L4. Reverse a DLL | Multiple Approaches

[@purushottam108]

17:21 "And the interviewer will hire you,no worry" best part🤑🤑🥰🥰

[@Thunder97654]

😂

[@aanandshukla]

😅

[@PCCOERCoder]

😂😂

[@qihse1531]

This simple problem is insanely hard to me...thank you Striver!!

[@keshavchoudhary8857]

this type of great content cannot be found elsewhere , Excellent!!!

[@AlwaysOntopoftheworld]

Is kunal kushwaha better or this one?

[@Ayush37262]

​@@AlwaysOntopoftheworldI think Striver, Kunal is more into Controversies

[@FaizShaikh-dt8wp]

You are really doing a revolutionary thing 🔥, Belive me you will definately going to have a huge audience just keep doing striver! Complete the DSA series from A to Z 🔥 before 2024 ends.

[@AmanSharma-xy1qm]

All the video lectures and the articles helped me a lot to gain confidence in DSA and will be helping me in the interviews. Thank you Striver bhaiya for bringing such amazing content for free.

[@vaibhavvm4147]

Bro how do u approach DSA, do u first watch videos and then solve it on ur own or how?

[@AmanSharma-xy1qm]

initially, i learn first the therory part like methods that are used in this, and i usually watch Striver's playlist of that topic , and learn the patterns of the questions then i go to leetcode and gfg and solve same patterns problems. With the approaches i learned from Striver. That's all. Hope you got this.@@vaibhavvm4147

[@Rahul_Mongia]

@@vaibhavvm4147 video dekh le bhai, dimag khol......., phir related ques khud try kar

[@harshitgarg2820]

Loved your dedication sir, completed this ques and loving this series like all others. Top-tier content🔥🔥

[@MdallShahria]

you are the boss .. please make a playlist about competitive programming.

[@shubhamanand8557]

It really helps me to understand in a very easy manner. Thanks a lot bhaiya.

[@Sunny-hg5oe]

Great thought of using the stacks!!!!

[@aravindsagar9433]

instead of using next ad back which is a bit ambiguous you could've used next and previous ( prev ) which will be more understandable

[@itsorcus649]

🤓

[@BeWarrior-dw4br]

The takeuforward website is not loading .. I'm trying to access the A2Z sheet for the last three days but it is not working.

[@guneeshvats46]

Dude your lectures are amazing

[@PCCOERCoder]

Lecture successfully completed on 26/11/2024 🔥🔥

[@RRR-qv1so]

Thank you for this amazing playlist bhaiya ❤❤

[@shanipratapsingh916]

Thank you sir this amazing lecture 🥰

[@Vvvvviieiei123]

lecture 4 done,love u bhaiya

[@ankit-jangid]

nice explanation brother, 1 thing I'd like to advice is please use the same variable names while coding which you use while explaining, it would be helpful. Good work and your videos are great!!✌🏻😇 thanks

[@lifehustlers164]

17:20 , dream come true!!!

[@aashwinkumar2242]

i guess there is an even more optimised solution now since the link is showing time limit exceeded in this approach . i did everything the same

[@aaryangupta4505]

if(head==NULL||head->next==NULL) return head; Node* temp = NULL; Node* curr = head; while (curr != NULL) { curr->prev = curr->next; curr->next = temp; temp=curr; curr = curr->prev; } return temp;

[@aashwinkumar2242]

yup thats what i did , made it much simple to understand you have to return temp->prev

[@karthik-varma-1579]

i taught of doing same approach. of changing and head and swap its pointers

[@shivamsays2861]

Nice videos thankyou

[@pratyushtripathi1728]

Understood 😊

[@oyeesharme]

thanks bhaiya for your efforts

[@adebisisheriff159]

Thanks Striver!!!

[@AkashGautam-k2u]

Thank You Bhaiya.

[@tle964]

nice video

[@saitejanedunoori5451]

great work and thank you !

[@DeadPoolx1712]

UNDERSTOOD;

[@TON-108]

Understood!!! Thank You.......

[@chetashreejagtap7585]

thank you striver👍👍

[@NazeerBashaShaik]

Understood, thank you.

[@striverdaaadi]

striver is best

[@RajNamdev_19]

Understood ;)

[@nishant4595]

the best!!!!

[@gourabrajak8132]

👍😊

[@YourCodeVerse]

Understood✅🔥🔥

[@sumitworld918]

Undertood 😌

[@charuprabha9696]

[@Ergoswami]

Completed!

[@artcodetravel]

If we point the head to last node, the behaviour would be same as revered doubly Linked list right??? Why do we need to do these processing???

[@techwithsam4302]

In that case we can't traverse in the forward direction instead we have to go backward to traverse!!

[@khalasianiket816]

understood ❤

[@subee128]

Thanks

[@ThakurIsGeek]

nice

[@firebout7675]

understood

[@MohitKumar-o3l1u]

Understood !!

[@abhishekprasad010]

Understood!

[@Zomb-zj4ip]

understooooooooooooood

[@rizonkumarrahi]

Great

[@XS_01]

i just swapped the data values, using two pointers as head and tail.

[@harshuke5831]

Understood

[@r_uhi05]

understood!

[@YashGaneriwal-je6rh]

done

[@ronakvaghela2714]

[@UjjwalMishra-xf9un]

why this code is not working,, its all same just using one extra variable; Node* reverseDLL(Node* head) { if(head==nullptr || head->next==nullptr)return head; Node * current=head; Node *back=nullptr; Node *front=nullptr; Node *temp=nullptr; while(current){ back=current->prev; front=current->next; temp=back; back=front; front=temp; current=current->prev; } return back->prev; }

[@harshitjaiswal9439]

🔥🔥🔥

[@varun1017]

why this is correct and last=curr->prev; curr->prev=curr->next; curr->next=last; this is not working? last=curr->next; curr->next=curr->prev; curr->prev=last;

[@kartikeygupta5998]

can we exchange head with with tail and tail with head

[@vishious14]

My approach looks something like this: Publlic class Solution{ public static Node reverseDLL(Node head) { Node p=null,curr=null,f=head; while(f!=null){ p = curr; curr = f; f=f.next; curr.next = p; if(p!=null) p.prev = curr; } head = curr; return head; } }

[@TarunKumar-km5ps]

👌👌👌👌👌👌👌👌👌👌👌👌

[@AkashKumarTiwary-u4b]

god

[@shalinikumari6050]

understood 29 sep 2024

[@himanshukaushik9223]

Bhaiya 0(1) ma bhi solution hai kya iska ??

[@vedikamishra009]

traverse toh krna hee h, so no O(1)

[@AyushRawat-v8m]

class Solution { public: Node* reverseDLL(Node * head) { if(head == NULL || head->next == NULL) { return head; } Node* pichli = head; while(head->next != NULL) { pichli = head; head = head -> next; } Node*mover = head; mover->next = pichli; mover->prev = NULL; Node*back = mover; mover = mover->next; pichli = pichli->prev; while(pichli != NULL) { mover->prev = back; mover->next = pichli; back = mover; mover = mover->next; pichli = pichli->prev; } mover->next = NULL; mover->prev = back; return head; } }; this is i ve done but ur one is more intutive and easy also

[@Rohith__Dev]

28-06-2024

[@harshilpatel3205]

Understood 🫡

[@ganeshvhatkar9253]

Understood

[@chiragbansod8252]

understood

[@pradipkumarmukhi]

Understood

[@ashishpradhan6250]

understood

[@adityapandey23]

Understood

[@himanshidafouty347]

Understood

[@satyanarayanbarik1701]

Understood