Coding Interview Tutorial 61: Word Ladder [LeetCode] 

Thank you, John. God bless you, too.

You're quite welcome!

Thank you, Jin. Happy coding!

no, you need that check on line 19 you imbecile.

@@nickscriv5141 why? There is no way there will be a duplicate since this is a hashset.

You're welcome, Siddharth!

Thank you. Happy coding!

You're welcome!

You're welcome, Avinash. Happy coding!

Hey Amell, when I played around, I found that if I change the way the queue is processed, for(int k=q.size();k>0 ;k--) to for(int k=0;k

@@avinashmv9686894578 Take q.size() out of the loop. The reason it is not working for you is because the size of the queue keeps on changing.

@@anurag_arora guys have you done 126. Word Ladder II

Thank you, Tapas. Happy coding!

Thank you, Karthik. Happy coding!

I believe because that's O(26) which is constant time, it's a non-dominant term and can be ignored as it won't have any effect on the run-time as the problem input grows.

Yes, Pushkar. I'll do that problem soon.

Keep up the focus, Ian. It's worth the effort. Happy coding!

There's a bug in this code. We need to store size of q ahead, then use that in the for loop because q's size is changing inside the loop.

@@zhuolunzhou6074 I felt werid when he typed queue.size() and was wondering the size would change along with we offer the new word, thanks for pointing out!

@@wigcrafter This is not a bug. Because the k is initialized with the q.size(), after that k is used for the loop.

The time complexity is O(L * N), where L is the length of each word and N is the number of words.

You're welcome! Yes, I'll do that problem soon. Thanks.

@@amellperalta Hi Amell, looking forward to your word ladder II. Thank you very much!

Thank you, Nayak. In this problem, it's necessary to use BFS because it finds the length of the shortest path. DFS will not work for this problem because it will not necessarily find the length of the shortest path. The time complexity is O(L * N), where L is the length of each word and N is the number of words.

Same question. There is only one element in queue at any time, why do we need a loop. Not sure if my understanding is correct

so you dont repeat the same word again. for ex: when you transform from DOG -> COG, you don't want to go from COG -> DOG again. I can expand on that if you want. =)

@K M when you reverse outer for loop you get expected output as 2 , but when i start i=0 in for loop then i get 1 , i also can't undeerstand y it's showing 2