2 NGR | Nearest Greater to right | Next Largest Element

[@kshmaverma5688]

Bhaiya aap jaise 1-2 utubers aur aa jaye..fir coding blocks aur ninjas to katora lekrr ghumenge.... U r a big support for the students like me who cannot afford those extremely expensive courses. Request h ki aap continue krtee rhoo... Plzz.... 🤣

[@thexhist3609]

We also have pepcoding which is good and everything is free.

[@tanned_cosines_]

lmfaoooooo

[@mishorpatra1670]

"Once you learn to approach problems, you'll stop running after the solution." - Sir Aditya Verma (bas ab ek backracking aur graph playlist banado aur kuch nahi chahiye)

[@amarnathprasad1]

Until he makes one, you can watch the first six lectures from here and solve questions from the FreeCodeCamp video(7th video in the playlist). I got better at graphs after practicing questions from the FCC video. kzbin.info/aero/PLmqutD5ta9PvsMYCyc0tVpmIl8-iIgxey

[@nitwaalebhaiya]

@@amarnathprasad1 Thanks

[@Adarsh-mn7pl]

bul ja, ab kuch nhi ane wala

[@nidhishprasad2506]

@@Adarsh-mn7pl why?

[@mantenapragnyabala1711]

​Hi do you know from where to do queue​@@amarnathprasad1

[@govindsharma6951]

can not explain how much happy i am after finding your channel. thank u so much for creating these videos

[@upasanaarora2641]

for almost 1 month searching for new fresh content to revise my concepts again,loved your channel.Its has all,what it needs to be done even in short time,To the point and great content.Great Job :)

[@Area-fd8ht]

❤

[@karanbhati7552]

More Concise code :: vector nextLargerElement(vector a, int n){ stack s; vector ans; for(int i=n-1;i>=0;i--){ while(!s.empty() && a[i]>=s.top()){ s.pop(); } if(s.empty()) ans.push_back(-1); else ans.push_back(s.top()); s.push(a[i]); } reverse(ans.begin(),ans.end()); return ans; }

[@pragyanshandilya4015]

legend

[@charugoswami5344]

Kindly help me to find error in this code.... vector nextGreaterElements(vector& nums) { stacks; vectorv; for(int i=nums.size()-1;i>=0;i--){ while(!s.empty() && s.top()

[@_AmbujJaiswal]

bhati bhai thankyou soo much

[@samikshayadav3475]

good good

[@manjeshpatil]

So basically at n-1 checking your code executes the while loop first even though the stack is empty that's y write the if condition 1st then else then the while loop. Correct me if iam wrong @@charugoswami5344

[@hereyoursdad]

sir more then that the little talk from your exeperience like what they want from us is really a diamond .. thank you so much for making these gems will help so many peoples

[@comps_52_mankritsingh84]

*Interesting thing to note* For those who are getting segmentation error even though your code is the same check whether you have done the seek or top operation on the stack in your code? for eg in the code while(s.top()0) In this code when the stack is empty the first condition will be checked first and give you segmentation error.Instead, keep the statement after the empty check and the error will go. Hope it helps ;)

[@abhijeeyt]

bhai 🙏🙏

[@Lucifer-jl9vh]

Thank you bhai

[@pranjalmukherjee]

Thank you bhai

[@tanya2586]

heyy.. i'm still getting error.. can you check why? vector v; stack s; for(int i=n-1;i>=0;i--){ if(s.empty()){ v.push_back(-1); } else if(s.top()>arr[i]){ v.push_back(s.top()); } else if(arr[i]>=s.top()){ while( s.size()>0 && arr[i]

[@comps_52_mankritsingh84]

@@tanya2586 try by removing that ! Sign in the last check you have put if(!s.empty()) .

[@TechnicalDrMusic]

Literally your DSA teaching approache is 🔥 than all others KZbins, big thank you ❤

[@pratikkedar4476]

Small Optimization : We can avoid using reverse method, just need to push elements from array in reverse order ! Make sure you define fixed vector length else will give compilation errors!!

[@pokeindia5361]

Why is this so we don't need size of vectors ryt?? I'm facing this problem

[@Demodulator7]

​@@pokeindia5361 we need size of answer vector and it will be equal to given vector

[@pokeindia5361]

@@Demodulator7 I'm asking why we need to fixed vector length vector is automatically adjusted on its own ryt??

[@Demodulator7]

@@pokeindia5361 but in optimization, you are filling vector from backward (so that you not need to reverse it) so you have to give the length that from where you are filling. And you can't use push_back because it adds value in last, but each time you want to add values at front because first you calculated value for I = n-1 (last element) and then for i= n-2 (second last element) and so on till the first element. So you have to do V[i]= .... So you have to give vector a size because vector can increase it's size in one direction only. If you directly do V[n-1] =.... Then it will give runtime error because vector is trying to access n-1 position but currently vector size is 0. One more point:- doing push_back doesn't require size declaration. But when ever you do V[I] = .... Then size of vector should be greater then I. If not want to declare size then you can use 'list' and do push_front Hope it helps...

[@pokeindia5361]

@@Demodulator7 thank u so much sir!!

[@shubhamjain54519]

14:18 for context. 15:19 2 can be safely popped because we have already found an element on its left which is greater than 2, ie 3. As any element (greater than 2 and smaller than 3) on left of 3 which needs next greater element on its right will always print 3, so 2 is of no use at this point.

[@williamgilfoyle3261]

Best explanation so far, but i would prefer "while loop" over "for loop" , for current problem and in the third if case we dont move to the next element in the array until we pop all the elements from array Like here Python : Language arr = [1,3,0,0,1,2,4] stack = [ ] res = [ ] i = len( arr ) - 1 while i>=0: if not stack: res.append(-1) elif stack[ -1 ]>arr[ i ]: res.append(stack[-1]) elif stack[ -1 ]

[@sushmitamandal1677]

my programme is not running can you please help to run the code?please

[@abhijeetgupta3461]

reverse(v.begin(),v.end()) after for loop ...

[@akashdawari4665]

continue bro, this good work ;)

[@TheAdityaVerma]

Thanks brother, Do subscribe and share, that keeps me motivated to do more !!

[@samirsaurabh5785]

@@TheAdityaVerma please make such great videos on all data structures and algortihms,you are the true mentor ❤️.

[@nandasaikishore478]

@@TheAdityaVerma please continue the course sir please 🙏🙏🙏

[@souravsaha3446]

@@TheAdityaVerma Brother please upload videos on some other topics please You are not only a teacher to me but also a coding guru who can teach with his experience please sir upload on backtracking and graphs

[@adarshanand176]

this is ur first impression on me... and its really looks like my dream teaching style. 🤩✌

[@adityachauhan9951]

best explanation available on KZbin thank you so much

[@RohithRock]

Crystal clear explanation bro. Really it helped me a lot in understanding the concepts perfectly. I was really afraid of datastructures and algorithms. I was not able to solve the questions thinking that ds is a really tuff subject. But now through ur way of teaching and explaining i am able to learn and understand easily. Thank you bro. Do more videos regarding the ds topics. From - rohith Nit bhopal cse

[@piyushaggarwal78]

Initial if and elseif condition are not required as it is redundant., Very good explanation of things, great content. Waiting eagerly for Trees and Graphs and Number theory.. I always prefer watching your videos on 2x speed , it sounds perfect that way

[@SurajKumar-bw9oi]

Right bro, Code written After while loop started is sufficient 👍

[@pranjalsrivastava1191]

true, no use for the first two conditions. Already covered by the portion after while!

[@Codermonk1997]

guys like u should be kicked on their arse, dumbo if u are so intelligent then go and try to get jon in mnc instead of watching his vidio. he is teaching how to convert ur idea into code, one if elsse will not going to increase time complexity to exponential. idiots like you exist every where.

[@arunavgoel5620]

@@Codermonk1997 Damm, the guy just pointed out another way to write the code, a way which reduces the number of lines and also makes the code run more efficiently...no need to get triggered by that

[@harjos78]

loved watching your videos.. Awesome .. crystal clear explaination. Was refreshing my DS concepts to teach my son who is in 7th grade. Now after he has watched your videos he says your explaination is way better than i can explain him.. Kudos Aditya. Keep up the good work. God bless you

[@ankitsharda1131]

With due respect to you I would suggest you to let your son enjoy his childhood. All this he can learn it when it will be the right time.

[@vanshshah7781]

7th grade is too late man

[@vathsan3906]

Bhai.. Sorry I would like to call you SIR.. Hats off.. I was scratching my head for the past 2 days trying to figure out why they are using stack, how to identify what to use and when to use and also to get the approach to solve... I was unable to understand it from GFG and thank god I found you!! Finally understood why we are using stack and what is it's purpose in this question...

[@rushikesh_chaudhari]

Just started watching your playlist and this is really mind blowing. Loved your way to find solution for problems

[@AyushGupta-ux4gq]

what are you doing now after 3 years of this comment ??

[@abhishekojha4540]

brother I was stuck in this question for so long time and no one has explained this question to me in so much detail. Thanks a lot.

[@SauravKumar-si7kt]

This is sth which i need in other youtube solutions. I never got satisfied with their solution because they are just discussing the solution. And those solution never helped me. Thank you for introducing these all concept. It would have taken a lot of time to think on our own.

[@theunusual4566]

Matlab, Kamaal Ka explanation bhai. Automatically understood next 3 questions. Bahut Khood Samjhaya Aapne. Thanks.

[@ShahNawaz-cx3pi]

*********Time complexity Doubt:-******** I got what you want to ask:- your main concern is about the inside while() loop, you are thinking that it will become O(n^2) , but this will never happen the worst case complexity can be O(2n) , in simple terms O(n) . Case 1 :- arr [] :- 6 5 4 3 2 1 In this case the while() loop will run only one time in each iteration of for() loop so, while() loop is contributing only O(1) complexity. In this case it is simple O(n). Case 2:- are[]:- 6 1 2 3 4 5 In this case every element in the array will considered 2 times. First time when we are traversing the array in backward direction through for() loop and another time when we traverse element in forward direction through while() loop. It is just like traversing array two times O(n) +O(n) =O(2*n). So, The overall complexity become O(2n) ~ O(n).

[@tiyashadas5247]

Thank you!

[@sanyam9712]

Thankyou

[@kingrajput6202]

thanks man

[@PankajSingh-vi3fy]

For me it's kinda tricky but got your point thanks

[@rajatagarwal6091]

Hi @Shah Nawaz. Thanks for your explanation. However I still have a doubt that worst case complexity would be O(n^2). let consider some array like 20,14,15,16,17,18,19, 10,7,8,9 , 6,1,2,3,4,5. What are your thoughts on this array?

[@bestsaurabh]

Instead of taking vector we can take array start filling it from end, this way we can save reverse function call.

[@aayush5474]

you can also do like vector v(n);

[@malkeetsapien4852]

Brilliantly taught Sensei !

[@nirbhay411]

Bhai kya hi samjhaya he apne . maja agaya

[@utsavseth6573]

Not all heroes wear capes. Some are named aditya verma. Superb.

[@amankumar-im8kk]

Awesome content. You will be hitting 1M subscribers soon. Far better than paid contents of coding blocks and coding ninjas.

[@niveditaprity]

Impeccable and great explanation. Please upload more videos on tree and graph also.

[@NotNotNithin]

Java implementation: private static List nearestGreatestToRight(int[] arr) { Stack st = new Stack(); List reverseList = new ArrayList(); List resList = new ArrayList(); int size = arr.length; for (int i = size - 1; i >= 0; i--) { if (st.empty()) { reverseList.add(-1); } else if (st.size() > 0 && st.peek() > arr[i]) { reverseList.add(st.peek()); } else if (st.size() > 0 && st.peek()

[@ayush7518]

thanks a lot bhai

[@salonipaliwal8946]

thank you

[@RohitKumar-uc4fw]

Use this code to directly store from right to left. public static int[] findNextGreater(int[] ar) { int[] result = new int[ar.length]; Stack stack = new Stack(); for(int i = 0; i < ar.length; i++) { while(!stack.isEmpty() && ar[i] > ar[stack.peek()]) { int index = stack.pop(); result[index] = ar[i]; } stack.push(i); } while(!stack.isEmpty()) { int index = stack.pop(); result[index] = -1; } return result; }

[@shivagupta138]

@@RohitKumar-uc4fw use this and you wont need to reverse it even " public static int[] nearestright(int []arr) { Stack stack=new Stack(); int result[]=new int[arr.length]; for(int i=arr.length-1; i>=0; i--) { if(stack.size()==0) { result[i]=-1; } if(stack.size()>0 && stack.peek()>arr[i]) { result[i]=stack.peek(); } if(stack.size()>0 && stack.peek()0 && stack.peek()

[@alexamish2020]

Ye channel bhot zyada grow krege. Gauranteed.

[@lavanyabalija699]

Before listening to your lectures , am unable to write the code at 1 stretch, it used to take lot of time to debug my code. Am glad i found ur channel broo...Now am able to write code with ease. Thanxxxx for ur methodology of making coding soo easyy...Plss make more n more videos. If possible make videos on all topics of CP.

[@animeshsingh2880]

Best videos in youtube for learning DSA

[@hackzyoboy2459]

165th like, 0 dislikes. That's the power of generous teaching. Great job bhaii

[@jhanvisaraswat6976]

Those three conditions putting things in perspective so much

[@akarshjaiswal1622]

Great explanation... Please start uploading videos again... Looking forward for mathematics, backtracking,graph videos !

[@priyaagrawal2968]

One of the very best explanation! Thanks Aditya !

[@aheribanerji7056]

Pls, sir make more videos... it's very helpful for students like us who can't afford the high prized course .

[@mystryb345]

Completed heap,dp,recursion now i m.feeling confident thanks bro ...

[@mohdhasnain3812]

you have studied from his channel or some where else please confirm that all the stutff you have studied in this channel is on point , effective nd easy to unxderstand

[@thinkingmad1685]

@@mohdhasnain3812 have you found some other channel, pls suggest btw i m feeling comfortable to understand from this channel only 🥲

[@r4ncreation686]

best explaination of stack till date

[@___vandanagupta___]

Your every video is a blessing!

[@RajveerSingh-yf1os]

THIS IS MY FIRST YT COMMMENT IN MY LIFE JUST LOVES THIS VIDEO

[@AnkitKumar-rh8il]

maza aa gya bhaiya , aapka explain karne ka tareeka OP hai

[@rishabsharma5307]

Little more simpler version of code // nearest greater to right vector ngr(vector arr, int n) { stack st; vector vect; for(int i = n-1; i >= 0; i--) { while(!st.empty() && arr[i] >= st.top()) st.pop(); st.empty() ? vect.push_back(-1) : vect.push_back(st.top()); st.push(arr[i]); } reverse(vect.begin(), vect.end()); return vect; }

[@rishabsharma5307]

Slightly better approach with no vector reversal ``` vector ngl(vector &nums) { stack st; int n = nums.size(); vector result(n, -1); for (int i = n - 1; i >= 0; i--) { while (!st.empty() && nums[i] >= st.top()) st.pop(); if (!st.empty()) result[i] = st.top(); st.push(nums[i]); } return result; } ```

[@avinashsaklani]

great explanation

[@abhishekanantharam645]

O bhai legend aadmi hai ye toh. Thanks✌️

[@_solopreneur321]

bhaiya u rocked its great to be learning from ur videos keep adding value. Very helpful videos.

[@vinayakrastogi890]

Bro I am waiting for your backtracking playlist... I know you have a busy schedule but please make videos on that

[@riaria5982]

Great Explanation! To improve runtime performance don't do reverse, it's better if you insert element : for(int i=input.length-1; i>=0; i--) { while(!stack.empty() && (stack.peek()

[@nishant5249]

your code is not absolutely correct

[@saumitrashukla591]

@@nishant5249 why? it seems correct

[@AdamRubiks]

oh damn

[@RohitKumar-uc4fw]

Use this code to directly store from right to left. public static int[] findNextGreater(int[] ar) { int[] result = new int[ar.length]; Stack stack = new Stack(); for(int i = 0; i < ar.length; i++) { while(!stack.isEmpty() && ar[i] > ar[stack.peek()]) { int index = stack.pop(); result[index] = ar[i]; } stack.push(i); } while(!stack.isEmpty()) { int index = stack.pop(); result[index] = -1; } return result; }

[@Newgen123Blogspotawaytosuccess]

These are the best video series that is available till Date. Also, make your account premium so that you can continue your work in a more focussed manner. Great work. Ready to pay thousands for such beautiful content.

[@akashverma5756]

Best Explanation of this problem online

[@ROHIT-gv7xk]

great work bro god bless you

[@shrutirani6828]

Please make more videos for placement prep... Ur videos are helping a lot.. Great mannn...

[@mohitsingh6717]

best channel on youtube 🙂

[@souravrajvi8193]

THIS IS FUNCKIN LEGENDARY

[@Akshat-tz2de]

Just a small doubt if we have used only 2 loops and used break in 2nd loop when condition is met then also we could have attained same complexity. As you are also using while loop inside for loop which is running for same time that the another for loop will run and for same amount of iterations. Moreover we are using stack here hence increase in space complexity as well.

[@rahulagarwal8059]

Wow.... Really happy that i found ur channel... The videos are very helpful👍

[@annuagarwal8438]

I am lucky to find your videos. Thank you for making such helpful videos. Your experience will be helpful to lot of people. Please make videos in graphs and backtracking also. I know you wont have much time. But ek mahine mai do video hi daal dena..We will wait. Thanks again😇

[@keshavdutttripathi]

Great work bhai. Really helpful

[@satyamshukla1418]

this is a gem of a content.thanks a lot for this bhaiya apne hisaab se time nikal krke agar aap backtracking graph par agar videos bana denge toh humari aur help ho jayegi.

[@shashijaiswal5014]

I will complete this playlist by tomorrow. 4 hr approx for this and 2hr per day. Will comment tommorow on last video of this playlist.

[@ajeetworking]

Thank you for the explaination. We can also approach this without reversing the array as vector nextLargerElement(long long arr[], int n) { stack stk; stk.push(0); vector ans(n, -1); for (int i = 1; i < n; i++) { while (!stk.empty() && arr[stk.top()] < arr[i]) { ans[stk.top()] = arr[i]; stk.pop(); } stk.push(i); } return ans; } in this, we push the element in the stack only if it is smaller than the previous element and if next is greater then we start popping until no further element is greater than the current element.

[@sayantaniguha8519]

non linear time complexity

[@375_rishavmishra9]

Bro ur great one of best teaching style.♥️♥️hope that you will continue with graph and trees also.

[@NitinKumar-sx6jl]

Please continue the series bhaiya great work

[@anveshkhambadkar6821]

Awesome content, Awesome Explanation Man.

[@Tanm0609]

very good explanation brother,keep it up and keep uploading

[@rohitjha971]

Please tree and graph bhi padhao

[@TheAdityaVerma]

Will try !!

[@nikhilnagrale]

@@TheAdityaVerma backtracking!!!!!!!!!1

[@manjeetkagada7534]

You are on next level man

[@shubhamdevrani7251]

Hey bro just loved ur solution for the problem...

[@garimadubey8970]

Thank you sir its very helpful vedio for us..nice explanation.

[@NidhiSharma-sw4oz]

How O(n) complexity ????? There is a while loop inside for loop. So how it is O(n) complexity ?

[@sandip_kanzariya8476]

Head's of brother 👌🙏

[@muhammedrameez652]

Great Work bro! It really helped me

[@jatinkumar4410]

amazing explanation....

[@tusharjain5658]

really this explanation is amazing

[@vrandakansal5940]

Bro!!...Great work...😄😄👌

[@ayushgupta8538]

*Thank you so much for this awesome video.*

[@abhishekrai4002]

bhai gajab maja aagya

[@UpscWithPavanOfficial]

Nice, Mza Aaya bhai.

[@kapishsingh]

its the first chanal in which i done like share and subscribe in less than 1 ms

[@NikhilTripathy]

17:00 this is gem!

[@koushik7857]

fabulous explanation!!!!

[@RicheshGuptaRichu]

Mann karta hai pdhne ka..Aesi videos dekh kar

[@vinayverma3099]

Isn't this also O(n^2) solution because of while loop inside a for loop.??

[@adityakainthola2953]

no every element is traversed atmost 2 times in both the loops so the complexity will be O(n)

[@varshithkemmasaram2115]

@@adityakainthola2953 i did not understand bro🥲🥲

[@shashankdaima8727]

@@varshithkemmasaram2115 The reason this method works is because he pops every element b/w arr[i+1] to arr[n-1] which are less than arr[i] until he reached a position that is greater than arr[i] or stack becomes empty. For eg:arr:{9,7,6,5,3,2} (Strictly Decreasing sequence). Initial Stack(right should be considered as top){-1}. For last element of arr. the ans is -1 and stack becomes {-1,2}.For second last element, we will check for element greater than current element. So, currentElement=3. Stack{-1,2}. (2Pop(2); stack contains -1 so return -1. now if you move to next Element you need to push 3 into stack making{-1,3}. So, we never have to deal with 2 again. That why the complexity is O(n). as Complexity is ranging b/w n to 2n so O(n)

[@aryannegi9414]

@@shashankdaima8727 but arent there two loops in this solution too?1st for i=n-1 to 0 and inside it while(!stack.empty&&stack.top()

[@VENDETTA_9]

For circular array: #include using namespace std; vector nextGreaterElement(int N, vector& arr) { stack s; vector ans; for(int i=2*N-1; i>=0; i--) { if(s.empty()) ans.push_back(-1); if (!s.empty() && s.top() > arr[i%N]) ans.push_back(s.top()); else if (!s.empty() && s.top()

[@saitejdandge2315]

Great lecture I got an intuition that we are using sub problem solutions to form solution current index i (Optimal substructure) and we have repetitive sub problems which is causing O(n*n). This is a dynamic programming, correct me if I'm wrong.

[@supritchafle3706]

best explanation on stack quns...

[@rajanoobrigaming1450]

Very helpful video vaiya ...help me lot

[@SanjuKumar-ye8xz]

Brother, tumne alag se vector define kiya jisase eski space complexity increase ho rhi hai.

[@Sithkar03]

Coding Ninja : Next Greater Element ( C++ Solution ) #include vector nextGreaterElement(vector& arr, int n) { // Write your code here vector ans; stack st; for(int i=n-1;i>=0;i--){ while(!st.empty() && st.top()

[@redwanniloy2068]

U did not show how stack's elements are storing in stack..in code...but i understand everything.. Thanks brother❤️

[@prerna.please]

can you tell me , how the elements are storing?

[@daddyjee6137]

U deserve more 🎉 ❤️❤️

[@bhawnabharti4369]

best explanation

[@wrushu]

If we have the same question for cyclic array like.. instead of having -1 for arr[n-1] we need to look at the 0th element. What would you suggest? What I came across is to traverse the loop 2 times. for(i=2*n;i>=0;i--) and reference ith element as i%n

[@pranjalmishra2602]

So un the worst case how is the complexity O(n) ? If has inner while loop as well... Should be N square in worst case right?

[@pranjalsrivastava1191]

correct ... even i have a doubt in this....the inner while loop can go till n and thus worst case complexity would be n square.... the same case would be when we use 2 for loops ....please explain someone!

[@pranjalmishra2602]

@@pranjalsrivastava1191 😶 👍🏻

[@harshashenoy8555]

Guys, while loop is over the stack And stack holds only the greatest element. Worst case happens only for one iteration where in the smaller elements gets popped. And for the next iteration the previous iteration value will be on top . So using this method the time complexity would be O(n). Try an example to understand more.

[@aniketsinha2826]

greatly explained !!!

[@AdityaLSharma]

Any possibility I can get all these notes in PDF form?

[@rohitmaheshwari9357]

bhaiya vector ko reverse krne ki jarurat nahi h ek pointer last se chla ke answer fill kr dete h tb bhi kaam chl jayega

[@whatwhat3435]

helps a lot love frm Nepal bro