👍🏻👍🏻

Same bro, hope you scored good in your exam

@@srujangurram yeah bro I scored 80 percent in my sem end examination

@@monishpidugu5134 that's awesome ! we have exam tomorrow. hope i does well too 🤞

I read I slept with my professor 🤣

My Professor Watches too but shortly explains algorithm 😂

You got me at "buffering" 😂😂

@@anupriyaranjan8966 Mam aap yaha 😄😀😄😀😄

DEAD

Possibly you will get the same question in the exam 😂😂

does not sound like a compliment

​@@kurru_kamiz3458why?it did sound like one

yes

yes bro

Hit like i have exam today 😂😂😂😂

If I watched this as my last resource the day before having an exam, I wouldn't pass anything. It's good to take a first glance, or to break apart a problem and solve it, but not as an academic foundation. Step up your game, don't be lazy.

@@carlosgallegos1265well it depends on person to person. I'm used to studying on the last minute and KZfaq videos are my last resort if I don't understand anything from books and Abdul Bari has been like a God to me literally. I wasn't much into algos and I'm really interested to get into this more. The teachers we have are too underpaid and are not as qualified as well as don't have their concepts clear some of the times. Having a community online helps us help one another and get info quickly without looking into 100s of pages lol.

+1

Sir have already explained how to solve the problem. Now it's your task of solving the whole problem.

Yeah, It seems so. Should have been ≥ m.

The 2nd bounding function is correct. It should be greater than m. You only apply the 2nd bounding function when descending down the right branch. When you descend down the right branch, it means that x[i]=0. In your example, after [1,5] and descending down the right branch x[2]=0 which means that 2 will not be included in the solution. The 2nd bounding function is telling you that you will not have enough values to reach the sum 'm' because the values going down the right branch will not be included in the solution. So for arr[1,2,3] and m=6, after taking 1 but not taking 2, you can't reach the value m=6.

No because if the rest of the weights can't add up to the number you are searching for then why continue? If the sum of your current sum plus the rest of the objects are less than that number then you'll never get it, so you stop.

I also notice.

yeah, noticed as well :D

Alaguseriel Azagu

Same 😅😅😅

Oh yeah, *Vedio.*

Same here also

Ya same here..even my professor gave the same example🤭

Same 😂

But sorting also takes some time O(nlogn) and here you are taking time to reduce time, which is meaningless

See the second bounding condition, if the considered weight + total remaining weight is less than target sum, then no need to search further

At first call, your example will be terminated

No, Sir, we cannot. That actually does not make any sense. Kadane's algorithm sees the sum as a whole, whereas this must be checked individually.

Thanks for liking

Yes bro u are right. That's why i come to the comment section. I am sure any one student are done comment on this silly mistake.

He does the same at 27,18 idk i think that is how you do it.

He is just taking track of remaining weight to become 0, that's why he is decreasing regardless of including or excluding the element

The left part of the node is the sum of all the x you included until now but the right part is the sum of all the remaining x you can add, so the right part changes repeatedly no matter if you include the particular x or not. So when x4 is excluded, the remaining choices you have are just x5 and x6, so sum would be x5 + x6 = 15+18 = 33 not 46. So think of the right part as something that has the sum of the remaining elements after a level you're at. At level x2, the values is x3+x4+.... and at x3 it is x4+x5+.... .