5 Best Ways to Find the Length of the Longest Anagram Subsequence in Python

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πŸ’‘ Problem Formulation: Given a string, our goal is to find the length of the longest subsequence that is an anagram of another subsequence within the same string. For instance, if our input is “abba”, the longest anagram subsequence is “ab” (or “ba”), and therefore the desired output is 2.

Method 1: Counting and Comparison

This method involves counting the occurrence of each character in the string then comparing the character counts to determine the maximum length of two identical sets of characters — in other words, the longest anagram subsequence. It’s a straight-forward approach that’s easy to understand and implement in Python.

Here’s an example:

from collections import Counter

def longest_anagram_subsequence(s):
    count = Counter(s)
    return sum(min(count[char], count[char]) for char in count)

print(longest_anagram_subsequence("abba"))

Output: 2

This function uses Counter to tally up the characters. The sum iterates over the character counts, effectively adding up the length of the longest anagram subsequence.

Method 2: Using defaultdict

Similar to Method 1, but it utilizes the defaultdict for a cleaner and more efficient code, which is particularly useful when dealing with large strings or when performance is key.

Here’s an example:

from collections import defaultdict

def longest_anagram_subsequence(s):
    count = defaultdict(int)
    for char in s:
        count[char] += 1
    return sum(min(count[char], count[char]) for char in count)

print(longest_anagram_subsequence("abba"))

Output: 2

By using defaultdict, we streamline the process of counting characters, avoiding key errors and improving the readability of our code.

Method 3: Frequency Array Method

For strings containing only lower-case alphabets, an array of length 26 can be used to keep track of character frequencies. This method is efficient as it avoids overhead from dictionary or counter objects and might be faster for small strings.

Here’s an example:

def longest_anagram_subsequence(s):
    count = [0] * 26
    for char in s:
        count[ord(char) - ord('a')] += 1
    return sum(min(count[i], count[i]) for i in range(26))

print(longest_anagram_subsequence("abba"))

Output: 2

Here, we map characters to indices in a frequency array, using the ASCII value offset by ord('a'). This method trades generalizability for speed.

Method 4: Sorting and Grouping

This method involves sorting the string and then using itertools’ groupby to group identical characters together. While not as efficient as the previous methods, it’s a conceptually simple approach.

Here’s an example:

from itertools import groupby

def longest_anagram_subsequence(s):
    s_sorted = sorted(s)
    return sum(min(len(list(group)), len(list(group))) for _, group in groupby(s_sorted))

print(longest_anagram_subsequence("abba"))

Output: 2

The code sorts the string, then groups identical characters together. The sum is calculated similarly to previous methods. This method, while simple, can have a higher computational cost due to sorting.

Bonus One-Liner Method 5: Using Counter with List Comprehension

This technique boils down the logic of the previous methods into a single line of code, using a list comprehension combined with Counter. It’s a quick and succinct method ideal for ‘code golf’ or Python enthusiasts.

Here’s an example:

from collections import Counter

def longest_anagram_subsequence(s):
    return sum(min(count) for count in Counter(s).values())

print(longest_anagram_subsequence("abba"))

Output: 2

The function uses a comprehension to process the counts directly, resulting in a compact yet readable line of code that achieves the same result.

Summary/Discussion

  • Method 1: Counting and Comparison. Strengths: Straightforward logic with good performance. Weaknesses: Might not be the most efficient with very large strings.
  • Method 2: Using defaultdict. Strengths: Clean and efficient code, good for performance. Weaknesses: Not as explicit as some other methods.
  • Method 3: Frequency Array Method. Strengths: Highly efficient for strings with only lower-case alphabets. Weaknesses: Less flexible as it’s limited to certain string constraints.
  • Method 4: Sorting and Grouping. Strengths: Conceptually simple, making it easy to understand. Weaknesses: Can be inefficient due to sorting, especially with larger strings.
  • Bonus One-Liner Method 5: Using Counter with List Comprehension. Strengths: Concise and elegant. Weaknesses: Might sacrifice some readability for brevity.