Word Ladder - LeetCode Hard - Google Phone Screen Interview Question

Summary

This video focuses on solving the 'Word Ladder' coding problem, where the objective is to find the shortest transformation sequence from a given start word to an end word by changing one letter at a time, using a provided word list. The instructor breaks down the problem, discussing the intuition and approach using Breadth-First Search (BFS) and a queue. They provide a step-by-step explanation of the algorithm, including code implementation in Java. The video aims to help viewers prepare for coding interviews by understanding advanced tree data structure problems and their solutions.

Takeaways

• 📌 The video is part of a D algorithms boot camp series focusing on advanced questions for the tree data structure.
• 📚 Viewers are encouraged to follow the playlist in the provided order for a structured learning experience.
• 🔥 The specific problem discussed was asked in a Google phone screen and is popular on LeetCode with close to 12,000 thumbs up.
• 🚧 Despite being labeled as a hard question on LeetCode, the presenter suggests it is not as difficult as it seems.
• 💬 The challenge involves transforming one word to another using a dictionary list, changing one letter at a time, under specific conditions.
• 📈 The solution employs a Breadth-First Search (BFS) approach utilizing a queue and a set for efficient lookups.
• 📝 A step-by-step solution process is explained, emphasizing the importance of incrementing the length for each word change.
• 📊 The time complexity is detailed as O(N^2 * M), where N is the number of words and M is the length of the word, highlighting the algorithm's efficiency.
• 🛠 The presenter walks through coding the solution in LeetCode, providing insights into handling base cases and iterating through character changes.
• 📦 Additional resources, including code and assignments, are available in the video description for further exploration and practice.

Q & A

What is the problem being discussed in the video?

The video discusses the 'Word Ladder' problem, where the task is to find the shortest transformation sequence from a given 'beginWord' to an 'endWord' by changing one letter at a time, and all intermediate words must belong to a given 'wordList'.

Can you provide an example to better understand the problem?

Yes, the example given in the video is: beginWord = 'hit', endWord = 'cog', wordList = ['hot', 'dot', 'dog', 'lot', 'log', 'cog']. The shortest transformation sequence is 'hit' -> 'hot' -> 'dot' -> 'dog' -> 'cog', which requires 5 steps.

What approach is suggested to solve the Word Ladder problem?

The video suggests using a Breadth-First Search (BFS) approach, where we start with the 'beginWord', add adjacent words (differing by one letter) to a queue, and continue the process until we find the 'endWord' or exhaust all possibilities.

How is the adjacency between words determined?

To determine if two words are adjacent (differing by one letter), the video proposes replacing each character of the current word with all possible 26 English alphabet characters and checking if the resulting word exists in the 'wordList'.

What data structures are used in the proposed solution?

The solution uses a queue (implemented using a LinkedList) to perform the BFS traversal, and a set (HashSet) to store the 'wordList' for efficient lookup.

What is the time complexity of the proposed solution?

The time complexity of the solution is O(N^2 * M), where N is the number of words in the 'wordList', and M is the length of each word. This is because for each word, we need to check all other words (O(N^2)) and replace each character (O(M)).

What is the space complexity of the proposed solution?

The space complexity of the solution is O(N), as we need to store the 'wordList' in a set for efficient lookup.

What are the base cases or terminating conditions for the BFS traversal?

The BFS traversal terminates if (1) the 'endWord' is found, in which case the length of the transformation sequence is returned, or (2) the queue becomes empty, indicating that no transformation sequence is possible, and zero is returned.

How does the solution handle visited words to avoid cycles or redundant computations?

The solution uses a set called 'visited' to keep track of words that have already been processed. When a new word is generated, it is added to the queue and the 'visited' set only if it exists in the 'wordList' and has not been visited before.

Are there any limitations or considerations mentioned for the proposed solution?

The video does not explicitly mention any limitations or considerations for the proposed solution. However, it is worth noting that the solution assumes the 'wordList' is finite and contains only valid words, and that the length of all words in the list is the same.

Outlines

00:00

🎥 Welcome and Context

The speaker welcomes viewers back to their YouTube channel, introducing a new video in the 'D Algorithms Boot Camp' series. They mention continuing to cover important questions related to the tree data structure, specifically an advanced question asked during a Google phone screen interview. The speaker also provides information about the video and code resources available in the description.

05:00

🗣️ Problem Statement and Intuition

The speaker explains the 'Word Ladder' problem, which involves finding the shortest transformation sequence from a 'begin word' to an 'end word' by changing one letter at a time, where each intermediate word must exist in a given word list. An example is provided to illustrate the problem. The speaker then discusses using a Breadth-First Search (BFS) approach and hints at the time and space complexity analysis.

10:03

✏️ Problem-Solving Approach

The speaker outlines the steps to solve the 'Word Ladder' problem using a BFS approach. This includes creating a set from the word list for efficient lookup, initializing a queue with the 'begin word', and iteratively exploring adjacent words (differing by one character) in the queue. The process involves adding valid words to the queue, marking them as visited, and terminating when the 'end word' is found or no solution exists.

15:04

💻 Coding the Solution

The speaker begins coding the solution to the 'Word Ladder' problem in Java. They handle base cases, initialize necessary data structures (set, queue), and implement the BFS algorithm as described earlier. The code involves iterating through the current word's characters, generating potential next words by replacing one character at a time, and checking if the new word exists in the word list and is unvisited. The process continues until the 'end word' is found or all possibilities are exhausted. The speaker also discusses the time and space complexity analysis.

Keywords

💡Word Ladder

A word ladder is a puzzle where you must transform one word into another by changing one letter at a time. Each intermediate word must be a valid word. This is the core problem being solved in the video, where the goal is to find the shortest transformation sequence from a given start word to an end word using a provided word list.

💡Breadth-First Search (BFS)

Breadth-first search is a graph traversal algorithm that explores all vertices at the current depth level before moving on to vertices at the next depth level. In the context of the word ladder problem, BFS is used to find the shortest transformation sequence by exploring words that differ by one character from the current word before moving on to words that differ by two characters.

💡Queue

A queue is a linear data structure that follows the First-In-First-Out (FIFO) principle. In the context of the video, a queue is used to implement BFS by adding new words to the end of the queue and removing words from the front for exploration.

💡Set

A set is a data structure that stores unique elements. In the video, a set is used to store the word list to efficiently check if a word exists in the list and to avoid revisiting words during the BFS traversal.

💡Time Complexity

Time complexity is a measure of how the runtime of an algorithm scales with the size of its input. In the video, the time complexity of the word ladder solution is analyzed as O(N^2 * M), where N is the number of words in the word list and M is the length of each word. This is because for each word, all possible words that differ by one character need to be checked, which is an O(N * M) operation, and this needs to be done for all N words.

💡Base Condition

A base condition is a special case in an algorithm that is handled separately from the general case. In the video, the base condition checks if the end word is present in the word list. If it is not, the function immediately returns 0 because no transformation sequence is possible.

💡Character Array

A character array is a data structure that stores a sequence of characters. In the video, a character array is used to temporarily store a word being modified by changing one letter at a time. This allows the algorithm to generate all possible words that differ by one character from the current word.

💡Visited Set

A visited set is used in graph traversal algorithms to keep track of vertices (or words, in this case) that have already been visited. In the video, a visited set is used to avoid revisiting words during the BFS traversal, which could lead to an infinite loop.

💡Space Complexity

Space complexity is a measure of how much auxiliary space an algorithm requires in addition to its input. In the video, the space complexity of the word ladder solution is mentioned as O(N), where N is the number of words in the word list, as a set is used to store all the words.

💡Optimization

Optimization refers to the process of improving an algorithm's performance, either in terms of time complexity or space complexity. In the video, the instructor mentions that the solution provided is not optimized for time complexity and that further optimizations may be possible.

Highlights

Introduction to a new video in the D algorithms boot camp focusing on advanced tree data structure questions.

Mention of a highly rated question on LeetCode that was asked in a Google phone screen, emphasizing its importance despite being labeled 'hard'.

Explanation of the problem 'Word Ladder' involving a transformation sequence from a 'begin word' to an 'end word' using a dictionary word list.

Clarification of the problem's rules, such as each adjacent pair of words in the sequence must differ by only one letter.

Example given to illustrate how to transform 'hit' to 'cog' by changing one letter at a time through a sequence of words.

Encouragement for viewers to pause the video and think about a solution before proceeding.

Introduction to the solution approach using Breadth-First Search (BFS) and a queue for efficient word lookup and transformation.

Discussion on the time and space complexity of the proposed solution, including an in-depth explanation of why it's O(n^2 \cdot m) for time complexity.

Demonstration of how to code the solution, starting with setting up base conditions and initializing necessary data structures.

Step-by-step coding walkthrough, showing how to implement BFS to find the shortest transformation sequence.

Explanation of how to check for valid transformations by changing one letter at a time and verifying against the word list.

Clarification on how to handle found transformations and when to add them to the queue for further processing.

Discussion on how to handle the end word and return the length of the shortest transformation sequence if found.

Final thoughts on the solution's efficiency and the importance of understanding the underlying algorithm for interview preparation.

Invitation for viewers to check out code links, assignments, and previous lectures in the video description for further learning.

Encouragement for viewers to engage with the channel through likes, shares, and comments for community support and learning.

Transcripts

hey everyone welcome back to my channel

another new video in this video we are

continuing the D algorithms boot camp

and we're doing a very important

question so I have started doing some

very important questions for the tree

data structure like some Advanced

questions we're continuing the playlist

in order so if you check out the

playlist Link in the description below

it's exactly in order you can you can

follow the exact same order without

shuffling it and

um yeah that's what we're doing so in

this video we're going to do a question

that was asked in the Google phone

screen so very important question 11,000

close to 12,000 thumbs up on lead code

hard question it says hard on curly

braces it is not a hard question it's

just a bit

uh it's not it's not hard

so let's do this question um what is the

question but yeah before we move forward

with that check out the links in the

description below for the code and

assignments and previous lectures and

everything um um I think my video was

paused for a second but it's fine I'm

not that pretty so you don't have to see

me the important thing is that you can

see the screen and the screen is not

hanging so that's the important thing

right okay um Let's Talk About It Word

ladder let's see what the question is a

transformation sequence from word begin

word to word end word using a dictionary

word list dictionary word

list is a sequence of words okay such

that these are the conditions every

adjacent pair of the words differ by a

single letter okay every s of I is okay

it should be in

the word list and the last one in the

series has to be the end word given two

words begin word and end

word uh return the number of words in

the shortest transformation sequence

from begin word to end word or zero if

no such sequence is

possible okay so for example you are

given the word hit and you are given the

end word cogg and you are given these

list of words so you have to make

hit from hit you have to make

cogg but you can only change

you can only change one letter at a time

so if you start from

hit okay if you start from hit you can

select hot because it's changing only

one letter which is I then you check hot

and then you can select

uh dot because it is only changing one

letter which is

D then you can select uh

dog which is only changing one letter

which is T then you can select

Cog which is only changing one letter

which is this D1 and the Cog is the end

word that we need so we reached Cog in

how many words 1 2 3 4 five

words okay simple that's the

problem how do we solve this what do we

do pause this video think about it think

about what we're doing and then we'll

get started so how are we going to solve

this question um let me first write the

question down so this same one we can

take hit uh Cog and the list entire list

that we have over here hope you can see

the screen so my

words

oh the words that I

have where is

it here we

go the words that I

have what are the words that we

have hot dot dog lot oh sorry that was

Hindi for etc

etc I keep forgetting that this is

English only Channel but anyway we had

dot we had hot we have dog we have a lot

we have a lot

log we have a cog so from Hit I have to

create Cog

Cog

okay this is my starting word and ending

word so what you're going to do is think

about it you have to select every every

adjacent word should only differ every

adjacent word should only differ in one

character we're going to use a BFS

approach bet for search because we'll be

using a q okay so starting put this

entire thing in a set so it's easy to

look

up easy

to look up and what I would do is I

would uh put in the

que I will put in the

que my first word which is hit and I

will take a length answer

initially that is zero I will put hit in

it I will remove hit from

the

Q and I will add all the

elements uh all the I will add all the

so when I added hit in it I will

actually increment it by one whenever

you add it increment by

one and uh when I remove move hit I will

add all the other items that differ by

one what all items in the set differ

differ by one character hot only so I

will add hot in it so that is sequence

two for now now I will remove hot and I

will add all the

items that differ one by hot so there is

Dot and there is lot so I will add dot

in it and I will add lot in

it size three then I will remove Dot and

I will add one that remains only one

character that is empty sorry not empty

only one character that differs from dot

dog dog you do know what differs means

right so what is the one character

difference so in D in D O T I can

replace t with G so it will make dog and

I have dog over here I cannot make it

and also I have to remove these items so

I can't add multiple items so I will

keep removing also so I I added hit I

added hot hot will be removed dot dot

will be removed lot lot will be removed

dog is added dog will be

removed okay then I will remove lot

so what is one uh character

difference word Cog is not one because

it has two character differences C and G

but log is so I will add log over

here okay incremented by

one all right then I will remove

dog and I will add Cog over here

actually I will not add Cog because I

have already found Cog so CG is my last

element so the destination element so I

found

Cog um or this will be five I found

Cog

so no this will be four only because I

did not add Cog so I found CG I will

just return this + one answer plus one

five is the answer yeah

correct

so create the set use the que uh using a

BFS approach because uh you know think

about the intuition we have to do Word

Lookup we have to check that only one

character difference should be there so

what is going to be the time

complexity space complexity is O of n

for the number of words because that's

the number of words we are storing

but if we

have the length let's say length L of

the word is called if is

M and uh for every word we are checking

in the set which is the word that is uh

one

difference so this into this so n into

n n

s and we are doing it for every

character also n² into M that's the

complexity now if this is not clear it

will be more clear when I code it

okay time

complexity it's very simple CU for every

word I'm

checking how many words are there

that differ by one so every word I'll be

checking in the worst

case so that's n n into n and whenever

I'm checking I'm checking each character

so multiplied by m which is the total

character

count o of n

is the

space let's code it it will be making

much more sense okay um so let's see how

we're going to solve this question here

I have lead code copy

this create a new file it's

called ladder uh what's the name of the

problem Word

ladder um

public in ladder length whatever

whatever okay by the way the code can be

found in the description below you can

run it on your system directly on your

browser and let's see how we can do this

so in ladder

length uh I have my begin word end word

and word list okay okay so let's start

to code this here I can say that first

of all base condition and I will have

my this thing side by side so you can

see the base condition if end word is

not in the

list

okay word list

dot

contains

what end

word

okay in that case just return zero I

have to create a

set you can call these all the visited

ones new hash set not a problem I will

create a

que of type

string just call it short Q

New Link

list and I will add the first begin word

in it

okay I will take a length zero for now

same breadth for search while Q

is not

empty I will take the size which

is Q do size and I will increase the

length by one and I will add all

the I will check for every element so

what are we doing over here uh I'm

removing that many number of items it's

it's breath for search nothing new for

in I

is equal to Zer I is less than size and

i++ I will get my current string which

is just remove from the Q so I remove

hit from the Q when I have removed hit

this current is hit for example I will

check in the remaining set all the items

that all the items that differ by one

character so what I can do

is what I can do

is

uh all the items for in J is equal to

zero J is less than for all the

characters for

example current dot

length and J Plus+ for every character I

will check so for car CH is equal to

let's say I start from a c is less than

equal to zed

C++ I will try

to change the character um that I have

taken for hit H I will try to change it

by let's say AIT and then I will check

if AIT exists over here if it doesn't

check for bit CIT dit EIT and then

similarly do h a hbt h c and then do h i

a h i b h i c h i d h i e so on and so

forth do you get

it

so temp of if I create a I have to

create a character of character

array

temporary okay

this will be current door to

C car array I will say temp of

J is equal to this character then

check new word is equal to what new

string

temp okay I hope you're able to

understand this I found hit I will

change the zero index of the hit

hit with all the characters so first I

will check for a it bit CIT dit if it

exists over here in my set after that I

will check for H A hbt after that I will

check for h a h i so every index in my

word I will replace with all 26

characters of the English

alphabet okay if this new

word

equals my end

word I have found my answer and that is

l+

one I can say otherwise if the word list

contains the new

word and it is not

visited like I've already not visited

it in that case add it in q

and remove also add it from

visited like I have visited this you can

also do it by Vera and start removing it

but this is

fine if nothing else works in the

end return zero means there is no

answer and that's it that's it so that's

why the complexity this will be constant

as as it is because uh this is like a

fixed number of characters 26 but we're

doing this for m so this m multiplied

by outer two Loops so n into

n so n so this is n cross M times n so

this this entire thing see is M it is

inside a for Loop so M multiplied by

number of times this for Loop is running

which is

n so n into M multiplied by number and

this Loop is running n n into M into n

N2 m is the time complexity oh that's it

problem solved okay well thanks a lot

for watching you can find the code in

the description below and if you have

any question let me know in the comment

section below and um yeah keep following

the playlist work hard good luck for

your interviews any questions you have

any suggestions you have let me know in

the comments uh links code everything

previous videos uh assignments can be

found in the description below if you

want to support the channel make sure

you like share and subscribe join the

learning public initiative you can take

a screenshot and share with #d with

Kunal on socials and um yeah check out

the previous videos are very good new

videos are coming and have

fun