Merge k Sorted Lists - (Leetcode - 23) - (Google, Amazon, Microsoft..) : Explanation ➕ Live Coding

[@JJ-tp2dd]

Unbelievable! How are you so so good bhai? Java Implementation: class Solution { private ListNode mergeTwoSortedList(ListNode l1, ListNode l2) { if(l1 == null) return l2; if(l2 == null) return l1; if(l1.val l2.val) { l2.next = mergeTwoSortedList(l1,l2.next); return l2; } return null; } private ListNode paritionAndMerge(int s, int e, ListNode[] lists) { if(s > e) { return null; } if(s == e) { return lists[s]; } int mid = s +(e-s)/2; ListNode L1 = paritionAndMerge(s,mid,lists); ListNode L2 = paritionAndMerge(mid+1,e,lists); return mergeTwoSortedList(L1,L2); } public ListNode mergeKLists(ListNode[] lists) { int k = lists.length; if(k == 0) { return null; } return paritionAndMerge(0,k-1,lists); } }

[@GghiggFfhhhf]

15 min ke baad khud se, code likh ke submit and got accepted the way he explain 🙏🙏😻😻

[@itsaayush7357]

Clear, concise and simple explanation!! India has some unique talents. When it comes to code so we have best of best teachers available. In every domain there is one king, and in DSA we have our mik.

[@guptajiivlogging8174]

Please bhaiya, it's my humble request to you please don't stop teaching dsa and cp and posting videos on solving question.. It help's a lot. After 5-6 months companies are going to visit our college so you and your channel is going to help me and my friends alot....Lots of love from jabalpur (MP)

[@souravjoshi2293]

Same bro

[@nawazthezaifre8870]

placement hoa kya bhai ??

[@22gauravkumar70]

@@nawazthezaifre8870 placement hua?

[@shweta1013]

You are just awesome !!.. Cant believe i wrote the code myself after listening to this video !!

[@Ramanujan-c5c]

Beautiful code .... creative way to solve it. Loved it. I heard that u are quite good coder, its true. U deserve the fame as striver... launch few short videos here and on linkedin. Hope u reach to great heights. I am really thankful to u for sharing so many great codes and solving problems with so ease. Love ur solution. Thank u so much for helping newbie coders and turning them towards Guardian of Ltc and red coder of CFr. Keep shining Sir. Tc.

[@codestorywithMIK]

This means a lot. Thank you ❤️❤️❤️

[@nik3889]

how ! i never seen that type of solution that is the best solution i ever watched

[@souravjoshi2293]

I think you are the best who can make things easier like a cake walk

[@akshaychavan5511]

This question felt hard for me when I initially attempted it. But, now it feels quite easy.

[@atifhu]

Bhai kya smjhate ho❤️❤️

[@gui-codes]

too good man. becoming your fan day by day. One stop for revision

[@vineetkumar2899]

awesome bhaiya😍

[@mayankagarwal6775]

really really good explanation. thanks for your efforts😃

[@ShivaaySingh20]

Thank you so much bhai for making this question very easy to understand 🙌❤.

[@tutuimam3381]

Amazing explanation🌹

[@elakstein]

Good one. Just want to point out that currently the time complexity is O(NlogK) and space complexity is O(logK) because of recursion stack, it can be solved iteratively with the same time complexity but with O(1) space complexity.

[@codestorywithMIK]

Indeed

[@alphadrones23]

yup.. we could use the iterative approach to merge the two lists

[@zgodgaming857]

the best explanation

[@khushijain3574]

you explained so nicely keep making such videos

[@codestorywithMIK]

Thank you so much 😇🙏

[@khushijain3574]

@@codestorywithMIK please create video on quick sort on linked list , i want your approach to solve this thanks

[@NamanSaxenaa]

GOD level! 🔥

[@shabananoor9423]

Best explanation

[@abhinay.k]

great video

[@Methre_]

very nice bhai;

[@nihalsingh6233]

Thank You soo much sir !! You really made it easy

[@codestorywithMIK]

I am so glad it helped ❤️❤️❤️

[@JJ-tp2dd]

Merge two sorted List -- Leetcode 21 class Solution { private ListNode mergeTwoSortedList(ListNode l1, ListNode l2) { if(l1 == null) return l2; if(l2 == null) return l1; if(l1.val l2.val) { l2.next = mergeTwoSortedList(l1,l2.next); return l2; } return null; } public ListNode mergeTwoLists(ListNode list1, ListNode list2) { return mergeTwoSortedList(list1, list2); } }

[@mudassarnazar1663]

You are genius ,

[@U2011-n7w]

very good explanation pls make video on today leetcode contest questions

[@codestorywithMIK]

I will soon start those as well . Just trying to make plans as per my time management. Soon

[@priyanshudubey2245]

Bhaiya weekly contests ke questions par bhi video bana dijiye please.

[@codestorywithMIK]

I will soon start those as well . Just trying to make plans as per my time management. Soon

[@RohitYadav-uo6uo]

can't we use a for loop in which we merge first two then we will merge the answer of that to next list and like that isn't that a good way?

[@saisathvikreddyloka7578]

🔥🔥

[@ishwarkokkili7646]

Masterful !!

[@codestorywithMIK]

Thank you so much 🙏❤️😇

[@shivamjadon2257]

mast video

[@codestorywithMIK]

Thank you 😇🙏

[@sunshineandrainbow5453]

What is the need of partition and merge, we can simply use a for loop from i =2 to n-1 and merge the result of lists[0] & lists[1] with lists[i] . At each iteration mergedList = merge(lists[i],mergedList) ...

[@mindpower185]

@codestorywithMIK ek baar just Merge K sorted array wala ...explain kardo ya github pe upload kardo please

[@harshlakra3260]

concepts ki video kab aarhi h

[@surajsidar3280]

If I submit Python code then There is TLE error if we submit using an iterative approach There is a maximum recursive depth error if I follow the recursive approach.

[@vishalgautam-f6v]

what will be the recurrence relation for the given algorithm? If it is T(k) = 2*T(k/2)+n then it will result in n*k or if it is not the correct recurrence relation then what will be the correct relation?

[@ShivaaySingh20]

interview me yadi questions puche to mai bhi interviewer se kahunga "Aao Story Se Code Kre "😁😅

[@codestorywithMIK]

Cool 😅❤️

[@nik3889]

bhai ek bar k sorted array wala kra do

[@MakeuPerfect]

bhaiya minimum height trees wala question karao plz

[@codestorywithMIK]

Can you please share leetocde link

[@MakeuPerfect]

@@codestorywithMIK meine link diya tha bhej diya but ye dikhai nahi de raha comment mein

[@AshishSharma-tf3fy]

just using vector property to skip binary search algo class Solution { public: ListNode* mergetwo(ListNode* l1,ListNode* l2){ if(l1==NULL) return l2; if(l2==NULL) return l1; if(l1->valval){ l1->next=mergetwo(l1->next,l2); return l1; } else{ l2->next=mergetwo(l1,l2->next); return l2; } } ListNode* mergeKLists(vector& lists) { if(lists.size()==0) return NULL; while(lists.size()>1){ ListNode* new_node=mergetwo(lists[0],lists[1]); lists.push_back(new_node); lists.erase(lists.begin()); lists.erase(lists.begin()); } return lists[0]; } };

[@drbullah1388]

Bhaiya, i thought of this on the first go ListNode* mergeTwoLists(ListNode* list1, ListNode* list2){ ListNode* head = NULL; ListNode* tail = NULL; if(list1==NULL) return list2; if(list2==NULL) return list1; if(list1->val < list2->val){ head = list1; list1 = list1->next; } else{ head = list2; list2 = list2->next; } tail = head; while(list1 && list2){ if(list1->val < list2->val){ tail->next = list1; list1 = list1->next; } else{ tail->next = list2; list2=list2->next; } tail=tail->next; } if(!list1) tail->next=list2; else tail->next=list1; return head; } ListNode* mergeKLists(vector& lists) { int k = lists.size(); if(k == 0){ return NULL; } if(k == 1){ return lists[0]; } ListNode* res = mergeTwoLists(lists[0], lists[1]); for(int i = 2 ; i < k ; i++){ res = mergeTwoLists(res, lists[i]); } return res; } Ye bhi sahi hai na?

[@codestorywithMIK]

Yes it should work too.

[@harsh.jain22]

I also thought of the exact same solution of iteration in first go 🤜🤛

[@yashsharma1774]

Your time complexity calculation is incorrect. It will be O(k*n). This is because you have perform calculation at each leaf here but in case of binary search you do calculation at each node of traversed path.

[@Whirlwind03]

Beautiful solution brother, I solved it using PriorityQueue now will solve using your approach public class Pair implements Comparable { ListNode node; public Pair(ListNode node){ this.node = node; } @Override public int compareTo(Pair o){ return this.node.val - o.node.val; } } public ListNode mergeKLists(ListNode[] lists) { if(lists.length == 0) return null; ListNode res = new ListNode(-1); ListNode t = res; PriorityQueue pq = new PriorityQueue(); for(int i=0;i 0){ Pair rem = pq.remove(); t.next = rem.node; t = t.next; if(rem.node.next != null){ pq.add(new Pair(rem.node.next)); } } return res.next; }