@Phantom mercy It exits the inner loop so w is out of scope.

lol

Pls bro, I'll like you the explain more on that of k[i-1] work without leading to error, of course I know that it is related to o base index but for example if we have a loop like this String hold="" ; for(int i=0; i

omg thank you so much!

can you please explain why it is an error?

@@KeyMajorSiddharth we want the weight of ith object for j=j-wt[i] so if you decrement i first it will point to i-1 object

please watch this playlist for detailed explanation of dynamic programming..kzfaq.info/sun/PLeF0b8iqbx4mTBJZ5ukIYj92_B4k2L1-8..

remember that sir has made all the first elements of both the array as zero..so the previous element will point to zero..so he has used this logic..for that equation..

@@AhamedKabeer-wn1jb no bro I think it still gives an error

you can think like generating the all the subsets of a set

Its also a possibility but complexity which we consider is worst case. So worst case complexity O(n*n)

Using dynamic programming, we may not call again functions for small values and take them immediately from a two-dimensional array and therefore the complexity is O (n * w), but if we called functions recursively, then the complexity would be O (2 ^ n)

You should put K[i][w] , and then a break line out of the loop , that way you get the table 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 1 2 2 3 3 3 3 0 0 1 2 5 5 6 7 7 0 0 1 2 5 6 6 7 8 Like this #include #define objects 4 #define capacity 8 using namespace std; int max(int a, int b) { if(a>b) { return a; } return b; } int main() { int Profit[objects+1] = {0,1,2,5,6}; int Weight[objects+1] = {0,2,3,4,5}; int Table[objects+1][capacity+1]; cout

​​@@xarenwoPls bro, I'll like you the explain more on that of Table[i-1] work without leading to error, of course I know that it is related to o base index but for example if we have a loop like this String hold="" ; for(int i=0; i

superb bro and also j>=0 at 15:53

thanks

no it is not. calculate by yourself for better understanding

He teaches how to mug up things in effective way. Even you don't know how the formula derived

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@@abdul_bari Thank you sir, for responding. I had to do something very similar in a test program. I don't think I got it right in that test. But later I had this idea that I could have calculated Profit Density. Can you please tell if the answer with that approach will be correct or not? Even though it will be a different problem.

@@AmitKumar-qh1ts No that will not guarantee the correct answer in all cases

Without getting too technical, "cout

I guess you would have rectified your problem but if not lemme help you out k[2][4] so item = i = 2, j = 4 column num i.e. instance or part of sack of total wt = 8, for item i = 2 weight is 3 (from above w[i] array ) basically weight for item 2 is 3 , as we check 3 less than j= 4 so we traverse case 2 , k[i][j] = max( p[i] + k[i-1][j-w[i] ] , k[i-1][j] ) i am using j instead of w in the table , to avoid the confusion of the w in above array of w[i] so i-1 = 1 , j-w[2] = 4-3 = 1, p[i] = 2, j = 4 so k[i-1][j-w[i] ] = k[1][1] = 0, p[i] = 2 hence p[i] + k[i-1][j-w[i] ] = 2 and k[i-1][j] = k[1][4] = 1 so as you see out of these two parts (2 and 1 ) max is 2 so the result or value of k[2][4] = 2 i am sure you would have been confused by the fact of using w in both places so use j in the loop pr table instead of w.

Pls bro, I'll like you the explain more on that of k[i-1] work without leading to error, of course I know that it is related to o base index but for example if we have a loop like this String hold="" ; for(int i=0; i

#include #define objects 4 #define capacity 8 using namespace std; int max(int a, int b) { if(a>b) { return a; } return b; } int main() { int Profit[objects+1] = {0,1,2,5,6}; int Weight[objects+1] = {0,2,3,4,5}; int Table[objects+1][capacity+1]; cout

the first "if" statement checks if either "i" or "w" equals to 0. there won't be a case where i-1 is negative because "i" won't be 0 due to the first check.

thanks bro

he explained the logic in previous video. which he says in the beginning of this video.

well i am beginner in programming and havent solved much of DP problems till now but can say , as of DP paradigm we see to achieve current step's optimisation , we compare and pick among prev step optimised solution and any prev solution part with current step addition, so here we apply max on these two ,1st prev step optimised version k[i-1][w] and current version is k[i-1][w-w[i]] + p[i] ,so here we are comparing with similar scenario thats why taking case from prev row and here k[i-1][w-w[i]] we are checking for rest of weight w-w[i] optimised part from prev solution with addition to the profit we are getting inn current state i.e. p[i] so thats what i think till now approach are being implemented

we consider only a single quantity of each element.

Thanks brother. I was having this issue for a long time and couldn't figure out what was the problem. Now i understood why was the answer wrong.

yes bro

The first values of vectors p and wt have to be zero. So, you have to use the following: p[] = {0, 15, 90, 160} and wt[] = {0, 2, 3, 7} Also int n = 3; int K[4][21]; Moreover, you can only take zero or one of every object. Hence the name 0/1 knapsack problem.

@madukesh kumar watched lecture no. 54

​@@VishalGupta-pm4bu In that video also he used pre-derived formula, he didn't derived it there he only showed results are similar that's all.

@@madhukeshkumar9783 sir has shown in previous video.

@@VishalGupta-pm4bu Bruh give us the link where Sir derived that formula...It would be of great help bhai