Lecture 49: Merge 2 Sorted Linked Lists || Sort 0s, 1s and 2s in Linked List

[@CodeHelp]

Do Visit Relevel: relvl.co/4sx5

[@myncom397]

the code is not correct for test case [5] , [1,2,3,4].

[@amitprakhar4802]

bhaiya ye TLE kyu de rha merge sort wala

[@avishkarpatil5871]

Approach 3 void sortArray(Node* head){ Node* temp = head; while(temp != NULL && temp->next != NULL){ Node* forward = head; while(forward != NULL && forward->next != NULL){ if(forward->data > forward->next->data){ swap(forward->data, forward->next->data); } forward = forward->next; } temp = temp->next; } }

[@nxtlvl2904]

Bubble sort se bhi kr skt hai n bhaiya

[@fgh427]

Babbar bhaiyya... I love you... Koii apna bhi itna help nhi karta jitna aap kar rhe ho.... Bhagwan aapko meri ummar bhi de de... Legend and GOAT....

[@iamvishnuranjan]

Bhaiya 1 month me (1- 49 )lecture 😍 what a course..

[@yashsonune4391]

Glad I have completed 1/3 of the series, and still continuing.

[@parthdeshwal4419]

how long did it take you ?

[@Anmol-h8z]

at 41:15 while inserting a node between two nodes we can use : curr1->next=curr2; curr2=curr2->next; curr1=curr1->next; curr1->next=next2; and also why we are checking for next1->data >curr2->data > curr1->data we could just check with next1 only as we know we start with element with smallest value any value will be greater than or equal from starting node and then going further in list if curr2 is inserted and then new curr2 value will definitely greater than previous one

[@harshitkumar197]

I coded the 2nd one merge two sorted array by myself. i looked at ur approach discussion and did by myself. thanks for the giving me the confidence

[@therealartist9]

in which company you are working ??

[@tjstar390]

BOTH Questions have unique concepts and bhaiya explain it very well,Thanks bhaiya.

[@RajveerKaur-ze1ll]

to check for the case when both elements have one one element, instead of adding a new if condition at beginning, we can also move the "if (next1==NULL)" condition outside the while loop, then also answer is correct, hope this helps.

[@prashantbpkumar510]

Exactly

[@tanishsinghal1879]

yeah boyh

[@priyanshpatel2078]

20:00 We can Simply Handle it by using this steps p3->next=NULL; p2->next=twoHead->next; p1->next=oneHead->next;

[@tusharmukherjeeju-cse5773]

Can't believe that someone dislike this video ! 🙂 💔 If you buy a course which takes from you 20k that course never give you that kind of explanation . Love from Kolkata ❤️🙌

[@divyareddy7622]

why cant zero and two 's list be empty here? he only took the case when 1's list is empty

[@visheshkaran7083]

@@divyareddy7622 because even if two's or zero's list are empty they will be null and onetail->next will be null so it won't make any difference.

[@yashparalikar05]

Your teaching and your explanations are absolutely exceptional!!!

[@unboxtheuniverse5336]

Wahhh bhaiya 💥😃😃 DSA ki ab Video na dekhu to din productive nhi lagta 😃 Bhaiya plz *"5 Month Internship Strategy Series lao with DSA busted course as resource"* (for 2nd year students)🥺

[@varaprasad8261]

Node* sortTwoLists(Node* first, Node* second) { if(first == NULL)return second; if(second == NULL)return first; Node* temp = first->data > second->data ? second : first; if(temp==first) temp->next = sortTwoLists(first->next,second); else temp->next = sortTwoLists(first,second->next); return temp; }

[@durgashreenv8426]

35:00 Explanation was just amazing, thank you so much!

[@anmolkumar01]

at 55:44 , i think we don't need to add extra case when there is only one element in list one instead what we can do is - we can simply write that condition of if(next1== NULL and curr2 != NULL){ curr1 -> next = curr2; } it will take care of case at 55:44 as well as previous case when first list finish and second list still has some element and please feel free to to correct me if i am wrong ,

[@Josuke217]

My approach: SC: O(n) , TC: O(n) For 0,1,2. class Solution { public: ListNode* sortList(ListNode* head) { vector res; ListNode* temp = head; while(temp != nullptr) { res.push_back(temp->val); // Store value of current node temp = temp->next; // Move to next node } sort(res.begin(),res.end()); // Sort the vector // Now convert the vector into the sorted linked list int index=0; temp = head; while(temp != nullptr) { temp->val = res[index++]; temp = temp->next; } return head; } }

[@Roshanhoro123]

thanks bhaiya for bringing all such good questions in this course and hats off to your efforts...aap itne busy hoke bhi regular videos daal rhe ho isse motivation milta hai ki chahe kitne bhi busy rhe humlog cllg exams me tb bhi hum pdhai krre... Thanks bhaiya...Keep smiling bhaiya bht suit krta hai aaapar..

[@devanshusahoo]

Thank you so much for your lovely videos, you really make understanding the concepts a cakewalk! Glad to be your student.

[@divyareddy7622]

why cant zero and two 's list be empty here? he only took the case when 1's list is empty

[@suranjandhara7535]

Wow 😳😳😳😳😳 nya video aa gya..... Maza aa gya .... Abhi hi dekh lunga....🥰🥰🥰🥰🥰🥰🥰❤️❤️❤️❤️❤️❤️❤️❤️❤️🌷🌷🌷🌷🌷🌷🌷🌷🌷 Thanku bhaiya .... ❤️❤️❤️❤️

[@mittalji919]

i used below merging algo and it is also working without any if condition nodeZTail->next = nodeO->next; delete(nodeO); nodeOTail->next = nodeT->next; delete(nodeT); head = nodeZ->next; delete(nodeZ); return head;

[@115GAJBHIYESHREYA]

Thank you for providing an amazing series.

[@StayAware9]

Itna cylister clear hogaya question and approach Oye hoye hoye Jug Jug Jio bhai!!!

[@anmolkumar01]

my approach for 1st question was that i just sorted it in two go without using extra space and in O(n) firstly : sort all the ones , if one present then remove it and link it to the head of list and update the head secondly : sort all the zeros , if 0 is present then remove it and link it to the head of list and update the head my code : void sort(Node* &head , int val){ Node* temp = head; while(temp->next != NULL){ if(temp->next->data == val){ Node* x = temp->next; temp->next = x->next; x->next = head ; head = x; } else{ temp = temp->next; } } } Node* sortList(Node *head){ if(head == NULL or head->next == NULL) return head; sort(head , 1); sort(head , 0); return head; } if there is any suggestion or correction then please feel free to correct me and above code is also accepted by the codestudio

[@vmstudy9965]

Mjaa aa gya bhiya aaj to🎉 Finally I solved this problem myself. After watching this video

[@chinmaykhemariya3816]

for 2nd question Node* sortTwoLists(Node* first, Node* second) { Node* a=first; Node* b=second; Node*k=new Node(-1);Node*tail=k; while(a!=NULL&&b!=NULL){ if(a->datadata){ tail->next=a;tail=a; a=a->next; } else{ tail->next=b; tail=b; b=b->next; } } if(a!=NULL){ tail->next=a; } else if(b!=NULL){ tail->next=b; }k=k->next; return k; }

[@ujjwalverma_]

Bhaiya boht mja aa rha is course me. Paid me bhi aisa mja ni aata. Discord pe discussion and all. Great 🔥🔥🔥

[@kinmeensura4283]

very good explanation . now i am understanding everything . you are a very good teacher and motivator also, please keep it up 😅😅

[@laksheybanga8106]

Bhaiya I m your biggest fan! bas jldi se phase 2 khtm krdo bhaiya!

[@DipsOfficial802]

smaj gya bhaiya clear hai concept 🔥🔥🔥🔥

[@RohitRana-tz2lr]

Thank you so much bhaiya for training our mind in solving a different variety of questions... hats off to you bhaiya...keep it up

[@anuragpatel7787]

hnji this is anurag patel Lec 49 completed successfully ✅

[@avibirla9863]

Maza aa gaya bhaiya kya question tha merge wala 👍👍Or last ka edge case 👍♥️

[@shubhammathur792]

amazing lecture bhaiya i think one of the best so far cleared every concept ;

[@hyphengamerandtech3064]

thank you bhaiya , placement ke baad aapko party meri taraf se .

[@udaypratapsingh8923]

bhaiyaa ye hashing or heap bhi cover krva dena ... jab bhi ye sunta hui dar sa lagta hai 😅😅😅

[@sounaksaha1455]

Present Bhaiya, consistency OP, videos OP, Reach OP.

[@trishalmandrik1295]

Hey! what does this OP mean?

[@060_jiteshtripathi6]

bhaiya maza hi aagya, kya mast samjhate ho

[@RajDas-uq6sh]

Sir in question 2 : If first -> next == NULL then answer will note come.. Please add another condition- if(first -> next == NULL) { first -> next = second; return first; }

[@tanbirsarkar2640]

sabse maza to linked list mai arha hai,,mast hai bahut,,basss pyar hojata hai

[@kanikagupta2806]

concept++; understanding +++++++++; thankyou so much babbar bhaiya maza hi aagya..

[@divyareddy7622]

why cant zero and two 's list be empty here? he only took the case when 1's list is empty

[@kanikagupta2806]

@@divyareddy7622 beacuse 1's list is connecting 0's and 2's list. If 1's list is empty then approach discussed without if condition won't work. Hope it helps.

[@Pawankumar-i8c]

thank you so much bhiaya for these precious lecture

[@theOmKumar]

Explanation + Minor adjustment to improve q1 while addressing memory leak! // // merge 3 sublist (assuming one's list is not empty) zeroTail -> next = oneHead -> next; oneTail -> next = twoHead -> next; twoTail -> next = NULL; return zeroHead -> next; //merge 3 sublist (Note: if 1st list is empty, then we can't connect zero list with two list using previous code) if (oneHead -> next != NULL) //one's list is not empty, so we'll connect 0 list with one { zeroTail -> next = oneHead -> next; oneTail -> next = twoHead -> next; } else zeroTail -> next = twoHead -> next; //one's list is empty, so we'll connect 0 list with 2s directly; twoTail -> next = NULL; //this will end the LL. return zeroHead -> next; // //to prevent memory leak, due to allocation of dummy node so we need to delete it. // Node* ans = zeroHead -> next; // delete(zeroHead); // delete(oneHead); // delete(twoHead); // return ans; }

[@lakshsinghania]

exactly! kudos

[@arpitghura]

this is what I am supposing as the oneHead does not have any one node, so if we put it outside it will create an additional LL which is of no use. So, we can put the line no. 60 inside the if condition to prevent the creation of additional LL.

[@tanishsinghal1879]

shi hai bhai agr tu comment m nhi likhta ye sb to mai ye sochta syad m hi glt hun

[@adityachauhan5913]

@codeHelp bhaiya ek dummy node le kr compare aur direct connect karte chale jaaye last me dummy ka nxt return kar dnge

[@KaushikSharma-c3q]

Awesome explanation.

[@kunalmodiart]

more easiest way to code Q2 would be to use 2 pointers pointer pointing smallest element gets added at the end of ansLL //============================================= Node* solve(Node* first, Node* second){ Node* ans=new Node(-1); Node* anstail=ans; Node* fhead=first; Node* shead=second; while(fhead!=NULL&&shead!=NULL){ if(fhead->datadata){ anstail->next=fhead; anstail=anstail->next; fhead=fhead->next; }else{ anstail->next=shead; anstail=anstail->next; shead=shead->next; } } while(shead!=NULL){ anstail->next=shead; anstail=anstail->next; shead=shead->next; } while(fhead!=NULL){ anstail->next=fhead; anstail=anstail->next; fhead=fhead->next; } anstail->next=NULL; return ans->next; } Node* sortTwoLists(Node* first, Node* second) { if(first==NULL){ return second; } if(second==NULL){ return first; } return solve(first,second); } //=============================================

[@adityaraj-zm7zk]

bhai it is very good question and you really very good explained , fanstically you are very good teacher

[@sachitdhamija1884]

bhaiya acha tha merge 2 sort L.L samjaha but sort 0,1,2 thoda problem hua but 2 baar aur dekho ga toh samjho ga lage raho

[@vaibhavraj2847]

bhaiya aapne jo array me merge and sort wala concept se padhaya tha ussejyada easy se ho gaya!! #include #include using namespace std; class node { public: int data; node *next; node(int data) { this->data = data; this->next = NULL; } ~node() { int value = this->data; if (this->next != NULL) { next = NULL; } delete next; cout next = curr2; tail = curr2; curr2 = curr2->next; } } while (curr1 != NULL) { tail->next = curr1; tail = curr1; curr1 = curr1->next; } while (curr2 != NULL) { tail->next = curr2; tail = curr2; curr2 = curr2->next; } head = head->next; return head; } int main() { node *head1 = NULL, *head2 = NULL; node *tail1 = NULL, *tail2 = NULL; insert(1, head1, tail1); insert(3, head1, tail1); insert(5, head1, tail1); insert(8, head1, tail1); insert(2, head2, tail2); insert(4, head2, tail2); insert(5, head2, tail2); insert(8, head2, tail2); insert(10, head2, tail2); print(head1); print(head2); node *ans = merge_two(head1, head2); print(ans); }

[@verma_jay]

Bhaiya bhot acha pdhate ho aap

[@aishavkhandelwal8271]

Mst ques tha dimag khul diya pura !! thx

[@alokranjan3185]

totatly different explanation bhaiya, thanks

[@himanshugaur6790]

Another Approach to merge two sorted linked list Node* sortTwoLists(Node* first, Node* second) { Node* dummynode = new Node(-1); Node* temp = dummynode; while(first!=NULL && second!=NULL){ if(first->datadata){ temp->next = first; first = first->next; } else{ temp->next = second; second = second->next; } temp = temp->next; } if(first!=NULL){ temp->next = first; first = first->next; } else if(second!=NULL){ temp->next = second; second = second->next; } return dummynode->next; }

[@presidentg6048]

Bhaiya -> using recursion ListNode* mergeTwoLists(ListNode* & list1, ListNode* & list2) { // base case if( list1 == nullptr ) return list2 ; else if ( list2 == nullptr ) return list1 ; // return the smallest node if( list1->val < list2->val ) { list1->next = mergeTwoLists( list1->next , list2 ) ; return list1 ; } else { list2->next = mergeTwoLists( list2->next , list1 ) ; return list2 ; } }

[@avishkarpatil5871]

Merge 2 Sorted Linked Lists - Approach 2 -> 99 % Faster void insertNode(Node* curr, Node* &tail){ tail->next = curr; tail = curr; } Node* mergeArray(Node* first, Node *second){ Node* addHead = new Node(-1); Node* addTail = addHead; Node* ptr1 = first; Node* ctr2 = second; while(ptr1 != NULL && ctr2 != NULL){ if(ptr1->data < ctr2->data){ insertNode(ptr1,addTail); ptr1 = ptr1->next; } else{ insertNode(ctr2,addTail); ctr2 = ctr2->next; } } while(ptr1 != NULL){ insertNode(ptr1,addTail); ptr1 = ptr1->next; } while(ctr2 != NULL){ insertNode(ctr2,addTail); ctr2 = ctr2->next; } Node* head = addHead->next; return addHead->next; }

[@rachit_joshi]

Thank You So Much BHRATA SHREE !!!!!

[@-akashkumarsingh]

Mere dekhne ki consistency se jyadaa toh video aane ki hai..

[@itz_arman_official__]

Thanks for your amazing content and your consistency is awesome bhaiya.

[@anuragpandey8165]

37:10 Just call sortTwoList(second, first) thats it.

[@justarandomguy6106]

How's that possible

[@AWAISKHALID-ze5cw]

the second approach could as to create a new head and make the new links after checking each data for the current pointer. it would create a new organised link with the previous one

[@Hit-xe5uh]

Can't we simply apply selection sort in the first question?

[@lalitbisht8381]

merge wala question krke maza aa gya!!!!!!!!!!

[@RitikSingh-oh3sj]

merge walai second question mai maga aa gaya bhaiya

[@B-NikhilRichhariya]

aap mehnat krte ho na wo dekh ke aur mehnat krne ka mn krta hai

[@arvindjindal1913]

ham second question mai ek nai list ans banakr aur usme koi value like -1 dekr phir jo first aur second mai jo chota hai usko ans list se link nhi karva sakte aur return karvate waqt ans ke next se return bhi toh karva skte h

[@abhishekhsinha1551]

glad for your kind help moye moye !!! love from IITG

[@prashantbirajdar9271]

ha bhaiyaa smja aya ache se ekdum bdiya❣😾❤

[@samsmith3961]

concept finally samajh aa gya

[@garimasharma276]

NEVER have I ever got such a crush on some one!! your way of teaching is quite captivating.

[@princekatare5399]

Bhaiya at 21:20....... instead of writing the if conditions like this..... if(oneHead->next != NULL) { zeroTail -> next = oneHead->next; } else { zeroTail -> next = twoHead->next; } oneTail->next = twoHead->next; We can write the code........ in the following order oneTail->next = twoHead->next; zeroTail -> next = oneHead->next; (we first link the twoHead to oneTail and then oneHead to zeroTail......) anyone please correct me if i am wrong...... And also thank you for the playlist bhaiya the DSA series is awsome and really helpful....

[@muneebjaved3940]

Thank you so much bhayia Concept ++++++; Explanation ++++++++;

[@anuptewary3016]

Present bhaiya with lot's of Josh

[@riyadhossain1706]

Became a fan of your learning style...........

[@StayAware9]

add a thanks button we want to pay you for your good work! ❤❤❤❤❤❤❤

[@RohitSingh-hc8yi]

2nd Problem was awesome!

[@RohitSingh-hc8yi]

Thankyou for such a quality content

[@rishikbisen4268]

it also one approach to solve problem 💥💥💥💥💥💥 Node* sortedMerge(Node* head1, Node* head2) { if (head1==NULL) return head2; if (head2==NULL) return head1; if (head1->datadata) { head1->next=sortedMerge(head1->next, head2); return head1; }else { head2->next=sortedMerge(head1,head2->next); return head2; } }

[@salmankhader1258]

small idea , if we put the condition which babbar wrote just before ending of while loop in the starting of the loop then it will handle the code for edge cases like if we have only one Node in our first list hence no need to write the seperate if condition for that edge case.

[@shishirkj]

can u explain this more plz

[@sarthakmoriya541]

what is Node can anyone please tell???

[@ritikamehta9791]

Great work sir!! mza aa rha hai ab pdhne me💯

[@kundanmali2405]

thank you for providing such great content.

[@Engineering_Unlocked]

thnx bhaiya :) completed on 8/7/24

[@theOmKumar]

Simpler Approach for Q2. , we can also do this without dummy node by putting head at right place first , and then simply traversing and joining smaller val data. ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) { if (!list1) return list2; if (!list2) return list1; ListNode *ans = new ListNode(-1); ListNode *temp = ans; while (list1 && list2) { if (list1->val < list2->val) { temp -> next = list1; temp = list1; list1 = list1 -> next; } else { temp -> next = list2; temp = list2; list2 = list2 -> next; } } while (list1) { temp -> next = list1; temp = list1; list1 = list1 -> next; } while (list2) { temp -> next = list2; temp = list2; list2 = list2 -> next; } return ans -> next; }

[@kunalmodiart]

i watched the solution given in the video then went on to code the Q by myself with my approach .. then i came in comments to add my code and i found your code which is literally similar to mine 😂 what a coincidence

[@antrikshgupta1964]

void sort(Node* &tail, Node* &head){ tail->next = head; head = head->next; tail = tail->next; } //Function to merge two sorted linked list. Node* sortedMerge(Node* head1, Node* head2) { Node* head = new Node(-1); Node* tail = head; while(head1!=NULL and head2!=NULL){ if(head1->data data){ sort(tail,head1); } else{ sort(tail,head2); } } while(head1!=NULL){ sort(tail,head1); } while(head2!=NULL){ sort(tail,head2); } return head->next; }

[@priyanshujain862]

Amazing bhaiya 🤩🤩

[@pahaljain9209]

Hat's Of bhaiya

[@neerajgarg9096]

EAGERLY WAITING FOR THIS LECTURE LOVE U BABBAR BHAIYA❤❤

[@ch0c0_1]

mza aagya bhai . Nice explanation

[@saurabhmalviya100]

Kya baat hai mazaa aa gaya .josh is high and it should be High 😂😂😂

[@saswatighosh3162]

It was lit 🔥🔥🔥🔥🔥🔥🔥

[@HemantNavlani]

In 1st question if we have made three sub lists of 0 1 and 2 in approach 2 would it not account for extra O(N) space ??

[@ritik8879]

attendance marked . bhaiya if possible video thoda jaldi jaldi daalo plz . my career depends on you

[@sundrmm0350]

Bhiyaaaa!!! Uh beauty🙇🏻‍♂️

[@divyanshupal1102]

if 0 empty or 2 empy then ?? there should be conditon for that also na???

[@Dharmikmonk]

Bohot aachi vdo hai bhaiya

[@ayushp4471]

BEST explaination as always 🔥🙌

[@jatingarg1897]

Maza Aaayaa bhai karke !

[@sushantgupta4757]

you are the best bhaiya

[@aadityagupta3720]

in first question we can use dutch national flag algo also

[@anuragpandey8165]

24:52 insertAtTail function kaha defined hai??

[@shripandey4120]

Bhaiya, 27:05 pr space complexity o(1) kse hogi, humne jo alag se three linked list bnayi hai uska space ?

[@muhammadasimsaeed2568]

Can someone explain to me why at 22:31 in line 60 we are connecting one->tail to twoHead->next as we have already connected zeroTail to 1's List in if condition and zeroTail to 2's list in else condition, Thanks in advance