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Hey, Guys in this video I have explained with code how we can solve the problem 'Merge K sorted Linked list '.
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Hope you like it. Comment if you have any doubt
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[@gigachad6844]

Best explanation so far, I've seen people using Priority Queues, Heaps and making it unnecessarily hard, why just now keep merging 2 lists until we're left with one only (answer). Thanks man

[@mickyman753]

the explaination was amazing ,i didn't thought there can be better approach then nk

[@anishprasad7024]

Second approach m apne Space complexity Log K batatya h but , wo K hona cahiye right because after every iteration at a time priority Queue/ Heap m at max K values reh rahe h ? Video time 8:13

[@CodeLibrary]

ya correct it will be O(K) only........by mistake i told O(Log k)

[@chitej7373]

Bro thanks for the great work and consistency

[@himanshunagar1510]

Thank You Bro , Nice Explanation

[@snehabaser3155]

Thanks man!

[@arushimathur7205]

Great Explanation !!

[@bhushanmahajan9107]

thanks man

[@sayantaniguha8519]

Can someone write the recurrence relation for the divide & conquer approach ?

[@1piecegaming622]

next video on clone a linked list please

[@it17harshahuja81]

bro no need for i++; only j--; is sufficient because on dry run it is taking same number of steps 👍👍👍👍

[@arnabroy2995]

Very nice explanation ❤

[@raviashwin1157]

clear explanation and clear code ❤️

[@CodeLibrary]

Thanks

[@lokanathpatasahani897]

Bro for after merging the linked lists you have to sort as you are not using priority queue in the last approach so it will take some extra time for sorting.
So I think in the last approach the overall time complexity will be O(nlogn).

[@HimanshuYadav-yj4to]

In the last approach, I think recursive call for smerge() function will also consume space in memory stack. So, I don't feel it's a O(1) space solution... Btw good explanation!

[@arnabdutta4662]

Same bro

[@tarunkumar.d8379]

Nope , we are not making recursive call with a whole array or any data structure but just a pointer that points to a address, so it's negligible.

[@sayantaniguha8519]

8:21 - Divide & Conquer Approach

[@harshyallewar5876]

Nice bro

[@palashagrawal2343]

are u solving every questions of babar?

[@rahulrggupta27]

two main approaches 1. heap (PQ), 2. 12.59s

[@yashsolanki2357]

can anyone explain me what is " linked lost ko dhaal do " 😂😂😂😂

[@kritekagrawal7340]

java Program
for(int i=0;i

[@harshidatanku6561]

Give the code in github

[@Riya-ly6uw]

very bad

[@arnabdutta4662]

Most optimized JAVA solution -
public ListNode mergeKLists(ListNode[] lists) {

int k = lists.length;
if(k == 0) return null;
if(k == 1) return lists[0];

// int i = 0;
int last = k-1;
// int j = last;
while(last != 0){
int i = 0;
int j = last;

while(i < j){
lists[i] = merge(lists[i],lists[j]);
i++;
j--;

if(i >= j){
last = j;
}
}

}

return lists[0];
}

ListNode merge(ListNode head1 , ListNode head2){

if(head1 == null)return head2;
if(head2 == null)return head1;

ListNode dummy = new ListNode(0);
ListNode temp = dummy;

while(head1 != null && head2 != null){

if(head1.val < head2.val){
temp.next = head1;
temp = temp.next;
head1 = head1.next;
}
else{
temp.next = head2;
temp = temp.next;
head2 = head2.next;
}
}

if(head1 != null)temp.next = head1;
if(head2 != null)temp.next = head2;

return dummy.next;
}
Hope it helps :)