Group Anagrams - Categorize Strings by Count - Leetcode 49
Summary
TLDRIn 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
🔍 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.
🚀 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.
Mindmap
Keywords
💡Anagrams
💡Sorting
💡Time Complexity
💡Hash Map
💡Character Count
💡ASCII Value
💡Default Dictionary
💡Tuple
💡Big O Notation
💡Optimal Solution
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
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