DP 18. Count Partitions With Given Difference | Dp on Subsequences

[@takeUforward]

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 ❤✌🏼

[@iamnoob7593]

understood

[@himalayadebbarma-we4pt]

Understood

[@shubhamshukla4215]

what if test case in dp17 is like {2, 0 , 0 , 4 , 2 ,2} and sum = 4 , then will this solution work ? i guess at the time whenever we found sum == 0 at index > 0 , then we have to calculate number of zero till index == 0 let's suppose it is z , and then retum 2^z . What everyone think about it ?

[@AbhishekKumar-cv1dh]

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 🔥🔥🔥🔥🔥

[@mayanksingh7501]

Those confusion and doubts and 2 hours will only help in longer run. Good work on not directly jumping to another video without clearing your doubts on your own.

[@after_dark_777]

for real should have moved on to this video those comments were so confusing

[@SamreenAftab-lg8tx]

Samee

[@solitucedagar4298]

sameeeeeeeeeeee

[@nikhilmadaan29]

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

[@Aniruddha_Chowdhury]

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

[@saketsoni2587]

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!

[@entertainmenthub373]

but aditya verma always tell you the in=dentification and where u can use this apparoach nd similar questions he is the god of cp ik he skip the case of 0 in array but aditya verma is for cp where u can use which approach and how to identify which approach by seeing solution

[@iamnoob7593]

@@entertainmenthub373 God of cp is tourist.

[@divyanshrawat2859]

its not like he was not able to figure out , he always gives an intution to solve the problem , not like come and tells you the whole solution.

[@saumyasharma9993]

aditya vermas approach works with zeroes as well, as in that we subtract dp[i-1][j-arr[i-1]]

[@PIYUSH61004]

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

[@pulkitchausali1354]

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

[@saurabhsaha6958]

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.

[@ankanbasu7381]

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; }

[@introvert9112k]

Then how do you take care of base case in Tabulation. As dp array cannot have -ve indexes.

[@RTXCR7]

@@introvert9112k for that i think we would need to make 1 indexed arrays of size n+1 having an extra 0 index free

[@cs26divyanshisachan37]

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

[@harshitsharma5647]

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

[@chetanthakral5322]

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.

[@takeUforward]

But this will increase the space req :)

[@chetanthakral5322]

@@takeUforward Oh Okay , I didn't think about that !

[@varunaggarwal7126]

@@chetanthakral5322 lol, I thought the same thing, but its not pure math its cs.

[@UCSParnashreeDas]

@@varunaggarwal7126 how this increases the space.. can u explain

[@varunaggarwal7126]

@@UCSParnashreeDas it's been 1 month,lol i have to revise this dp, but I guess you can see + sign while striver is subtraction, means less

[@parthsalat]

Guys! For DP 17, Memoization code, striver's approach is correct (6:38), but here's a cleaner alternative for the base case: if(ind

[@abhishekgururani6993]

It'll be hard to convert this base case to tabulation dp states as it's not well defined wrt the indices.

[@parthsalat]

@@abhishekgururani6993 You're right. But the tabulation base cases are straightforward and simple. One can write that by applying basic logic.

[@amitmahato6404]

Yes, I also tried this. got correct in both tabulation and memoization. Striver has complicated this solution a little bit.😅

[@amitmahato6404]

@@abhishekgururani6993 Simple base case, if sum == 0, return 1; means fill all the 0 indices row with 1. The change is that you have to traverse j=0 to j

[@blackblood181]

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; }

[@anuragprasad6116]

this makes the code really simple to understand. perfect base case.

[@theterror9080]

Best Bhai @@anuragprasad6116

[@manojnandhan8677]

but how will you handel -1 index in tabulation?

[@narendramaurya4823]

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

[@stith_pragya]

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

[@adebisisheriff159]

Understood! Striver. The best Software Engineer himself

[@gauravgogoi5240]

After this video, they updated the problem! Real influencer

[@balajisrinivasan6618]

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

[@santoshpokhrel7693]

but for me it creates problem while writing the tabulation form. Esp. with all the subsequences questions.

[@chase.2595]

yeah man, we have to start from n-2 if we do 0 to n-1 in recursion@@santoshpokhrel7693

[@ganeshktvb9234]

yaa bro really it creates confusion in tabulation @@santoshpokhrel7693

[@hashcodez757]

"UNDERSTOOD BHAIYA!!"

[@tarunrao1505]

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.

[@aryarajendra1088]

same bhai

[@architgupta3551]

@@aryarajendra1088 #include int countPartitions(int n, int d, vector &a) { // Write your code here. long long sum=0; for(int i=0 ; i

[@kevalkishan3824]

@@aryarajendra1088 try doing mod with 10^9+1 dp[row][col] =(dp[row][col]%mod+ dp[row-1][col]%mod)%mod; int val = num[row-1]; if(col>=val){ dp[row][col] =(dp[row][col]%mod+ dp[row-1][col-val]%mod)%mod; }

[@itsmepratham2712]

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

[@molymusic1188]

thankyou my guy🙌

[@soumik5854]

solved this problem on my own very proud of this

[@dharmeshpoladiya9047]

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

[@aryanagrawal4794]

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.

[@takeUforward]

Yes u can, won’t be an issue :)

[@aryanagrawal4794]

@@takeUforward yes i did that in the count subsets problem, but when I did that in the current problem of partition i.e. looping from 1 to sum it's giving WA, but when i changed to 0 to sum it's giving correct and. Why so?

[@takeUforward]

@@aryanagrawal4794 because in partition at 0 there are other cases to handle. Here at 0, its take and notTake, hence

[@aryanagrawal4794]

@@takeUforward ok bhaiya got it.

[@madhurnarang2440]

@@aryanagrawal4794 Hey can you explain it to me?

[@vaibhavrai7042]

To deal with the 0 case, we can also take the dp array to be of size [n+1][sum+1] and initialize the dp[n][0]=1 int[][] dp = new int[n+1][sum+1]; dp[n][0]=1; for(int index=n-1;index>=0;index--)

[@Lyrics_dunya]

For Problem 17 , I struggled a bit in Tabulation while writing the base cases for the new changes. All 3 cases covered, also notice the bases cases are written in reverse order, so that the priority is given to the last one if its true. One more thing since we are considering the case for i = 0, please start the 2nd loop of target from 0 till k(inclusive). Happy learning! dp[0][0] = 1 if arr[0]

[@TyagiAviral]

Hey, thank you. Great comment.

[@MukeshKumar-cc3uh]

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

[@gauravgogoi5240]

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

[@manishisaxena5657]

actually we can make the base case as if(index == -1) { if(tar == 0) return 1; return 0; }

[@anuragprasad6116]

underrated! taking this as a base case really simplifies the solution.

[@ScienceSeekho]

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

[@yashshukla1637]

Done. Understood the concepts so well.

[@adityasaini7165]

we can also do like this, private static int count(int i, int[] num, int k){ if(i == num.length){ if(k == 0){ return 1; }else{ return 0; } } int take = 0; if(k >= num[i]) take = count(i + 1, num, k - num[i]); int notTake = count(i + 1, num, k); return (take + notTake) % MOD; }

[@akr1831]

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

[@anveshkarra3861]

understood , the space optimization is amazing sir

[@tasneemayham974]

UNDERSTOOODD!!! Thank you, Striver!!

[@SaiPhaniRam]

Python implementation: def countPartitions(n: int, d: int, arr: List[int]) -> int: ''' Carefully analyze the problem. Let S1, S2 be the sum of two different subsets S1 + S2 = total_sum of array elements Goal: Count the subsets, such that (S1 >= S2) and (S1 - S2 = D) total_sum = S1 + S2 => S1 = total_sum - S2 From S1 - S2 = D total_sum - S2 - S2 = D S2 = (total_sum - D) // 2 (or) S1 = (total_sum + D) // 2 Problem boils down to finding number of subsets whose sum is (total_sum - D) // 2 (or) sum is (total_sum + D) // 2 ''' ''' Space Optimization ''' if not arr: return 0 total_sum = sum(arr) target = (total_sum - d) // 2 # print('Target: ', target) if (total_sum < d) or (total_sum - d) % 2 != 0: return 0 mod = 10 ** 9 + 7 prev = [0 for t in range(target + 1)] # Base Cases # 1. At index 0, if val == 0 => 2 (pick/ unpick doesn't matter) # 2. Else, At index 0, val 1 way (pick it) prev[0] = 2 if arr[0] == 0 else 1 if arr[0] != 0 and arr[0]

[@GManiKarthik]

#Understood #DP18 #Hatsoff #Striver

[@champu5645]

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;

[@nithish_raina]

Yes, and it would also work even if the array doesn't contains any zeroes.. The base cases are also so simple to handle over here.

[@VinayKumar-xs6el]

can u share the code here please that starts from 0 & ends at n-1

[@champu5645]

@@VinayKumar-xs6el will share soon, i also have to check where i have written the code. It's been 2 years

[@AmanBawane-o9q]

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

[@techietech7073]

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

[@YatriSpecial]

How do I write the base case in tabulation, Could you please tell me?

[@nithish_raina]

Method 1 is time intensive if let's say the no of zeroes were 15-20 in an array...Hence method 2 with handling zero in the base cases is required at such cases.

[@ritikshandilya7075]

Great Explanation Striver , Thanks

[@andrewcenteno3462]

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

[@chaitralishinde7815]

Understood better than ever!

[@rohandevaki4349]

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?

[@TheDebuggers-fx9vp]

i thnk a sligh change in base case can handle all the case well,instead of returning when we encounter the target to 0; if we keep exploring until the end of the tree then we can get our desired result class Solution { public: int helper(int arr[], int n, int sum, vector& dp) { // Base case: If there are no elements left if (n == 0) { return (sum == 0) ? 1 : 0; } // If the subproblem has already been solved, return the stored result if (dp[n][sum] != -1) return dp[n][sum]; // Compute the result for this subproblem int notPick = helper(arr, n - 1, sum, dp); int pick = 0; if (arr[n - 1]

[@ShahNawaz-cx3pi]

*********** Iterative Code for the count number subsequence whose sum is k ( & 0

[@nehakaushik3602]

OMG bhaiya ...simple "genius"

[@shoryapawar2219]

Understood Thanks Striver for this Dp series

[@parthsalat]

DP 18 starts from 7:25

[@Sneha-gb7iq]

For tabulation, we can do dp[0][0]=1 and dp[0][nums[0]]+=1 Without checking for all cases

[@sonicboom6635]

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?

[@Parthj426]

us But i got confused as in DP 18 ques , it is written that two partitions have their union as Whole array . It should have been given that both are exclusive of each other , for being self-explanatory well.

[@sahilarya9510]

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

[@dharsan.s7937]

March

[@kathanvakharia]

Understood...Completed 18/56

[@ntgrn-pr5yx]

understood , thank you striver

[@ranasauravsingh]

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

[@musichub2168]

Understood 💯💯Great Explanation

[@arpitrajput6424]

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

[@riddhibandyopadhyay584]

Yes, but it will take extra time complexity

[@arpitrajput6424]

@@riddhibandyopadhyay584 ya you're right but can be done ✅ with this approach .👍

[@priyankabagal9269]

You have literally made dp look so easy !

[@tallalhabib4526]

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

[@beginnertopro7265]

tabulation -; previous question include 0 in question;- void Vikash() { int n, k; cin >> n >> k; vector ans(n); F(i, 0, n) cin >> ans[i]; vector dp(n, vector(k + 1, 0)); // cout

[@undefined_exception_0]

There is no need of the first loop. int countSubsets(int arr[] , int size , int k) { vector dp (size , vector (k+1 , 0)); // base case if(arr[0] == 0){ dp[0][0] = 2; }else{ dp[0][0] = 1; } if(arr[0] != 0 && arr[0]

[@ayushidalal5488]

@@undefined_exception_0 This works perfectly. Thank you!

[@VikasGupta-ok9lh]

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

[@inderjotsingh5868]

you don' t even need this , just go from 0 -> n , and if at n , target == 0, return 1 else 0 , all other cases are already been taken care off , if you do in this way . How ? our code will do this will we are making the recursive calls at n-1

[@shriRadheHariom]

Understood, well explained.

[@shubh625]

ok

[@uvs_6032]

I have a doubt. The question says the the union of two subsets will give the original array , that means there can be repeating elements and its not necessary that S1 + S2 = totalSum E.g -> arr = {1, 2, 3, 4, 5} totalSum = 15 S1 = {1, 2 ,3, 4} S2 = {1, 5} sum1 = 1 + 2 + 3 + 4 = 10 sum2 = 1 + 5 = 6 The union of the above two will give original array but sum1 + sum2 != totalSum. So how is the code working ? Is the problem statement wrong?

[@kushbudhiraja4927]

you are right code is only passing 6 test cases not all

[@paracetamol9116]

the question's wording is wrong, look at the same question on gfg, won't see "union" written anywhere.

[@ankishkhandelwal1061]

In case arr =1,2,3,4 Diff =2 S1>S2 Target is =(10-2)/2 =4 Give 2 as answer 4 3,1 But 1+2+3-4=diff How this case will handle

[@ShahNawaz-cx3pi]

In question it is mentioned that the union of both the set is equal to the given array.

[@ankishkhandelwal1061]

@@ShahNawaz-cx3pi got it

[@TanviPoddar-f1k]

At 15:15, shouldn't the condition be ((nums[0] != 0)&&(nums[0] == target)) ??

[@TON-108]

Understood Bhaiya!

[@akashkumarsingh3595]

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

[@swagcoder]

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

[@affanrahman5711]

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

[@takeUforward]

I discussed that only, but that adds a log n

[@sourabhchoudhary7289]

@@takeUforward Does 1ll

[@sanskarjaiswal4394]

Instead of changing the code we can just add a if statement before "take" variable that if arr[index]== 0 then skip it...

[@tami9154]

wouldn't it be better to go an ind

[@sakshamsengar9798]

if we keep base condition like this then we dont have to check for individual case ....this is giving right answer..... if(i==n){ if(s1==0)return 1; return 0; } if(s1

[@prabhakaran5542]

Understood ❤

[@abhisheks7450]

why doesnt the for(int i = 0;i

[@kushjoshi9716]

Nicely Explained! Understood!

[@anshugangwar7344]

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 .

[@AnanditaTiwari-c7m]

Create an array dp of size n, where n is the length of the input array A. Initialize all elements of dp to 0. Sort the input array A in ascending order to simplify the comparison process. Iterate through each element A[i] in the sorted array A. For each element A[i], update the maximum sum at index i in the dp array based on the following condition: If A[i] is the first occurrence in the sorted array or A[i-1] and A[i+1] are not present in the sorted array, then set dp[i] to dp[i-1] + A[i]. Otherwise, set dp[i] to the maximum value between dp[i-1] (excluding A[i]) and dp[i-2] + A[i] (including A[i]). After iterating through all elements in the sorted array, the maximum sum will be the maximum value in the dp array.

[@shreyashtech8556]

bro doing gods work

[@harisrashid0773]

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;

[@the_shridhar]

just add this in previous question: ``` if (i < 0) { return !target; } ```

[@SatyamKumar-bw4vi]

Hare Krishna..!! understood.

[@mrinmoyhalder7293]

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 ?

[@atulkumarsingh6507]

TARGET1: MATRIX- COMPLETED, TARGET2-ARRAY/STRING: COMPLETED, eagerly waiting for TARGET3:?

[@shiershdwivedi742]

"Hi Striver, This is Thriver here hope you are doing extremely well "😅

[@shauryatomer1058]

thanks for another great video

[@namanchandak1532]

at 14:03 why you removed for() dp[i][0]=1;

[@madmaxgaming5864]

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

[@stain0p101]

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

[@ramyasree4986]

completed subsequences set problems great learning

[@riturajseal6945]

Is the condition num[0]

[@neerajgarg9096]

The condition is necessary because we are declaring dp vector of size n×(tar+1) so lets if our arr[0]th element is greater then tar (arr[0]>tar) then if we do dp[0][num[0]] it will give us TLE as we fixed its size to (tar+1) but we are storing value at a index which does not exist so thats why that condition is important. Hope u understand ;)

[@mmbmmbmmb]

@@neerajgarg9096 THANK YOU SO MUCH! my brain was hurting thinking why but now i see

[@sujalgupta6100]

Understood. Best on whole yt.

[@mayankmalhotra21]

if someone is going from 0 to n if(i==arr.length) { if(tar==0) return 1; else return 0; } this will handle the case for zero (1, 0,0 ) will give -> 4 (if tar = 1)

[@hrushikesh-1914]

Understood. Thankyou sir.

[@RaghavSharma-nt3hr]

CHECK:- 9:30 - 12:14

[@Jainam-17-18]

space optimization code for previous question(DP-17): class Solution{ public: int perfectSum(int arr[], int n, int sum) { vector prev(sum+1,0); if(arr[0] == 0){ prev[0] = 2; }else{ prev[0] = 1; } if(arr[0] != 0 && arr[0]

[@arinmodi8884]

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.

[@rajatsingh4286]

Striver can you explain the approach when there are negative elements also in the array

[@sarthakkrishna3492]

use a map

[@rajatsingh4286]

@@sarthakkrishna3492 how ? can you explain with example

[@shubhrajyotipoddar1684]

if(k==0) dp[ind][k]= 1; if(ind==0) return (arr[ind]== k)?1:0; I feel like not returning and instead adding 1 to the dp should do the trick And in case of tabulation: dp[0][0] = 1; this should do instead of the for loop of ind= 0 to n-1{dp[i][0] =1}

[@DevashishJose]

understood. Thank you so much