Nice explanation. One finding though, Set will remove duplicates. For example if we have same values for given (x, y) coordinates, we will endup missing one value in output. Am I missing something?
what does iterating over m1.second and m2.second actually mean? Is m2.second having the values of the tree in the ascending order? And we are pushing a vector() into the ans 2d vector, is it just to create some space in the answer vector?
everything is good but you have named the coordinates of node values wrong, provided the root is the origin (0,0). Go to timestamp 12:43 for your reference.
In the Map Map Set, if you use Unordered_map then time will be NlogN in total. If you use Map then NlogNlogNLogN. 2 extra logs for ordered map key search. So. It depends on your choice of DS.
Will you able to make videos on binary lifting? Which will be very helpful to solve problems like lowest common ancestor of given two nodes and kth ancestors of a given node in a tree.
Code in Python for DFS(It becomes helpful with sorted function): class Solution: def helper(self, placement,level, root, dic): if(not root): return dic[placement].append((level, root.val)) self.helper(placement-1, level+1, root.left, dic) self.helper(placement+1, level+1, root.right, dic)
def verticalTraversal(self, root: TreeNode) -> List[List[int]]: dic = defaultdict(list) self.helper(0,0, root, dic) result = [] for i in sorted(dic.keys()): temp = [] for j in sorted(dic[i]): temp.append(j[1]) result.append(temp) return result
This problem can be solved using HashMap & PriorityQueue in Java. `class Solution { public List verticalTraversal(TreeNode root) { Map map = new HashMap(); int[] width = {0, 0}; verticalTraversal(root, 0, 0, map, width); List result = new ArrayList(); for (int i = width[0]; i
![[#2024MAMA] BIGBANG (빅뱅) - 뱅뱅뱅 (BANG BANG BANG) + FANTASTIC BABY | Mnet 241123 방송](http://i.ytimg.com/vi/EQsYfiPR9uM/mqdefault.jpg)
The solution is really good, but I would like to point out that using set doesn't handle duplicate values. So to pass all the cases use multiset!
👍
wait set doesn't handle duplicate values means!!!!? Set is meant to take just unique values right?
@@neversettlejay yes, hence you'll end up missing duplicate values.
use multiset
@@neversettlejay ya
Please keep this series on going and free of cost. Please
😳
Your teaching style is rock solid thanks for this wonderful explanation of the topic I stumbled upon.
Thanks 👍🏼
Ye tech ki dose bahut imp h.Ye sab coders ko leni chaiye.....
wonderful explanation, solved this leetcode problem in one go just by following your explanation .
Great ❤️
Your intuition to explain is so good , and I have a doubt on for loop.... Would you please explain how the for loop work.?
It really helpful to me
best explaination for this problem and dfs is easier than bfs in this tbh
I was getting runtime Error, but you solved my doubt. Thanks a lot for your Awesome Explanation!!
Welcome :)
thank you sir , your lectures help very much I just watched little of it and understood what to do next, thank you
Nice :)
Solved it one try within 20 mins using treemap of treemap of priorityQueue. I must say I learned well from you lol. Thank you so much legend!!!
Welcome
great explanation, ans.push_back(vector()) , I dont understand what is these and why we use "vector()" ? Would you please explain me
Too good explaination.This video is really good and all contents are easily understable .
Thanks 😊
I was not able to solve the problem due to 1st constraint, thank you for the solution.
Thanks :)
The content of this channel is growing exponentially surely this will become a repository to crack faang interviews thanks for your effort tech dose
Thanks for your appreciation. I hope it helps every student preparing for programming interviews.
Seriously very amazing solutions of this problem you provide 😊.. thanking you so much sir
Welcome :)
finally understood the problem lots of the Thank you!!❤❤
Better than striver! But you could have used unordered map and mutilset instead of map and set.
Nice explanation. One finding though, Set will remove duplicates. For example if we have same values for given (x, y) coordinates, we will endup missing one value in output. Am I missing something?
But it is assumed that there is no duplicates in Tree,if there is duplicates then go for multiset
Use multiset
what does iterating over m1.second and m2.second actually mean? Is m2.second having the values of the tree in the ascending order? And we are pushing a vector() into the ans 2d vector, is it just to create some space in the answer vector?
Awesome solution dude. Hats off to you.
Thanks :)
Instead of using set using multiset will solve the problem of having duplicates in same row and col.
Thay asked the same today in my interview
Lucky me
I watched this video after the interview
Thank you youtube
Ur recommendation algo sucks!
How does youtube know that you want to see this 😂 Google knows about your interview too!!
@@techdose4u scary. My meeting invite 😱
@@vijayjayaram606 Haha :P
Inside the first for loop, the first statement of pushing the vector() means are we pushing the entire vector?? I am confused with the representation.
For duplicate values in Python simply use List and make sorting arrangements accordingly (DFS approach)
Thank you so much! I was stuck at this question.. you made it clear.
Welcome :)
Can someone tell why "ans.push_back(vector())" is used in 1st approach);
i don't think we have similar DS in nodeJS, map and set both follow insertion order here, so we manually need to sort the keys/values.
everything is good but you have named the coordinates of node values wrong, provided the root is the origin (0,0). Go to timestamp 12:43 for your reference.
Brilliant approach.
Thanks :)
bhaiya explanation is awsm
Really good explanation !
Thanks :)
Bro your explanation is truly good. Bro could you please make a video for 1st year students in order to get into product based companies.
For students from tier 3 college.
I will try to make it.
@@techdose4u we will be waiting
..
@@praveenprakash6336 forget bro...we NITS IITANS deserve them..you go for TCS🤣
@@depression_plusplus6120 Don't judge a guy by his college okay.....
Nicely explained🙌..keep up with this awesome work.👍
Thanks
Striver said that the time complexity is O(N*logN). You are saying that time complexity is O(N*logN*logN*logN). Which one is right?
In the Map Map Set, if you use Unordered_map then time will be NlogN in total. If you use Map then NlogNlogNLogN. 2 extra logs for ordered map key search. So. It depends on your choice of DS.
@@techdose4u I asked him he agreed that time complexity will be O(N*LogN*LogN*LogN). He told the time complexity of java code which is O(N*logN)
Will you able to make videos on binary lifting? Which will be very helpful to solve problems like lowest common ancestor of given two nodes and kth ancestors of a given node in a tree.
Its difficult to accomodate with graph videos as many of them are pending as well. So, I will try after graph.
@@techdose4u thank you!
Huge thanks💜
Welcome 😀
awesome explanation
Welcome :)
Excellent explanation sir
Thanks 😃
Code in Python for DFS(It becomes helpful with sorted function):
class Solution:
def helper(self, placement,level, root, dic):
if(not root):
return
dic[placement].append((level, root.val))
self.helper(placement-1, level+1, root.left, dic)
self.helper(placement+1, level+1, root.right, dic)
def verticalTraversal(self, root: TreeNode) -> List[List[int]]:
dic = defaultdict(list)
self.helper(0,0, root, dic)
result = []
for i in sorted(dic.keys()):
temp = []
for j in sorted(dic[i]):
temp.append(j[1])
result.append(temp)
return result
Nice explanation!
Thanks :)
can you let me know why we have inserting value on basis of col * row and not in row * col.
That was epic explanation!
Thanks :)
Thank you so much
Welcome :)
Hello,
Could you please help how to solve this question using JavaScript?
Because map and set are little different in JS.
I will ask people to share their code in comment in different languages. It will be much more helpful.
Which will run faster BFS or DFS?
but if we have duplicates elements in same row and column and set will not work it would be better to use vector instead of set
What should be the space complexity?
what is the time complexity of this code?
This problem can be solved using HashMap & PriorityQueue in Java.
`class Solution {
public List verticalTraversal(TreeNode root) {
Map map = new HashMap();
int[] width = {0, 0};
verticalTraversal(root, 0, 0, map, width);
List result = new ArrayList();
for (int i = width[0]; i
BRILLIANT
😀
Approach = 🤘
🤘🏼
How to print top view of bi art tree in o(n)
good problem
how were we supposed to figure this out on ur own without looking at the solution?
Thanks sir
Welcome :)
Not working for test case->
[1,2,3,4,5,6,null,null,7,8,null,null,9,null,10,null,11,10],hence instead of set I took vector in map
can we solve it recursively??
Saviour!!
:)
what is the difficulty level of this one
Medium I think
where you work Google or Meta?
Code link is not reachable..
I will check
TreeMap
Hard to understand
Please watch it again
@@techdose4u Thank you so much, I understand pretty well and be able to code out by myself now, your visualization and explanation helped tremendously
Nice explanation !
Thanks :)