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Hi, everyone, welcome to this new video.

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So in this video, we are going to discuss the most important question of dynamic programming, which

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is 01 knapsack.

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So let me try to explain.

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What is the question?

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So there is one thing I've learned over and the thief wants to steal the item.

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So Thief has of it has a knapsack and the weight of the knapsack is engaging.

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And these are the food items that is present.

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Right.

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So a number of items are food.

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And this is the weight of each item.

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So this is one weights.

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So these are bits of each item and these are the prices of the item.

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Right.

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So this is price.

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So our aim is to do what he wants to do.

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He wants to maximize profit.

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Right.

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He wants to maximize profit given he has a knapsack with ten kids only.

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So what we need to do, we need to help Steve here and we need to maximize the profit of Steve.

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So I hope you have understood this discussion.

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And these are items to focus on his ball, second on his apple, third on his banana, and the fourth

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one is, let's say, orange.

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So how are we going to solve this question?

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So it's very simple.

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We're told we will use recursion.

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So how it will help us here.

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So let's talk about this item first.

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Let's talk about this item.

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Let's try to see how we can solve this.

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So the weight of this item is 15 gidgee.

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The weight of knapsack is 10 gidgee.

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So we cannot pick this item rate.

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The item weight is greater than the knapsack weight.

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So you cannot pick this item.

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So we are discussing one very important condition here, that if the weight of the item.

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So we have so this is four and the index is three.

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So we are discussing one important condition of the weight of the item is then the knapsack with.

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Then you cannot pick this item, right, in that case, it will do we will call the question on the

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rest of the items and the question will give us the maximum profit.

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So our answer will be our answer will be called recursion.

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So let's say the name of the function is knapsack.

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So you need to call the rest of the items basically three items, so called the recursion on the rest

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of the items.

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Fine.

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Now let's change the weight.

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So let's say instead of 15 gidgee, if the weight of the item is, let's say, six gidgee light, the

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weight of item is six gidgee.

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Now let's discuss about this item.

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So what will happen now?

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So what will happen now?

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This item has a very thick skin and the weight of the knapsack is engaging, so I can pick this item

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right.

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So I have two options here.

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Pick this item and try to maximize your profit.

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Do not pick this item and try to maximize your profit.

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So this is the condition we are, including the item we are picking the item and we are not picking

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the item, we are excluding the item, right.

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So we are excluding the item and everything myself.

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See, the weight of this item is sketchy and the weight of knapsack is engaging.

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So I can pick this item, but it is not necessary that I will pick this item.

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Right.

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It's my choice.

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It's my choice.

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But I can pick this item or do not pick this item right?

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Our ultimate goal is to maximize the profit.

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So what I'm going to do is I will consider both the cases.

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I will consider both cases the case when I am, including the item when I'm picking the item, the case

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when I am not, including the item.

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Right.

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And we will see which case gives us the maximum profit.

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So in case, if I am including this item, what will be my answer?

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My answer will be I'm including this item.

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So the price that I will give, the money that I will gain my profit will be rupees sixty dollars 60

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plus.

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After picking this item, after you have picked this item, what you need to do, you need to call recursion

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on the rest of the items.

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Right.

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To find out the maximum profit for the rest of the items.

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So we need to call recursion.

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On the rest of the items, but one more thing now, the weight of your knapsack has become full because

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you are including this item and the weight of this item is cagy.

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So 10 minus six, that will become four.

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So what you need to do so here, in this case, the weight will not change because you cannot pick this

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item.

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Weight of your knapsack will change.

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This is very different.

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So in this case, the weight of knapsack will not change, but in this case, the weight of knapsack

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will change.

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So weird.

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Minus six right now.

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Your knapsack has a very tough for now in the case.

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When I'm excluding this item, when I do not when I'm not picking this item and I'm trying to maximize

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the profit.

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So in that case, what you will do, what will be your answer?

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Your answer will be zero, because you are not picking this item and you will call Dacogen on these

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items.

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Right.

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You will collect on these items rest of the items.

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So knapsack, which is our recursive function, the number of items get reduced by one because you are

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not picking this item and the weight of knapsack will remain same.

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Right.

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So what we will do after finding the value for these two, including and excluding what we will do,

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we need to take the maximum out of them.

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We will take the maximum of include and exclude.

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Right will take the maximum because our ultimate goal is to maximize the profit.

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So I am calculating the profit when I am including this item, I am calculating the profit when I am

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excluding this item.

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So that's why I will take the maximum of them.

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And that will be my final answer.

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Right, so we know all the records.

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This is the first recorded evolution when the item where is good and upset with this is the second recorded

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revelation when you can pick this item and you are picking the item.

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This is the third quarter when you can pick the item, but you are not picking this item.

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Right.

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And finally, that makes some of these to include gays and exclude a case.

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That will be your final answer.

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Right, working with a biscuit's.

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Let's talk about this case.

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So in this case, if the number of items is zero, base case number of items is zero.

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We do not have any item.

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What would be your profit?

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Your profit will be zero if you do not have items to pick.

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Then how we can pick the item.

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Item does not exist.

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So your answer will be zero.

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Your profit will be zero.

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Similarly, if the weight of the knapsack is zero, the weight of knapsack is zero, you cannot pick

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any item.

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So in that case, also your profit will be zero.

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So these two are the best cases.

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Simple, right?

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So we now have base case.

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We now have recursive case.

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Let's start writing the code.

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So this problem is basically present on a bit rate zero one knapsack.

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Oh.

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So one thing that I forgot to explain is what do we mean by zero and one so.

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Zero one means here we have written, so zero one means you cannot break diatom, either you will pick

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the complete item or you will not pick the item, right.

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So it will not happen.

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Like you have an apple.

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And the cost of Apple is basically, let's say price of item is twenty two hundred dollars and it is

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five Gadge.

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So and the knapsack rate is two point five Kaju.

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So it cannot happen.

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Like you will break the apple into half and you take half of it.

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No that is not allowed.

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So either you pick the complete item or do not pick the item, but you cannot break the items.

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Right.

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So that is the meaning of a zero one here.

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So let's see, let's try to ride the recursive call for this problem.

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So these are a few examples given at a ribbon, so the naming is difficult, so is the weight of the

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knapsack.

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This is the weight of each of the item and this is the price of each of the item.

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So the first thing is, let's rename the vector is nothing.

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But this is your price at the price of each item.

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What is vector be your vector B's veered off each item, right, and what is in Jersey?

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So this is the total weight of the knapsack.

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Right.

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So, as discussed, what we are going to do is we are going to write one knapsack function, so knapsack

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function.

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This network function, what it will take, it will take how many number of items out there?

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That is price, not size or village, not size.

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So that will be same.

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That is the number of items.

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Second is it will take what is the weight of the knapsack?

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It will take your place area and it will take your word said.

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Fine, so let's write.

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Their definition of the function knapsack, this function will return as the maximum profit rate, the

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maximum profit.

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So how many items are there?

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Second, what is the weight of the knapsack?

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That is, what is the price of each item?

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What is the price of each item?

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So vector and price price.

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And what is the weight of each item?

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So we're adding.

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Fine, best case, so the best case that we discussed was if the number of items is zero.

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Number of items is zero, Battalia, maximum profit, if you do not have item to pick, then you cannot

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pick anything.

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So in that case, your profit will be zero simple and other businesses can be if the weight of knapsack

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is zero, if the weight of knapsack is zero, you cannot pick any item.

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So your profit will be zero.

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Simple.

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Well, what we can do, we can grab these two conditions here.

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So if the number of items is zero or the weight of knapsack is zero, then your profit will be zero.

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Now, let's talk about the recursive case, so the first case is basically if the weight of item is

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the weight of item is greater than the weight of knapsack, then you cannot pick this item.

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So in that case, where you going to do is we will simply call the recursion on the rest of the items.

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We are simply calling recursion on the rest of the items.

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So a number of items will reduce by one because we cannot pick this item right.

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So the knapsack will remain same then the price area, then the witchetty.

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So what is this condition?

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So if you see here I am talking about this condition, if the weight of the item is lower than the knapsack

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weight, for example, if let's say this is 20, so if the weight of the item 20 is greater than the

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weight of knapsack, then you cannot pick this item and you will collocation on this part.

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So call the equation on a number of items will get reduced by one because you cannot pick this item

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and the weight of the knapsack will remain the same.

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So that is what I am doing here.

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Right.

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So if the weight of item is lower than the weight of knapsack, then call that equation on the rest

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of the items.

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So number of items will get reduced by one.

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And this is your price area and you're very steady.

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Now, if this is not the condition in the part.

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In the early part, you see we have two options here.

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If the word is, let's say, six kids, so I have two options here, I can include this item and try

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to maximize my profit, exclude this item, and I will try to maximize my profit.

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So after including and excluding the maximum profit that I will get will be the maximum of both of them.

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So what we need to do, I have option here that I did include this item and find out the maximum profit,

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I did include this item and find out the maximum profit or exclude this item and find out your maximum

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profit.

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So exclude the item and find out the maximum profit, right, and what will be your answer, your answer

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will be make them off the maximum profit that you can get by, including the item or by excluding the

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item.

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Now, what we need to do, we need to find out the value of include and exclude.

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So what will be the value of include?

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Let's first talk about include, right, including the item.

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See if you include the item, for example, I am including this item, so 60, my profit will be 60.

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I am taking this item so I will get a profit of 60.

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So what is this, 60?

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This is your price at.

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Right.

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So the price of the item, the price of the item, plus the collocation on the rest of the items.

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Right.

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So.

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Place of and minus one, great plus, we need to call the on a list of the items so knapsack number

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of items will get reduced by one because you have picked this item also since you have picked this item,

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since we are picking this item, the weight of this item was six CGY and the weight of knapsack was

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langauges or 10 minus six for the weight of a knapsack will decrease, the weight of your knapsack will

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decrease.

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So the weight of knapsack will become the total weight minus the weight of this item find.

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But said will remain same, where today will remain same.

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So this is the record for including item.

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Let's talk about excluding the item when you are not picking this item.

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So when you are not picking this item, you will get zero profit because you are not picking item.

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Plus, you are calling the equation on the rest of the items, the rest of the three items.

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Right.

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Calling the equation on just of the three items and your knapsack rate will remain same because you

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are not picking the element or picking the item.

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So simply you are calling the recursion.

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So a number of items will get reduced by one because you are not picking this item, the knapsack which

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will remain same because you are not picking the item and that is your price area and that is your way.

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It's very simple, right?

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That is the complete code.

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So let's test our code and then we will submit.

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So our goal is passing basic test cases.

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Now let's try to submit our code so most really what will happen, our code will encounter random sorry,

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not runtime error.

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The time limit exceeded because our solution is recursive and right.

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So we are getting time limit to delay.

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Right.

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So and that was pretty obvious.

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So this is recursive solution.

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So that's why we are getting excited.

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Now, what do you want to do is in the next video we will convert this recursive solution into a solution,

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right?

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It will, right.

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The Dynamic Programming Board will convert this recursive coding to binding programming code.

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So one more thing that I want to discuss before ending this video is basically so here.

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What I'm assuming we are assuming that we have only one Apple, we have only one banana, we have only

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one orange.

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We have only one bowl.

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And if I'm picking the apple, then I cannot pick Apple again later.

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So here our inception in this question is we have one copy of each item made, one copy of each item.

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Now let's see if we have infinite apple, we have infinite orange.

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We have infinite bananas and we have infinite balls.

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So what will happen, let's say, if your TV is basically going into an Apple store?

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So there will be many laptops of exact same configurations.

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So basically in this question that I want to say, it is let's say I am modifying this question, do

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we have infinite we have infinite copies of each item, right?

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We have infinite apples.

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We have infinite range between finite bananas.

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We have infinite balls.

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So what I want to say here is if I am picking the item once, then here I am doing I cannot pick the

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item twice the rate.

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I can pick the item only once, but since now we have infinite copy of each item so I can pick one item

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more than one time.

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Right.

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So how we are going to modify this code for solving this problem when we have infinite copy.

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So it's pretty simple.

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What we are going to do is this is basically when you cannot pick this item, this is basically when

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you are not picking this item and this is the only recursive relation where you are picking the item.

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So what you are doing after picking the item, you're calling it because the rest of the items we are

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calling recursion on the rest of the item, and that is wrong.

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This condition has to be changed when we have infinite copy.

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So what do we do?

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We will not call the question on rest of the items.

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We will call recursion on all of the items.

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So for infinite copy, if you encounter one question and the question mentioned that you have infinite

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copy of each item, you can pick one item more than one times, basically infinite number of times.

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So the only change that you need to do in this code is basically, instead of calling that equation

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and minus one, you will call recursion on.

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And that is the only change that you need to do in this question for basically this infinite copy.

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Right.

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So, yeah, that's the last thing that I want to discuss about this question, the recursive code.

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So in the next three days or we will try to write the dining programming code for this problem.

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So this is all about this video.

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I will see you in the next one.

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Thank you.