWKUw-jw.html try this

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Right bro

i think 8-6 =2 used because max profit for bag is 8 and 8 is located in 5th row which is wight 5 which selected for including in bag. and then maximum profit is 8 5th row profit is 6 then 8-6 = 2 , and now we have to search in which row profit 2 is located which is 3rd row of weight 3 so in bag{wi 3,5} selected with max profit 8. please correct me if i am wrong

@@fardinalam7290 - I think that's incorrect, you want to subtract weight because that's the bounded variable. You want the most profit per unit of weight, and there's no limit to potential profit. Conceptually, you're solving sub-problems until you reach the final query. If ƒ(8) says the optimal item to take is weight 5, then you add its value, and query with the remaining bag capacity for ƒ(3), repeat until done. You don't want to take off the price, because that's a different unit than your input variable of weight. Note: I haven't watched this in a while, so I'd have to review it to be sure. I recommend solving the problem on paper, it helps give a better understanding of how it works. Especially if you explain it to someone else. However, that gave me an idea - it could be interesting to add another bound for this to limit the price as well - rather than taking items worth as much as possible like some kind of heist, maybe you're instead buying so you want to keep the price low, haha.

Sir please reply

if the profits are not in increasing order with weights then this method will fail

He is filling it by 3 because knapsack capacity is much large and weight is the same that can be placed inside it, see weight is same for each row..