I need your support, and you can do that by giving me a like, and commenting "understood" if I was able to explain you. Keeping a like target of 500 ❤✌🏼

I'll be honest, I was bamboozled with the 0's in array edge case since DP17 and I was simply unable to find a clear answer from the comments. Had I simply closed my eyes and went ahead with DP 18, I would have legit saved ~ 2 hours of confusion!! Thankyou so much Striver, this lecture cleared all my doubts 🔥🔥🔥🔥🔥

hats off man,, spent 2 hours on DP17 to fix that but after watching this DP 18 everything went super smooth

I studied DP from aditya verma, but was never able to figure out how to handle the zeros in the array, you made it super easy, Thanks!

Understood! Hats off to ur dedication, u are still teaching while suffering from fever.

Striver's concept explanation is so cool, easy and easily get stuck into the head. I wish to meet him one day and say a lot of thanks to him.

The deduction of (totalSum - D) / 2 was amazing. Understood!

Understood 💯💯Great Explanation. Thank you very much for all you efforts🔥🔥

understood until 14:00 ❤ . Will learn the optimization later

Hello, The tabulation code is not passing all test cases in both videos dp-17 && dp-18. Is anyone facing the same problem. Pls HELP.

Us, but the way you transformed the problem into previously solved problem is amazing , that's the way we have to think ... Thanks ❤

understood. Also thanks for explaining Space Optimization so good. I am from tier-3 mechanical never knew all these stuff

UNDERSTOOD.........Thank You So Much for this wonderful video...............🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

UNDERSTOOODD!!! Thank you, Striver!!

After this video, they updated the problem! Real influencer

Understood 👍. Hats off to your dedication for us. ♥

I was confused with the previous video 0 case, but now understood. Thanks.

Another modified target could be (difference + totalSum)/2; To explain this how this came is : s1= sum of elements in subset 1 s2 = sum of elements in subset 2 s1 - s2 = d (We need to find 2 subsets with difference d ) s1 + s2 = totalSum (We know the sum of 2 subsets would be equal to the total sum of array) Adding these 2 equations , we get s1 = (d + totalSum)/2, thus we only need to find a subset with sum s1.

Already understood before starting of video that's what Striver teaches us

Great Explanation Striver , Thanks

Understood! Striver. The best Software Engineer himself

UNDERSTOOD... ! Thanks striver for the video... :)

understood , the space optimization is amazing sir

You have literally made dp look so easy !

Understood Thanks Striver for this Dp series

understood. Thank you so much

bro doing gods work

Understood Bhaiya!

Understood. Thankyou sir.

Understood, sir. Thank you very much.

Understood! Thank you so so so much!!!

Understood 💯💯Great Explanation

Nicely Explained! Understood!

In the space optimized solution why the loop sum=1 to target doesn't work as we have been doing previously and setting curr[0] =1 since for any index with target as 0 there is only single subset. Can someone explain please?

we can also use : if(ind < 0){ if(sum==0) return 1; return 0; } apart from these conditions: if(ind == 0) { if(sum==0 || num==arr[0]) return 1; if(sum==0 && arr[0]==0) return 2; return 0; }

Thank you so much!

Superb Bhaiya guys watch at 10:10 Amazing Concept Again and again Thankyou bhaiya Aka Striver

Thank you striver ! Just a note : writing recursion from 0 to n-1 looks far easier to handle bases cases than writing recursion from n-1 to 0 on subsequence problems

hey can anyone please clear my doubt. I'm applying DP-16 concept.like after filling the tabulation with boolean, check and try for every s1, that it's possible to get standing on last index, if possible then calculate s2 and check the condition given in the qsn and increment count, Now this method is not working, I think it's because the tabulation is not recording duolicacy, e.g, In array if the sum 7 is made by more than one way, then it's not taking that record, and while checking for (s1>s2 && s1-s2==d), only once the count is increasing, is it the actual reason of failing ?

4:48 or I think, we can go 1 step deeer into indx = -1 then base case would be simpler if (indx == -1) { if (sum == 0) return 1; else return 0; }

Thank you so much I've understood

Understood Sir.

Understood ❤

Understood!

completed subsequences set problems great learning

Hey, Striver I understood the logic in this but how are you able to think of using logic like this very quickly ? , i didn't get that by looking at the question.

Understood, well explained.

Understood 🔥🔥

UNDERSTOOD

Understood !!

In Lecture 17 , we can sort the vector and reverse it, then no need to change base case it will work absolutely fine.

If we have handled the case num[0] == 0 previously then why we are checking it again for num[0]

understood!

Understood🎉

at 15:53 , the target should run from 1 to sum right? we have run from 1 to sum in the count subsequences with sum k also, why did you take from 0 to sum?

understood

another base case(Better and Concised): if(ind

Thank you

I thing the simplest way to get rid of so many conditions of 0's we can simply start the recursion for 0 to n-1; then we will get correct ans i.e.4;

Calculating Max Sum of Array with the following condition, If A[i] element contributing in max sum,than you can't consider A[i]-1 and A[i]+1 element for max sum contribution, if present in array . Note: Elements of the array can be repeated multiple times. It's unsorted. It can have 1 to n number of element. eg: int A[ ]={1,6,7,1,2,3,3,4,5,5,5,6,9,10}; If I consider 10 as contributing max sum, I can consider 9 and 11. (Here 11 is not present but 9 is present still we can’t consider it). If I consider 5 (total max sum contribution 5*3) but can’t consider 4 and 6. I have tried sorting and storing index and count in the ordered map. Also try to compare with unbounded knapsack criteria. Still, I was struggling with an optimal and clear idea to calculate. Any idea how to approach it .

understood😃

understoooooooood

OMG bhaiya ...simple "genius"

Understood. Best on whole yt.

For Those who did not understand why total sum-d has to be even so imagine totalsum=15 and d=2 now count s2=(15-2)/2 it will give s2=6; ans s1-s2=d so s1=2+6=8 now here it is said that s1+s2=totalsum but s1+s2=14 which is not equal to total sum to total sum-d has to be even

Understood.

what we do in case of negative elements if its given?

Love it ✌🏻☺️

amazing

int gfg the case where 0 can be in the begining can be solved by sorting given array in decreasing order but this will increase time compl,but will still pass test cases;

Thank you striver, made it super easy and interesting to learn DP. "UNDERSTOOD" 💓💓

Can anyone please explain why target is starting from 0 here? in all other problems it's starting from 1. If I take 1, some test cases are failing

Thankyou very much for explaining base case of last question. I was stacked for last 6 hours. ❤️❤️

bhaiya in count subset problem which u had taken earlier , in tabulation the loop started from 0 to sum but when sum==0 it has been handled previously right? then it should start from 1 right? as in problem subset sum==k u started the sum loop from 1 to sum as for sum==0 we have handled it previously before the loop.

That algebra at 10:15 is crazy I would never see that under the time constraints of a real interview. It's brilliant tho

Understood

Understood!!Thank you

Hare Krishna..!! understood.

Can someone please explain me the case if(nums[0]

What if we find with target = totalsum Then at n-1 iteration of tabulation find S1 and S2 by totalsum - S1. Then If S1>=S2 and S1-S2=D we can return True else False

Below code is failing on 7th test case, giving wrong answer. Can someone help in finding what is the mistake I am making. int countParts(vector &arr, int sz, int d, int index, int s1, int s2, vector &dp){ if(index == -1){ if(s1>=s2 and s1-s2 == d) return 1; return 0; } int mod = 10e9 + 7; if(dp[index][s1] != -1) return dp[index][s1]; int pick = 0, notPick = 0; if(s1>=s2) pick = countParts(arr, sz, d, index-1, s1-arr[index], s2+arr[index], dp); if(s1>=s2)notPick = countParts(arr, sz, d, index-1, s1, s2, dp); return dp[index][s1] = (pick%mod + notPick%mod)%mod; } int countPartitions(int n, int d, vector &arr) { int totSum = 0; for(auto &val: arr) totSum += val; vector dp(n+1, vector(totSum+1, -1)); return countParts(arr, n, d, n-1, totSum, 0, dp); }

just mind blowing how you came up with the modified target, kyse kar lete ho bhaiya?

In that case my recursive code was going from 0 to n-1 and to avoid case of zero I sorted my array hence all the zero cases were sorted

Is there any easy way to write base case as I am getting really confused

You have expressed S1 in terms of S2 then solved this problem. Can we do vice versa then solve this problem?

Thanks Striver. Understood.

In lecture 17, can we handle the cases having zeroes by just multiplying our previous answer with pow(2,n) where n is the no. of zeroes

Can anyone explain why are we always initializing the DP[0][i] with val[0]: for(int i = weight[0];i

Understood...Completed 18/56

15:00 the most important point to be noted if you are in an intervie

why return 0 case is not considered in base case of tabulation?

new welcome line

Understood thanks :)

why line number 23 if(num[0] != 0 && num[0]

Thanks a Lot Striver 🥰

understand

please tell us the time when this amazing course will get end......expected time??

for DP 17 for [ 7,1,0,2,5] with tar=7 Method 1: originalAns*(pow(2,n)) Method2: Changes in base case will that work? I can see from recursion tree there would be redundant calls at level where 0 is considered

can we not put if sum == -1 then return 1;

as we are handing separately so now we need our TARGET LOOP should start from 0...correct ??