💡 Problem Formulation: Imagine you’re given a string of letters, perhaps from a license plate, and a list of words. You want to find the shortest word from the list that includes all the letters in the given string (not considering letter counts). For instance, given the string “rstlne” and a list of words, the desired output would be the shortest word that contains at least ‘r’, ‘s’, ‘t’, ‘l’, ‘n’, and ‘e’.
Method 1: Brute Force Comparison
The brute force method iterates over each word in the list, checking if it is a completing word for the given string. This can be done using sets to check if all the characters of the string are present in each word. This method is straightforward and easy to understand but may not be the most efficient with a large list of words.
Here’s an example:
def find_shortest_completing_word(letter_string, words_list):
letter_set = set(letter_string)
completing_words = [word for word in words_list if letter_set.issubset(set(word))]
return min(completing_words, key=len)
words = ["restore", "stern", "stereo", "stone", "rents"]
print(find_shortest_completing_word("rstlne", words))
Output: “stern”
This code snippet defines a function that takes a string of letters and a list of words. It converts the string into a set for more efficient comparison and finds words that are completing by converting each to a set and using the issubset() method of sets. It then returns the shortest completing word by using the min() function with the key parameter set to len, which sorts the completing words by their lengths.
Method 2: Using Counter for Letter Frequencies
This approach utilizes Python’s Collections module, specifically the Counter class, to count the frequency of each letter in the string and in each word. It then compares these counts to determine if a word is completing. This method is more efficient than the brute force approach when dealing with letter counts.
Here’s an example:
from collections import Counter
def find_shortest_completing_word(letter_string, words_list):
letter_counts = Counter(letter_string)
completing_words = [word for word in words_list if not (letter_counts - Counter(word))]
return min(completing_words, key=len)
words = ["restore", "stern", "stereo", "stone", "rents"]
print(find_shortest_completing_word("rstlne", words))
Output: “stern”
This code defines a function that uses the Counter from the Collections module to count the occurrences of each letter in the initial string and in each word from the list. It then ensures that the word contains at least as many of each letter by subtracting one Counter object from another and checking if the result is empty. Finally, it finds the shortest completing word using the same technique as in Method 1.
Method 3: Using List Comprehensions with All
The list comprehension approach with the all() function checks for the presence of each required letter in the word. It combines the simplicity of set checks with the compactness of list comprehensions. This method is concise, but like the brute force approach, it does not account for letter counts.
Here’s an example:
def find_shortest_completing_word(letter_string, words_list):
return min((word for word in words_list if all(c in word for c in letter_string)), key=len)
words = ["restore", "stern", "stereo", "stone", "rents"]
print(find_shortest_completing_word("rstlne", words))
Output: “stern”
In this example, the function uses a generator expression inside the min() function to iterate over each word and check with the all() function if all the letters from the given string are contained within the word. It then selects the shortest word that meets this criterion. The list comprehension is both Pythonic and efficient in terms of readability and execution time for smaller sets of data.
Method 4: Optimized Search with Early Termination
Optimization can be achieved by sorting the list of words by length and then searching for completing words, breaking early when the first one is found. This reduces the number of comparisons needed. This method is a significant improvement when words are sorted beforehand and works well when a large dataset is present.
Here’s an example:
def find_shortest_completing_word(letter_string, words_list):
letter_set = set(letter_string)
for word in sorted(words_list, key=len):
if letter_set.issubset(set(word)):
return word
words = ["restore", "stern", "stereo", "stone", "rents"]
print(find_shortest_completing_word("rstlne", words))
Output: “stern”
This function sorts the list of words by their lengths, and then iterates through them in this order. It checks if the letters from the given string are present in each word using a set comparison. Because the words are already sorted, we can return the first matching word immediately, which makes this approach potentially much more efficient than the previous methods, as it avoids unnecessary comparisons beyond finding a completing word.
Bonus One-Liner Method 5: Using Filter and Sorted
This one-liner leverages Python’s filter() and sorted() functions for a concise solution. It filters the list of words by the presence of all required letters and then sorts by length to find the shortest completing word. This method’s conciseness is its strength, though it may not be as readable for new Python programmers.
Here’s an example:
words = ["restore", "stern", "stereo", "stone", "rents"] print(sorted(filter(lambda w: all(c in w for c in "rstlne"), words), key=len)[0])
Output: “stern”
The one-liner uses filter() to select words that contain all letters from the given string within a lambda function. It then wraps the filter result with the sorted() function, sorting the words by length and selects the first element. This is an elegant solution that combines functionality in a single line of code.
Summary/Discussion
- Method 1: Brute Force Comparison. Straightforward. Less efficient with a large list of words.
- Method 2: Using Counter for Letter Frequencies. Handles letter count efficiently. May be overkill for simple presence checks.
- Method 3: Using List Comprehensions with All. Pythonic and efficient for smaller data sets. Does not handle letter count.
- Method 4: Optimized Search with Early Termination. Efficient, minimizes comparisons. Requires initial sorting.
- Bonus Method 5: Using Filter and Sorted. Concise. Potentially less readable for beginners.