Thanks for clean code. sudo code have minor issue , please correct it . if ( t1 -> data < t2 -> data ) You missed equal elements . it should be if ( t1 -> data data )
yes. I wondered the same. I came here from leetcode. there some values in the two sorted lists which are same. but what an explanation. I know only basics of linked lists. I was able to undersand well enough
letsss fuckin gooooooo did both the approaches by only just getting one hint and last handling of edge case where i is not null where i tell temp ki tera next is i and vice versa yaaay
I'm confused how the space complexity is O(1)? We are creating a dummy node which extends to hold the nodes of both linked list 1 and linked list 2. Don't the space complexity be O(n1+n2) where n1 are number of nodes in linked list 1 and n2 are number of nodes in linked list 2.
Can't we use dummy node to make it as a copy of the lowest head value of either of the lists (which acts as the head of the new merged list) , and repeat the same process . Will the solution be different ?
There's a little mistake at the end for handling remaining nodes after either one list has reached null, there should be a another while loop to handle those cases. Also there isn't a java code to the above problem kindly provide one
No,we don't need the while loop here. As we are using the exiting nodes itself,not creating new ones. Suppose n1 is null, so n2 is left. we pointed temp->next as n2. so, from here, link is maintained up to n2 becomes null. Hope you got this.
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */ //Method 1 (Using extra space) //Naive Implementation class Solution { public ListNode mergeTwoLists(ListNode head1, ListNode head2) { MySinglyLinkedList mergedSinglyLinkedList = new MySinglyLinkedList(); ListNode currentNode1 = head1; ListNode currentNode2 = head2; while (currentNode1 != null && currentNode2 != null) { // Retrieve the values of the current nodes from both the singly linked lists int val1 = currentNode1.val; int val2 = currentNode2.val; // Append the smaller value to the merged singly linked list if (val1
Bro woke up and dropped the whole playlist
Turu
Definittely this is one of the best playlist on linked list in youtube.
best series i have ever seen especially the dry run thank u so much for an exceptional series
Striver made the solution as cake walk... best teacher !!
Byfar the best linked list series
Thank you very much Raj Bhaiya
Great video, -->
This video is not updated in the SDE sheet and in the Article as well.
we missed youuu here ! KEEP UP THE GREAT WORK YOU'RE DOIN MAN
Bhaiya ,I can definitely say, you are god of DSA❤❤
dude you're better than that neetcode guy thank youuuuuuuuu
Thanks for clean code. sudo code have minor issue , please correct it . if ( t1 -> data < t2 -> data ) You missed equal elements . it should be if ( t1 -> data data )
yes. I wondered the same. I came here from leetcode. there some values in the two sorted lists which are same. but what an explanation. I know only basics of linked lists. I was able to undersand well enough
There is no need for this Equal to condition bcz it already handled by else part
But it is good to think about this condition👍👍
@@ujjwalsingh6889 yes. I realized it later. when I did a dry run. Thanks for replying ujjwal
@@anirudhv0062 You need to improve your DSA These are basics
why this problem is not in the A2Z sheet?
Bhaiya when can we expect stack queue playlist
Lecture successfully completed on 30/11/2024 🔥🔥
meri tarf se ik papii supper good explanation
🏳️🌈
6:33 - good soultion
letsss fuckin gooooooo did both the approaches by only just getting one hint and last handling of edge case where i is not null where i tell temp ki tera next is i and vice versa yaaay
amazing explanation ✨✨
Very good explanation. 😊😊
Epic bhaiya!!
Loved it, bro!
Welcome Back sir❤🎉
Thanks ... nice explaination
Awesome 😎 17:36
On which topic you will be making your next playlist??
Nice explanation ❤
Understood✅🔥🔥
shouldn't time complexity be min(n1,n2) coz this is where while stops and then it's just one extra link that we've to do?
if the smaller values were in the n2 then in this case we will traverse till the end of n2. so min(n1,n2) will not work.
But we have to consider the worst case in which we traverse both the linked lists up to the last node
Happy New Year 🎉
I'm confused how the space complexity is O(1)? We are creating a dummy node which extends to hold the nodes of both linked list 1 and linked list 2. Don't the space complexity be O(n1+n2) where n1 are number of nodes in linked list 1 and n2 are number of nodes in linked list 2.
ACTUALLY ITS LIKE VARIABLE DECLARATION .
we are not creating a new node we are just declaring it like a temp node
@@aneeketvispute5063bro
If we change temp.next = t2 then how the next node of t1 is not lost
Since we're changing the link
In which lec you've explained the concept of dummy nodee?
8:15 Can anyone tell in which video dummy node concrpt is explained by him😅
Can't we use dummy node to make it as a copy of the lowest head value of either of the lists (which acts as the head of the new merged list) , and repeat the same process . Will the solution be different ?
Best bhai Thank you
What if I use a multiset instead of an array in brute force approach??
I couldn't find this question in the a2z playlist.
class Solution {
public ListNode mergeTwoShortedLists(ListNode l1,ListNode l2){
ListNode ans=new ListNode(-1);
ListNode temp=ans;
while(l1!=null&&l2!=null){
if(l1.val
hey folks, does anyone know what tool he is using to write and draw?
its an ipad
is this a dynamic programming problem?
UNDERSTOOD;
Thanks Striver!!1
ty
understood!!!
There's a little mistake at the end for handling remaining nodes after either one list has reached null, there should be a another while loop to handle those cases. Also there isn't a java code to the above problem kindly provide one
No need for while loop , see the video 11:40
No,we don't need the while loop here. As we are using the exiting nodes itself,not creating new ones. Suppose n1 is null, so n2 is left. we pointed temp->next as n2. so, from here, link is maintained up to n2 becomes null. Hope you got this.
Wait what 😂 he did solved that with if else. Watch it completely before commenting.
Understoood
Thanks
To be exact, time complexity would be O(min(n1 + n2)).
Understood
Understood!
Nyc
understood
Undertsood
UnderStood
understood :)
done and dusted
Done and dusted😅
God or what
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
//Method 1 (Using extra space)
//Naive Implementation
class Solution {
public ListNode mergeTwoLists(ListNode head1, ListNode head2) {
MySinglyLinkedList mergedSinglyLinkedList = new MySinglyLinkedList();
ListNode currentNode1 = head1;
ListNode currentNode2 = head2;
while (currentNode1 != null && currentNode2 != null) {
// Retrieve the values of the current nodes from both the singly linked lists
int val1 = currentNode1.val;
int val2 = currentNode2.val;
// Append the smaller value to the merged singly linked list
if (val1
where the hell is the dummy node concept been spending hours to find but couldn't find it Someone plz pin it
mila?
free(dummy node):
god
us
understood
Understood!
Understood
Understood!
Understood