Group Anagrams - Categorize Strings by Count - Leetcode 49

Summary

In this coding tutorial, the presenter introduces a method to group anagrams from a list of strings efficiently. By recognizing that anagrams can be identified by sorting or by counting character occurrences, the video suggests using a hash map with character counts as keys for better performance. The presenter explains the process of counting characters, using ASCII values to map characters to indices, and then using these counts to group anagrams. The solution is optimized to O(m*n) time complexity, where m is the number of strings and n is their average length, making it an efficient approach to solving the problem.

Takeaways

• 😀 The video discusses a coding challenge to group anagrams from a list of strings.
• 🔍 An anagram is a word or phrase formed by rearranging the letters of another, such as 'tan' and 'nat'.
• 🔄 To identify anagrams, the video suggests sorting the strings and comparing them for equality.
• ⏱ The naive approach of sorting each string has a time complexity of O(n log n), where n is the average string length.
• 🚀 A more efficient method is introduced, utilizing a hash map to group anagrams with a time complexity of O(m * n).
• 📚 The hash map uses a count array to tally the occurrence of each character from 'a' to 'z' in the strings.
• 📊 The count array is transformed into a tuple to serve as a hash map key, as lists are mutable and not hashable in Python.
• 🔢 The ASCII values of characters are used to map them to indices in the count array, subtracting the ASCII value of 'a'.
• 📝 For each string, the character counts are incremented in the count array, which then becomes a key in the hash map.
• 📦 The strings with the same character counts are grouped together in the hash map under the same key.
• 📈 The final output is a list of lists, where each sub-list contains anagrams grouped by their character counts.

Q & A

What is the main topic of the video script?

The main topic of the video script is about writing code to group anagrams together from a list of strings.

What are anagrams and how can they be identified?

Anagrams are words or phrases that contain the same characters but in a different order. They can be identified by sorting the characters of each string; if the sorted strings are equal, they are anagrams.

Why is sorting each string not the most efficient way to group anagrams?

Sorting each string has a time complexity of O(n log n) for each string, where n is the length of the string. This becomes inefficient when you have to sort m strings, resulting in a time complexity of O(m * n log n).

What alternative method is suggested in the script to group anagrams?

The script suggests using a hash map (or dictionary) to count the frequency of each character in the strings and then group strings with the same character count together.

How does the character counting method work in terms of time complexity?

The character counting method has a time complexity of O(m * n), where m is the number of strings and n is the average length of the strings, making it more efficient than sorting each string.

What is the maximum number of unique characters considered in the script?

The script considers a maximum of 26 unique characters, which corresponds to the lowercase letters of the English alphabet from 'a' to 'z'.

How is the 'count' array used in the hash map?

The 'count' array is used as a key in the hash map, where each index represents the count of a specific character from 'a' to 'z'. The value associated with each key is a list of strings that have the same character count pattern.

Why are lists changed to tuples when used as keys in the hash map?

Lists are mutable and cannot be used as keys in a hash map. Tuples, on the other hand, are immutable and can be used as keys.

What data structure is used to avoid dealing with edge cases when a count does not exist in the hash map?

A default dictionary (or default hash map) is used, where the default value for a new key is a list, thus avoiding the need to check if the key exists.

How is the final result of the anagram grouping presented?

The final result is presented by returning the values of the hash map, which are the lists of anagrams grouped together based on their character counts.

What is the significance of using a hash map for this problem?

Using a hash map allows for efficient grouping of anagrams by leveraging the constant time complexity for lookups, insertions, and updates, making it an optimal solution for this problem.

Outlines

00:00

🔍 Grouping Anagrams with Efficient Algorithm

This paragraph introduces the problem of grouping anagrams from a list of strings. An anagram is a word or phrase that is formed by rearranging the letters of another, such as 'tan' and 'nat'. The speaker explains that sorting the strings and comparing them can identify anagrams but may not be the most efficient method. The paragraph concludes with a hint at a more efficient solution that will be revealed in the following content.

05:01

🚀 Optimizing Anagram Grouping with Hash Maps

The speaker presents an optimized approach to group anagrams using a hash map, which significantly reduces the time complexity compared to sorting each string. The method involves counting the frequency of each character from 'a' to 'z' in every string and using this count as a key in the hash map. Each key maps to a list of strings that share the same character count pattern, effectively grouping anagrams together. The speaker also discusses the use of a default dictionary to handle new counts and the conversion of the count list to a tuple to serve as a hash map key. The paragraph concludes with the final step of extracting the values from the hash map to get the grouped anagrams, resulting in an efficient solution with a time complexity of O(m*n), where 'm' is the number of strings and 'n' is their average length.

Keywords

💡Anagrams

Anagrams are words or phrases that contain the same letters but are arranged in a different order. In the context of the video, the main theme revolves around grouping anagrams together. The script uses 'tan' and 'nat' as examples of anagrams, which can be identified by rearranging the characters to form equivalent strings.

💡Sorting

Sorting is the process of arranging items in a particular order, typically ascending or descending. In the script, sorting is mentioned as a method to determine if two strings are anagrams of each other by arranging their characters and comparing the sorted versions. For instance, sorting 'tan' and 'nat' results in 'ant' for both, confirming their anagrammatic relationship.

💡Time Complexity

Time complexity is a measure of how long an algorithm takes in terms of the amount of input. The script discusses the inefficiency of sorting each string (O(n log n)) and introduces a more efficient solution (O(m * n)). It's a critical concept in the video, as it drives the search for a faster method to group anagrams.

💡Hash Map

A hash map, also known as a dictionary or hashmap, is a data structure that stores key-value pairs and allows for efficient retrieval. In the video, a hash map is used to group anagrams by counting the frequency of each character in the strings, with the character count serving as the key and the list of anagrams as the value.

💡Character Count

Character count refers to the frequency of each character within a string. The script explains that by counting the occurrences of each character from 'a' to 'z', one can determine the anagram groups. For example, both 'tan' and 'nat' have one 't', one 'a', and one 'n', leading to their grouping together.

💡ASCII Value

ASCII value represents the numerical representation of a character in the ASCII encoding system. The script uses ASCII values to map characters to indices for counting character occurrences, subtracting the ASCII value of 'a' from the character's ASCII value to get the index in the count array.

💡Default Dictionary

A default dictionary, or hashmap, is a variant of the standard dictionary that provides a default value for nonexistent keys. In the script, it's used to simplify the process of adding strings to the anagram groups without having to check if the key exists, with the default value being a list.

💡Tuple

A tuple is an immutable data structure that can store a sequence of elements. The script mentions changing the count array to a tuple because tuples can be used as keys in a hash map, unlike lists, which are mutable and cannot be used as keys.

💡Big O Notation

Big O notation is used to express the upper bound of the time complexity of an algorithm in terms of the size of the input. The script concludes with the optimal time complexity for the anagram grouping solution being O(m * n), where 'm' is the number of strings and 'n' is the average length of each string.

💡Optimal Solution

An optimal solution is one that provides the best performance or outcome for a given problem. The video script describes finding an optimal solution for grouping anagrams with a time complexity of O(m * n), which is more efficient than initially sorting each string.

Highlights

Introduction to the problem of grouping anagrams in a list of strings.

Explanation of what anagrams are and how to identify them through character rearrangement.

Example given with 'tan' and 'nat' to illustrate the concept of anagrams.

Sorting strings as a method to determine if they are anagrams of each other.

Discussion on the time complexity of sorting each string in the list.

Introduction of a more efficient solution using a hash map for grouping anagrams.

Utilization of character count array to represent the frequency of each letter in a string.

Explanation of using a hash map with character count as a key for grouping anagrams.

Clarification that there are at most 26 unique characters (a-z) to consider.

Description of the hash map's key-value structure for anagram grouping.

Coding demonstration of creating a hash map to group anagrams efficiently.

Use of a default dictionary to handle the list of anagrams automatically.

Conversion of the character count list to a tuple to use as a hash map key.

Final step of extracting the values from the hash map to get the grouped anagrams.

Discussion on the time complexity of the optimal solution, O(m*n).

Conclusion and encouragement for viewers to like and subscribe for more content.

Transcripts

hey everyone welcome back now let's

write some more neat code today so today

let's look at a good

question to practice the fundamentals

group anagrams

so we're given just a list of strings

and we want to group

all anagrams together so for example we

have tan

and we have nat and they are anagrams

together

so in the output we're gonna group them

together

into one sub-list and how do we

know these two are anagrams of each

other the first one

is tan the second one is na

and you see if we swap the n and the t

then we get we get tan right

and that's the same as this so by

rearranging

the characters we can get equivalent

strings so these two are equal

so another way of looking at it is two

strings

are anagrams of each other if we take

each of them and sort them right so if

we sort both of these

we're gonna get ant right for both of

them

because that's the sorted version so if

they're anagrams of each other

when they are sorted they should be

equal right so

so one way to group anagrams together

would be to take

each one of these strings in the input

and then sort them

but the time complexity of that is going

to be

n log n where let's just say n

is the average length of each of the

input strings

so that's how much it takes to sort each

of the strings

and we have to do that m times where

let's say

m is the length of

how many like basically how many input

strings were given in the first place

so you can see that this is going to be

the overall time complexity

so my question is can we do better than

this and actually

the simple solution in this case happens

to be more efficient and let me show you

that solution right now so one also

condition that we're given

is that each character is going to be

from lowercase a

to lowercase z so at most we have 26

unique characters right so let's just

have

an array called count so for

each one of these strings we want to

count

the characters from a to z

right how many does it have of each so

we know it has

one e one a and it has

one t and for this one

we would see that the exact same thing

is true it has one e one a

and one t so if we use a data structure

called a hash map in this case and in

this case

our key is going to be

this over here so this is what we can

use to

identify anagrams and then our value is

going to be the list

of anagrams so in this case how many

strings or which strings have

this pattern of count one e

one a and one t well we see there's one

each there's one t and there's one last

one

a so those are gonna be the values we'll

have a list of them

e t and then a right so we'll have three

strings

and as you can see in the output it's a

little messy but that's what we have

here right so these three

are grouped together so we're going to

use a hash map

to group them together since all we're

and since we're using a hash map and all

we're doing is counting

the characters of each and

we know that we have a limit of 26

lowercase characters

the overall time complexity is gonna be

big

o of m where m

is the total number of input strings

that we're given

times n where n is the

average length of a string because we

have to

count how many of each character it has

so we're gonna have to go through

every single character in a string and

since we

are using count right or this array

in our hash map we're using it as a key

in our hash map we kind of also have to

multiply this by 26

because that's what's going to be the

length of our count array

but you know that this reduces anyway so

the actual time complexity is big

o m times n

so now let's code it up as you can see

though this is basically one of those

problems that can be solved

with a hash map very efficiently so

let's create our hashmap we'll call it

result and so remember what we're doing

is we're mapping

the character count of each string

and we're mapping that to the list

of anagrams so to do that we of course

have to go through every string that

we're given in the input

and we want to count how many of each

character it has so we'll have

initially a array so it's we're going to

have

0 in it we're going to have 26 zeros

right because

one for each character so from lowercase

a

all the way to lowercase z

and now we want to go through every

single character

in each string and we want to count how

many of each character so we want to map

in this case a to index 0

and we want to map z to index 25 so how

can we do that well we can

one way at least is just take the ascii

value of the current character we're at

c and then subtract the ascii

value of lowercase character

a so as you can see lowercase a

minus lowercase a is going to be 0 z

lowercase z minus lowercase a is going

to be 25 so

we do have it correctly because for

example let's just say

a is ascii value 80 and i don't know if

this is actually correct but it's just a

random value then

b is going to be ascii value 81 and so

on

but if we want to map this to zero

we know one way to do that is take 80

minus 80.

if we want to map lowercase b to 1

we can take 81 minus 80 right

so that's basically what i'm doing here

in case you haven't seen something like

this before

and so we're just going to increment

this by one

we're just counting how many of each

character we have and now

in our result we want to

add so for this particular count

we want to add we want to append

we're going to append this string s

so we want to group all anagrams with

this particular

count together now what if this count

does not exist

yet well i'm actually going to change

this dictionary to

a default dictionary or a default

hashmap

where the default value is a list so we

don't have to deal with one edge case

and also right now our count is a list

but we know in python lists cannot be

keys so we're actually just going to

change this to a tuple

because tuples are non-mutable

but mainly this is python stuff you

might not have to worry about this

in other languages and so that's

actually it

so now in our dictionary we have grouped

the anagrams together so we just can

take

result.values we don't want the keys

anymore we just want to return

the values the anagrams grouped together

so we can

return that and we're actually finished

so this

is the optimal m times n solution

where m is the number of strings we're

given and

n is the average length of each string

how many characters are

in each string so i hope this was

helpful if it was please like and

subscribe it supports the channel a lot

and i'll hopefully see you pretty soon