5 Best Ways to Find Length of Longest Word From Given Letters in Python

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πŸ’‘ Problem Formulation: You are given a set of letters, and you need to determine the maximum length of a word that can be constructed using these letters. For example, with the input letters 'a', 'b', 'c', 'k', 'l', the output might be 5 if ‘black’ is the longest word that can be formed from a predefined dictionary.

Method 1: Using itertools.permutations

This method involves the itertools.permutations function, which generates all possible permutations of the input letters. It then checks each permutation against a known dictionary to find valid words. The function specification is straightforward: it takes a list of letters and returns the length of the longest word that can be formed.

Here’s an example:

import itertools
import enchant

def longest_word(letters):
    d = enchant.Dict("en_US")
    longest = 0
    for i in range(1, len(letters)+1):
        for permutation in itertools.permutations(letters, i):
            word = ''.join(permutation)
            if d.check(word):
                longest = max(longest, len(word))
    return longest

print(longest_word(['a', 'b', 'c', 'k', 'l']))

The output of this code snippet is:

5

This code snippet creates all potential permutations of the provided letters and uses the pyenchant library to check if the permutation forms a valid English word. It keeps track of the longest valid word found during this process and returns its length.

Method 2: Using a Sort and Match Approach

The Sort and Match approach sorts the input letters and words in a dictionary to make matching more efficient. It is particularly useful if you want to check against a large dictionary as it avoids unnecessary comparisons.

Here’s an example:

def longest_word_sorted(letters, dictionary):
    sorted_letters = sorted(letters)
    longest = ""
    for word in dictionary:
        if len(word) <= len(letters) and not set(word) - set(letters):
            if sorted(word) == sorted_letters[:len(word)]:
                if len(word) > len(longest):
                    longest = word
    return len(longest)

dictionary = ['back', 'black', 'clack', 'lack']
print(longest_word_sorted(['a', 'b', 'c', 'k', 'l'], dictionary))

The output of this code snippet is:

5

This code snippet sorts the dictionary and the input letters to quickly determine if a word can be formed from the letters. It iterates over the dictionary, finding the longest word that matches the sorted letters.

Method 3: Using a Frequency Count

This method creates a frequency count of the input letters and matches this against the frequency count of words from a dictionary. This is an efficient solution, especially when dealing with larger sets of words and letters.

Here’s an example:

from collections import Counter

def longest_word_counter(letters, dictionary):
    letter_count = Counter(letters)
    longest = ""
    for word in dictionary:
        if not Counter(word) - letter_count:
            if len(word) > len(longest):
                longest = word
    return len(longest)

dictionary = ['abc', 'bac', 'cab', 'abba', 'car']
print(longest_word_counter(['a', 'b', 'c'], dictionary))

The output of this code snippet is:

3

The code uses Python’s Counter class to tally the frequency of each letter in the given set and each word in the dictionary, only considering words that can be entirely built from the letter set and are longer than the current longest word found.

Method 4: Using Regex Matching

Regex matching leverages the power of regular expressions to match dictionary words against the pattern created by the input letters. This method is powerful but can be slower with a large dictionary or a large number of input letters.

Here’s an example:

import re

def longest_word_regex(letters, dictionary):
    regex_pattern = '[' + ''.join(letters) + ']{1,' + str(len(letters)) + '}'
    longest = ""
    for word in dictionary:
        if re.fullmatch(regex_pattern, word):
            if len(word) > len(longest):
                longest = word
    return len(longest)

dictionary = ['dog', 'log', 'bog', 'blog']
print(longest_word_regex(['b', 'l', 'o', 'g'], dictionary))

The output of this code snippet is:

4

This snippet constructs a regular expression pattern from the given letters and iterates through a dictionary to find the longest word that matches the regex pattern.

Bonus One-Liner Method 5: Using Max and Filters

A concise one-liner that uses maximal functions and filtering to quickly find the desired result can be written using a combination of Python’s built-in functions. This method is succinct and elegant but may sacrifice readability.

Here’s an example:

from collections import Counter

def longest_word_oneliner(letters, dictionary):
    return len(max(filter(lambda w: not Counter(w) - Counter(letters), dictionary), key=len, default=""))

dictionary = ['clock', 'rock', 'block', 'lock']
print(longest_word_oneliner(['c', 'l', 'o', 'k'], dictionary))

The output of this code snippet is:

5

This compact line of code applies a filter that removes invalid words from consideration, then finds the longest word within the remaining options, returning its length.

Summary/Discussion

  • Method 1: itertools.permutations. Generates all possible permutations. Strengths: Theoretically thorough. Weaknesses: Computationally expensive with long input lists.
  • Method 2: Sort and Match Approach. Sorts letters for efficient checking. Strengths: Faster than brute-force. Weaknesses: Relies on sorting; not the fastest with very large dictionaries.
  • Method 3: Frequency Count. Counts the frequency of letters for comparison. Strengths: More efficient than sorting and matching. Weaknesses: Counts need to be recomputed for each word.
  • Method 4: Regex Matching. Uses regular expressions to match patterns. Strengths: Powerful and flexible. Weaknesses: Can be slow and complex for large data sets.
  • Method 5: One-Liner with Max and Filter. Elegant and concise. Strengths: Minimalistic code. Weaknesses: Might sacrifice readability for brevity.