L12. Find the intersection point of Y LinkedList

[@hardikpatel352]

intuation behind the optimal approach🔥🔥, legend ....

[@amanshrivastav0405]

You need to be a superhuman to discover the third approach on your own! Thankssssss alloooooottt striver 😇

[@BholaSingh-m8z]

Sir you, covers brute-force, optimized approaches, and even a killer third option! Well done, keep up the good work!

[@vibhavsharma2724]

A big salute to striver and his third approach 🫡

[@codingsoham]

The last algorithm man. After learning about it, it feels like why I didn't I think about that possibility. Thank you Striver for such a great explanation.👌👌

[@tanish_gupta]

Best video for this problem on youtube till date. Covered everything from brute-force to the best-optimised way to solve this problem. Kudos Striver!! Enjoying this path and the playlist.

[@technosujeet]

third approach literally blown my mind

[@bruvhellnah]

"literally"

[@shaikhsoheel8673]

Quality content brother I was waiting for Linkedlist God bless you brother

[@RajeevCanDev]

teachers like him comes in a decade or more, bhaiya please try to upload stacks and queues and strings as well

[@preethilatadanguria6946]

saw many videos to understand the 3rd approach but then landed on this video which has ended by search.thank u stiver for brilliant explanation.

[@blurbmusic1507]

optimal approach just GENIUS!! hats off captain!!

[@pratyushbhatt1712]

Beautiful Optimal approach. Thanks Striver!

[@shashankyadav8977]

The hashing one can be done in one iteration as well. We can insert in map acc to both LL at once. And check at that moment only if the freq is >1. Code - C++ int intersectPoint(Node* head1, Node* head2) { Node*p1=head1; Node*p2=head2; unordered_map hashmap; while(p1||p2) { if(p1) hashmap[p1]++; if(p2) hashmap[p2]++; if(hashmap[p1]>1) return p1->data; if(hashmap[p2]>1) return p2->data; p2=p2->next; p1=p1->next; } return -1; }

[@LinhHoang-ml1qo]

but it takes more space complexity while the last solution of Striver only takes O(1)

[@ujjawalchourasiya3988]

Aayla Jaado in 3rd approach Maja arha hai ab problems krne m . Thanks striver bhaiya for delivering such approaches things gets better and understandable ❤

[@AkashRaj-vj6br]

Thank you Striver for explaining so well...

[@ashwaniagrawal27]

Timestamps Hashing Approach - 1:20 Length Difference Approach : 8:10 Optimal Approach- 16:25

[@hat_awesome21]

striver can we expect stack and queue from u next ?

[@janaSdj]

Yes please

[@tonylee1868]

String much needed Stack and queue already there

[@thorxn1382]

😢

[@stith_pragya]

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

[@AmanVerma-iu8jp]

the way you explain the concepts is lit🔥

[@tanyaahuja7424]

Thank you striver !!

[@hareshnayak7302]

Wow Just wow what a mind blowing approch striver , Cool.thanks for this amazing video.

[@TON-108]

After following playlist till here, now i'm able to do these question myself 🌚 Thank You Striver Bhaiya!!!

[@navneetuppal9753]

Bro how are you able to think the approach and till which approach could you think of like till brute better or optimal and how?

[@LukmaansStack]

i was able to solve bruteforce but the 2nd and 3rd approches were out of my mind☹☹@@navneetuppal9753

[@ankush8243]

Thank you bro!💙 keep teaching like this💯🤗

[@ShaikBanu-j2o]

Third approach is mind blowing

[@MohammadEhshan-ve6uz]

Done it in 5 lines: Node* first=head1;Node* second=head2; while(first!=second){ first=(first!=NULL) ? first->next : head2; second=(second!=NULL) ? second->next : head1; } return first;

[@darshilmodi6851]

third approach is the real goat

[@oyeesharme]

thanks bhaiya for providing such an amazing content

[@ritikshandilya7075]

Thankyou so much for great explanation @Striver

[@knowthrvdo]

NICE LECTURE AND PLZ UPLOAD REMAINING LECTURES OF A TO Z SDA SHEET PLZ THEY ARE VERY HELPFULL FOR US

[@kushalswarup2662]

mind blowing method. cant believe woahh

[@kushalkollu8628]

damn man! what an idea for the optimal approach

[@anastasiiamaslova4220]

great explanation!

[@tejaspatel6965]

damn the third approach is beautiful

[@prajapati-suraj]

if anyone does figure out the third approach on his own, he is a genius.........

[@MohammadEhshan-ve6uz]

me

[@piyushrawat4668]

damnn🔥🔥the third approach!!!

[@utkarshpundeer07]

3rd approach was a 🤯🤯

[@vaibhavbhatt734]

i have taken a course of dsa and in that course one teacher tell the 2 approach till this date I think this approach is optimized but today (wake up to reality ho gaya)

[@ranjeetkumaryadav1837]

Thanks sir Last one is 🔥🔥

[@SHIVAMSINGHPARIHAR-w1i]

Was asked to me in Oracle Intern Technical Round 1 Interview

[@prathameshjadhav2942]

So help full video

[@hashtagcc]

The first time who have invented this algo must be from another dimension😙

[@prathameshjadhav2942]

Nice teaching

[@rohandas6298]

Why the time complexity of the last approach is not O(2 * (n1 + n2)) ??

[@BhavanaReddy15]

Same doubt

[@Fe-ironman]

@@BhavanaReddy15 because you're traversing both linked list simultaneously not separately. for example while traversing first linked list you traverse m step and reach null then traverse at maximum n step at worst case when there's no match and simultaneously you're traversing through the other linked list first n step and then m step at worst case when no match so the complexity is O(m+n)

[@NazeerBashaShaik]

Understood, thank you.

[@selene8721]

Thank you so much!!

[@pratyushtripathi1728]

Understood 😊

[@kale-lb5pr]

node* intersectionY(node* head1, node* head2) { if (head1 == NULL || head2 == NULL) return NULL; node* temp = head1; map mpp; // Insert nodes of the first linked list into the map while (temp != NULL) { mpp[temp] = 1; temp = temp->next; } // Traverse the second linked list and check for intersection temp = head2; while (temp != NULL) { if (mpp.find(temp) != mpp.end()) { return temp; } temp = temp->next; } return NULL; } int main() { vector arr={3,1,4,6,2}; vector b={1,2,4,5,4,6,2}; node *head1=convertarrtoLL(arr); node* head2=convertarrtoLL(b); node* head=intersectionY(head1,head2); cout

[@manga_mania.]

thank you striver

[@AlaskaAlaska-xq4gk]

zeherrrrrrr teaching

[@saswatrath4646]

3rd approach was awesome haha

[@DeepakPatel-d5v]

Thanks a lot bhaiya

[@harigs72]

Understood.🎉❤

[@YourCodeVerse]

Understood✅🔥🔥

[@DeadPoolx1712]

UNDERSTOOD;

[@ManishLakkavatri]

Understood!

[@Learnprogramming-q7f]

Thank you Bhaiya

[@SiddhesDas]

small correction if (n1 < n2) { return collisionPoint(headB, headA, n2 - n1); } else { return collisionPoint(headA, headB, n1 - n2); } baaki first class

[@ramkrishnapatil2105]

dimag hil gya bhai 🤐🤐

[@khalasianiket816]

understood ❤

[@shelkeonka]

In 2nd approch when N1 > N2 then in collision funtion we have to bring the t1 at alin postion which matches the t2 but the code is just work for when N1 < N2 ??

[@HeetMungra-g9l]

hey!! what if we start traversing from end of the both list and return the last node where they are equal??

[@hajeeramohamad7641]

in the second approach - what if the length of both the linked lists is same? how will the collision function work then?

[@NARUTOUZUMAKI-bk4nx]

Understooood

[@Manjith-1887]

mapm; while(headA!=nullptr) { m[headA]++; headA=headA->next; } while(headB!=nullptr) { if(m.find(headB)!=m.end()) { return headB; } headB=headB->next; } return nullptr;

[@knight-z1x]

beautiful

[@DEVANSHVYAS-z1b]

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode *p1 = headA, *p2 = headB; while(p1 != p2){ p1 = (p1 ? p1->next : headB); p2 = (p2 ? p2->next : headA); } return p1; }

[@manikanta2000]

My simple java code public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { ListNode temp1=headA; ListNode temp2=headB; while(temp1!=temp2){ if(temp1==null){ temp1=headB; temp2=temp2.next; } else if(temp2==null){ temp2=headA; temp1=temp1.next; } else{ temp1=temp1.next; temp2=temp2.next; } } return temp1; } }

[@rishipatwa6823]

can we do like reverse both ll and then check them until they are equal ..once unequal return the prev node

[@kale-lb5pr]

how does storing nodes directly make difference if we encounter same node value form second linked list why hashmap was showing no???/

[@factswithjenny9]

thanks

[@chiragbansod8252]

understood

[@ArpanChakraborty-do6yz]

Understand 31:53

[@harshitjaiswal9439]

understood.

[@akashumre22]

Bro thanks

[@nehapimpalkar396]

understood!

[@harshuke5831]

Understood

[@SurajSharma-z9p]

Please correct me if I'm wrong. If the two linked lists are never intersecting and are of different lengths, this will be a infinite loop because t1 will never be equal to t2 and we'll keep on interchanging them with heads in case of null

[@shivangairan6904]

Bro even if they are non-touching, after the first exchange for both they will come in parallel so the loop will terminate when they reach null together

[@SurajSharma-z9p]

@@shivangairan6904 Understood. Thanks

[@harshuke5831]

Understood 30lakh

[@fsyk1306]

Hey Striver, for the last algorithm, wouldn't it work too if when T1 or T2 reach NULL , they can be put back to Head1 or Head2 respectively? T1 goes back to Head1 and T2 goes back to Head2, it's working the same.

[@aryasharma69]

Try with different example like N1=5 & N2= 7, then you will get to know why it fails

[@RAHULSHARMA-on6tm]

in this code if you remove the if t1==t2 return t1, it should work fine right? Because the while loop will break in case both of them become equal, but the problem is, removing this condition will cause an error to appear. Can someone please answer this issue. Thanks

[@rahulnegi4027]

i would be simpler if these were circular h1->tail = h2 & h2->tail = h1;

[@Chlo_Chle_Hum]

is circular linked list important?...if yes,from where to do..someone pls suggest

[@LukmaansStack]

striver bhai we need string please

[@NonameNoname-f2t]

SUUUUUUUUPERRRRRRRRRBBBBBBBBBBB

[@VineetChelani1304]

can we even compare each node of one LL to Other LL and get the first intersection ??

[@dhrumilparmar1505]

that's brute force approch..still works but time complexicity will be of O(n^2).here we will get in O(max(len(t1),len(t2)) + distance of intersection from smallest LL)....somewhat linear complexity

[@SuvamoySamanta-n2x]

why use this line mpp.find( temp) != mpp.end(); please explain this line

[@SuvamoySamanta-n2x]

!= use why

[@roshnisingh7364]

please share the code for last approach

[@ekjotsingh9609]

Why wasn't a set used in the hashing approach? Wouldn't that be better?

[@aryasharma69]

That will still take O(n) space

[@kale-lb5pr]

pls can anyone explain me what is the difference in storing a node in hashmap and a value ? if the same value is present in second LL also why doesnot it indicate yes? pls give me hintts...

[@lot2say590]

Because Node is a class which consists of the address of the data too. Hence while comparing the same value, we are not checking if it's in the same address, so we can't conclude on that. So to check if it's in the same address, we compare the whole Node and not just the value.

[@djpro6413]

@@lot2say590 Thanks bro same question arise to me

[@stellar3496]

waatha idhan da coding 🤯🤯

[@naturesrevolution]

anyone pls answer in second approach what happens if t1 and t2 contains same element before linking?

[@qwerty20917]

No problem in that case too bro.. cause we are comparing the pointers themselves (whether they point at same position/node or not),not their values.. Hope this helps..

[@naturesrevolution]

@@qwerty20917 thank you bro

[@utkarshsingh3319]

❤❤

[@jagadeeshp1163]

what if we take 3->1 and 1-> it fails ??

[@Shivi32590]

thank you!

[@dhakerk094]

agar dono linked list me koi intesection point nhi hai to

[@swapnil7142]

third option ke liye thought process rahega, ye dimag me kaise aayega

[@kale-lb5pr]

pls explain my concern im just a beginner dont break my spirit like that!!!!!!!!!!!

[@moksh455]

hey striver it's showing an error if i submit the exact code on leetcode on the test case [ 1, 2 ] and [ 2 ] , could you explain why

[@moksh455]

ahh got it we need to add a check after we interchange temps with heads as well

[@yourhonestbro217]

This question was not that difficult, my first approach was the optimal one

[@SamyakSharma-oy1bv]

respect++;