Very well explained. but there is one issue in 2 min 12 sec the path of length 7 is wrong. you have changed two chars instead of 1, while transforming "lot" to "dog".

@techdose4u

3 жыл бұрын

Yea. I did that in graph and I have flagged a correction there.

@chiragmali6752

3 жыл бұрын

@@techdose4u Cool.. keep it up!!

@techdose4u

3 жыл бұрын

@@chiragmali6752 thanks :)

Hey Man, I got placed today at Western Digital Corporation. Thanks a lot for your videos. Keep going🔥🔥

@techdose4u

3 жыл бұрын

Woww great 🎉 Congratulations 💥 Would love to hear from you in LinkedIn. Please ping there.

@raviashwin1157

3 жыл бұрын

how did you apply for the company,it was on campus or off campus??

this is gold, especially for the ones who have questions on when should we use DFS and when we should use BFS

legendary explanation! Good job buddy!

Hello sir, I got placed at LinkedIn, thank you for your videos and constant guidance!!!.

@techdose4u

3 жыл бұрын

Great ❤️

Great explanation..Really enthralled by the way of explaining every bit

Thank you so very much.Best explanation for this problem.

your explanations are so good that i just listen to them and go ahead and solve the problem

@techdose4u

3 жыл бұрын

Nice :)

Hello Tech Dose, Amazing and detailed explanation. What will be the space complexity?

Simply Makkhaan(Butter). What an amazing explanation....... Hat's Off.

@techdose4u

3 жыл бұрын

Thanks :)

Hi, thanks for the problem explanation. Question: why don't we put "wordList" to a Dictionary instead of Set so to make word comparison O(1)?

Your explanation is to the point. Keep it up!

@techdose4u

3 жыл бұрын

Thanks

Thank you uncle ☺️❣️

excellent dry run!

What tools do you use to draw on the screen? @TechDose?

Superb Thank you so much

Greatt explanation sir

Thanks you very much, I only needed the graph hint to solve this

@techdose4u

3 жыл бұрын

Nice :)

Very Nice Explanation! Thank you.

@techdose4u

2 жыл бұрын

Welcome 🤗

Awesome explanation, thanks you so much :)

@techdose4u

3 жыл бұрын

Welcome :)

If we check popped string with the end word at the starting only not in the for loop then tym complexity will we O(n*26*w)? Am i right?

Well explained, thanks

Very well explained.

Can someone please explain why DFS doesnt work? I tried implementing with DFS and using a HashMap to store computed results already. But I am getting the wrong answer, getting 41/49 on leetcode.

Like before video as I know this will be awesome 😀 BFS approach I guess? Where at each step check new word can be formed with char. Replacement from A to Z... For better time complexity, put all words in set.

@techdose4u

3 жыл бұрын

Yep. You already solved it :D

@ayushthakur2896

3 жыл бұрын

why using set gives better complexity?

@LGL1999

2 жыл бұрын

@@ayushthakur2896 Constant time access to elements in Sets vs Lists I think?

you really draw out all the words such an eyesoar man

@techdose4u

3 жыл бұрын

😅

Well explained, but if you use set(for find()) then I feel complexity will be O(N2.WLogW).

so the main idea was to use O(n^2) to make the vali edges between words and then simply apply BFS?

Can't we use dfs & dp to reduce complexity?

You explained in an Excellent way

@techdose4u

3 жыл бұрын

Thanks

Thanks, it was amazing.

@techdose4u

3 жыл бұрын

Welcome :)

Just Curious, why you disliked this ques? at 0:08 :P btw I always find your videos easy to understand. Thanks :)

@techdose4u

3 жыл бұрын

😅

Fabulous! Very clear

@techdose4u

2 жыл бұрын

Thanks 😊

could anybody please tell me that why DFS does not work in this problem? Need help thank you~

Design Search Autocomplete System can you make an video on this ?

What is the space complexity?

What will be the complixity of this code?

I think there will be an error we have to return 0 at the end as if the endword is also present but there is no path in the graph to end word so we will return 0 then

What's lsize doing? Please someone explain

Instead generating new string and find in in set we could compare two words def is_one_word_change(w1,w2): if len(w1)!=len(w2): return False wc = 0 for i in range(len(w1)): if w1[i]!=w2[i]: wc+=1 if wc>1: return False return wc==1 this will reduce the time

@techdose4u

3 жыл бұрын

👍

basically, we're assuming all edges are weight 1 and doing a djikstra.

@prakharagarwal6237

2 жыл бұрын

in simple words BFS

Great content 🔥🔥

@techdose4u

3 жыл бұрын

Thanks

hey bro apka brute force approch mind bloing he but kya iska optimal solution nhi he ?

Hey, well explained. Can you do it for word ladder 2 problem also ?

@techdose4u

3 жыл бұрын

Already done.

Pls, discuss the 2-way Bfs approach also for this qs.

@techdose4u

3 жыл бұрын

Sure I will try

Sir in graph we used visited matrix then why not here. The remove method here will charge O(n) thereby increasing the time complexity

this question was asked to me at amazon..

@techdose4u

3 жыл бұрын

Nice :)

Sirji pranaam 🙏🏻

@techdose4u

3 жыл бұрын

🙏

amazing viedoe

@techdose4u

3 жыл бұрын

Thanks

Thank you sir.

@techdose4u

3 жыл бұрын

Welcome :)

U are great man

@techdose4u

3 жыл бұрын

Thanks

are we using 0(zero) as true here?

Nice explanation 👍

@techdose4u

3 жыл бұрын

Thanks

Why didn't you consider find and erase function time complexity in the Final Time complexity?? Btw great explanation

@techdose4u

3 жыл бұрын

Maybe I missed 😅 It's easy to miss something when calculating time complexity.

Thank you sir.very good explanation .please good the same one in graph too sir .it is very useful to know multiple approaches

@techdose4u

3 жыл бұрын

Welcome :)

Can u please provide solution and explanation for the below : Write a program to find Largest 4 digits Prime Number whose Sum of Digits is also Prime. 1.Prime Number is any number that is divisible only by 1 and itself. Examples are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,........ 2.Sum of Digits means sum of all the digits in that number. Example is number is 1234 then the sum of digits is 1+2+3+4=10

how you can make dog from lot? this connection is invalid.

Please make a video on the Manacher's Algorithm..

@techdose4u

3 жыл бұрын

Sure bro. But Currently I am covering graph :)

sir for checking string we are using set which gives us the result in average O(1) time then why have you written there that for string compare it is O(N*26*N) it should'nt be O(N*26) at 15:39?

@techdose4u

3 жыл бұрын

26 permutation for each letter. For a word, you will have 26*N. You can have N words at a level. So a total of (26*N*N).

@parteeksatija5536

3 жыл бұрын

@@techdose4u Sir lets say there is a word "hit" now to try every combination time will be O(26*N) for one word and we can have total N words so time complexity would be O(26*N*N). Sir then What is W for in O(26*N*N*W)?

@paraschawla3757

3 жыл бұрын

@@parteeksatija5536 * TC O(26*N*N*W) * N - Number of characters in String * 26*N - Number of transformations happening in a string of N characters * 26*N*N - Comparison with endWord string ( String comparison is O(N)) * W - Number of Words in the BFS call Second multiplication by N is because all transformations i.e. 26*N occurrences of strings are being compared by endWord String comparison is O(N) , check stackoverflow.com/questions/55330338/time-complexity-of-string-comparison/55330371 W - total number of words added in queue

thank you so much..awesome..I am following your videos since 1 yr. I can't find your coding profiles anywhere.(want to see)😁

@techdose4u

2 жыл бұрын

Thanks :)

Hi, I just had a doubt regarding the following test case. beginWord = "a" endWord = "c" wordList = ["a","b","c"] can anyone clarify why its output is 2 and not 1?

@lokeshkoliparthi9268

3 жыл бұрын

we have to start counting from first node

@devyeag3096

2 жыл бұрын

First start "a" change to it a to z if it is present as a same then we ignored it( suppose after convert a to a then it is same then we ignore it then check for b to z and depth = 1) and check for other character and it is find then add it (convert a to c it is find then depth = 1+1) and returns it...

Nice explanation

@techdose4u

3 жыл бұрын

Thanks

I learn a lot from your videos. Can you guide us of how you manage to stay dedicated for coding while being working in corporate ? I am working in corporate, I too want to go too far in coding, but not able to due. Please guide us

@techdose4u

3 жыл бұрын

Sure

@rohanseth1392

3 жыл бұрын

I liked your planning videos very much. Can you make a video telling us about the plan for those who are working in corporate

Nice

@techdose4u

Жыл бұрын

Thanks

code link is broken, can you fix it.

@techdose4u

3 жыл бұрын

I will check

I wish you are my csc sir

@techdose4u

3 жыл бұрын

🤗

❤

I looking for bidirectional bfs approach in the video :(

@techdose4u

3 жыл бұрын

Don't worry. I will cover biderectional BFS in other questions. I thought it will be easier for most people to understand simple BFS. It will be helpful if you can recommend a most asked question in interview regarding birectional BFS :)

@yeswanthbonagiri7142

3 жыл бұрын

@@techdose4u I think a simple question is : Given an undirected uniform graph, a source node and destination node, find minimum no of nodes exists in the path btw them. This is similar to word ladder 😅 I guess.

@techdose4u

3 жыл бұрын

@@yeswanthbonagiri7142 Yea 😅 I needed a proper asked question.

@yeswanthbonagiri7142

3 жыл бұрын

@@techdose4u leetcode.com/problems/open-the-lock/ leetcode.com/problems/jump-game-iv/solution/

I tried using dfs+memorization but getting 44th testcase wrong anyone else

I think you don't need to do line 14 to 21 in the latest version of the question

@techdose4u

3 жыл бұрын

:o Dint check it

i was able to do this using BFS bidrectional

@techdose4u

3 жыл бұрын

Great ❤️

Why used set??????????

Is this code is in java or c++ ?

@techdose4u

3 жыл бұрын

Cpp

👍

@techdose4u

3 жыл бұрын

👍

I have tried this similar question from leetcode today...but the expected answer is 5 mine is 6.I was doing this some different algo..Bu I donot know that graph will be used in this question..Meaning this question is of graph

@techdose4u

3 жыл бұрын

👍

🙏

@techdose4u

3 жыл бұрын

🙏

😀

@techdose4u

3 жыл бұрын

😁

You mentioned curr.compare(temp) is taking O(N) time complexity, why not just check `if curr==temp` this would be O(1)

@manokumar89

2 жыл бұрын

curr==temp is not O(1). its o(len of the string). its does a deep check.

I have a doubt. in the given example how u transformed lot to dog as it is given that only one letter transformation is valid.

@techdose4u

3 жыл бұрын

That was incorrect. I had flagged the correction when I was showing the graph. Dint realize it earlier in the previous slide.

*** JAVA VERSION OF YOUR ALGORITHM *** class Solution { public int ladderLength(String beginWord, String endWord, List wordList) { boolean isEndPresent=false; Set St=new HashSet(); for(String s: wordList){ if(s.equals(endWord)){ isEndPresent=true; } St.add(s); } if(!isEndPresent) return 0; int depth=0; Queue Q=new LinkedList(); Q.offer(beginWord); while(!Q.isEmpty()){ depth+=1; int lSize=Q.size(); while(lSize-->0){ String curr=Q.poll(); int len=curr.length(); for(int i=0;i

uhh u can use bfs...

Github not found

@techdose4u

3 жыл бұрын

It should work now. Try it.

Anyone noticed that TechDose has downvoted this question... 😂

Leave everything else, At 0:17 you also disliked the problem 😂

Why there is so dislike on the question, even you also disliked it?

@techdose4u

3 жыл бұрын

Which question ?

This is not the optimal way to solve this question. An nlogn solution exists

@coder5518

2 жыл бұрын

could u please send that solution link?

Dond dell me whad do do

Thanks for the explanation.Just that too many ads , as I don't have premium.It is quite distracting.

your code is not compact enough.

@jacobstech1777

3 жыл бұрын

# compact solution wordList = set(wordList) queue = collections.deque([[beginWord, 1]]) while queue: word, length = queue.popleft() if word == endWord: return length for i in range(len(word)): for c in 'abcdefghijklmnopqrstuvwxyz': next_word = word[:i] + c + word[i+1:] if next_word in wordList: wordList.remove(next_word) queue.append([next_word, length + 1]) return 0

@techdose4u

3 жыл бұрын

👍🏼

Great explanation..Really enthralled by the way of explaining every bit

Great explanation..Really enthralled by the way of explaining every bit

Great explanation..Really enthralled by the way of explaining every bit

Great explanation..Really enthralled by the way of explaining every bit

Great explanation..Really enthralled by the way of explaining every bit

Great explanation..Really enthralled by the way of explaining every bit

Great explanation..Really enthralled by the way of explaining every bit

Great explanation..Really enthralled by the way of explaining every bit

Great explanation..Really enthralled by the way of explaining every bit

Great explanation..Really enthralled by the way of explaining every bit

Great explanation..Really enthralled by the way of explaining every bit