Intersection of Two Linked List | EP 18

Lecture Notes: leadcoding.in/...
Watch the complete Linked List Series • Linked List Series
Coding Ninjas Courses 15 % OFF aff.codingninj...
Question (codestudio) bit.ly/39k7U1h
Question (leetcode) leetcode.com/p...
Placement Preparation Roadmap • Roadmaps for Placement
Hi, I am Fraz.
I am Software Engineer working at Curefit and I love teaching.
This is my Channel where I upload content related to Interviews and my personal experiences.
My contact details
Instagram :- / frazmohammad
Connect with me on LinkedIn :- / mohammad-f-bb4886157
Telegram Group t.me/LeadCoding

[@kakadiazeel]

Leetcode discussion section is one of the best. Last solution is one of the most voted and appreciated solution. Kudos for your mathematical explainations also. 🌟

[@kulkarnisoham]

100% true

[@TomShelbyXD]

My mind is blown after seeing all these solutions for a single problem 🤯 and the last solution 💯

[@Snehanomics]

Bhaiya one thing I will appreciate in you that you are so so so humble and a calm teacher! You teach so well !

[@mohammadfraz]

Thank you so much

[@pranabnandy450]

You can add one more approach.............. >> first traverse the List1 to end..... >> then make temp1 -> next = head1 >> Then use slow Fast pointer concept to detect loop in a linked list >> If it exist store the node in a separate pointer ( target = slow ) >> If it does not exist then ( target=NULL) >> Lastly again traverse the List1 and make last node point to NULL ( temp1->next = NULL) Correct me if I am wrong.

[@pawansingh7430]

In the unordered set Approach We have to work on nodes and not on the values. So even if values inside the node is same the nodes maybe different because of their next pointer, so we don't check for values So we will create Set and not Set

[@krishnaradhey2814]

Bro do cover all the DS topics step by step

[@mohammadfraz]

Will try

[@chirag488]

After 15 minutes of trying , I found the most optimized approach as told in the video. Thanks Fraz bhaiya :)

[@rishisrivastava4878]

Bhai aap bht acha video bnate Dil se bol rha hu abb jb bhi recommend krta hai aaka ya striver ya babbar I always choose you❤ itna asan code rhta hai ki fat se yaad ho jaata hai❤

[@keshavgaur4535]

the last approach was awesome 💥

[@mohammadfraz]

Thanks a lot Keshav for supporting

[@keshavgaur4535]

@@mohammadfraz welcome bhaiya 😇

[@sanjanachandravanshi9864]

@@keshavgaur4535 what is time complexity of last approach?

[@mdnadeem6343]

Op bhaiya last approach was lit 🔥 I was thinking that bs Itna hi likhna tha 😅😅.....I like the way you convert every solution into a simple and sober equation thanks bhaiya ❣

[@mohammadfraz]

Thank you so much 😀

[@DipsOfficial802]

Thank you so much bhaiya for such a great solution thanks🙏🙏🙏

[@deepurana6010]

You are amazing bhaiya❤. Loved the way u teach

[@tech_wizard9315]

By when you will upload entire playlist?😍

[@mohammadfraz]

As soon as possible

[@surendrababu223]

Understood

[@anishbishnoi29xD]

Sir thanks for video. Please Continue More Java DSA

[@maaltez]

Bhaiya aapka content ek number h, i just want one more help i am looking for good headphones and not able to make decision. Can you please suggest which headphones are you using.

[@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; } cout

[@thomanani6826]

Nice explanation brother

[@vagabondfoodie5788]

Bhaiya what is the tc and space complexity of last approach?

[@shamimsarker839]

Best

[@desihiphop4998]

Bhaiya 1 approach strike krgyi thi wo length waali baaki uske agli waali to katiye 🔥🔥 thi maja aagya !!!!!!!!!!

[@mohammadfraz]

❤️❤️ bhai apna naam to batao Talha Yunus

[@desihiphop4998]

@@mohammadfraz bhaiya Himanshu Joshi telegram pe महाराज ke naam se hun dono jagah pareshaan krta hun aapko Comment kr kr ke 😂

[@mohammadfraz]

Maharaj is for muhfaad?

[@desihiphop4998]

@@mohammadfraz Haan bhaiya

[@18devpratapsingh33]

C++ Solution using unordered_Set class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { unordered_set s; ListNode* temp = headA; while(temp!=NULL){ s.insert(temp); temp = temp->next; } ListNode* temp2 = headB; while(temp2!=NULL){ if(s.find(temp2)!=s.end()){ return temp2; } temp2 = temp2->next; } return NULL; } };

[@yashbahuguna8135]

you didn't discussed the stack approach

[@18devpratapsingh33]

Brute Force Approach class Solution { public: ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode* tempB = headB; while(headA!=NULL){ while(tempB!=NULL){ if(tempB==headA) return headA; tempB = tempB->next; } tempB = headB; headA = headA->next; } return NULL; } };

[@vaidanshkukreja8970]

BRUTE FORCE APPROACH : ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode * temp1 = headA; ListNode * temp2 = headB; if(temp1==NULL || temp2==NULL) return NULL; while(temp1!=NULL || temp2!=NULL) { temp1 = headA; if(temp1 == temp2) return temp1; else { temp1 = temp1->next; while(temp1!=NULL) { if(temp1 == temp2) return temp1; temp1 = temp1->next; } temp2 = temp2->next; } } return NULL; } ---------------------------------------------------------------- MAYBE: Time Complexity : O(n*m) Space Complexity : O(1) Any doubts related to the code ping me happy to help 😀

[@shyaamaltripathi5373]

Bhaiyya jo aapne doosra wala solution bataya tha usmein unordered set wala usmein agar kisi ek point pe dono list mein same elements hain but lists are not intersecting tab kaise hoga ?

[@pawansingh7430]

We have to work on nodes and not on the values. So even if values inside the node is same the nodes maybe different because of their next pointer, so we don't check for values So we will create Set and not Set

[@shyaamaltripathi5373]

@@pawansingh7430 Ok thank you

[@SunnyGupta00]

Reach++

[@pranavkorhale5089]

//2nd approach public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { ListNode temp1 = headA; ListNode temp2 = headB; int count1 = 0; while(temp1!=null) { count1++; temp1 = temp1.next; } int count2 = 0; while(temp2!=null) { count2++; temp2 = temp2.next; } temp1 = headA; temp2 = headB; if(count1count2) { int diff = count1-count2; while(diff!=0) { temp1 = temp1.next; diff--; } } while(temp1!=null) { if(temp1==temp2) { return temp1; } temp1 = temp1.next; temp2 = temp2.next; } return null; } }

[@Sonu-tg6tg]

Please make videos using JAVA

[@nishantparaskar2048]

public ListNode getIntersectionNode(ListNode headA, ListNode headB) { ListNode a = headA ; ListNode b = headB ; while(a!= b) { if(a == null) { a = headB ;} else {a =a.next ; } if(b == null) { b= headA ;} else { b= b.next ;} } return a ; // or b } i took the reference from this video and did this

[@parthrastogi572]

Isse kharab bhi koi padha sakta hai kya!! Poor explanation Don't watch this video!!