Solution Explanation: Finding the Shortest Completing Word

This problem involves finding the shortest word from a given list that contains all the letters present in a license plate, respecting letter case and frequency. The solution employs a counting approach for efficient comparison.

Approach

1. License Plate Processing:

   • We first process the licensePlate string to count the frequency of each letter (ignoring case and non-alphabetic characters). A hash map (or array for simplicity) is ideal for this; we'll use an array indexed by the letter's position in the alphabet (a=0, b=1,...).

2. Word Iteration:

   • The code iterates through each word in the words list.

3. Word Frequency Counting:

   • For each word, we count the letter frequencies similarly to step 1.

4. Comparison and Shortest Word Selection:

   • We compare the word's letter frequencies with the licensePlate's frequencies. A word is a "completing word" only if its frequency for each letter is greater than or equal to the frequency in the licensePlate.
   • If a word is a completing word, we check its length against the current shortest completing word found so far. We update the shortest completing word if a shorter one is found.

5. Return Value:

   • The function returns the shortest completing word.

Time and Space Complexity

• Time Complexity: O(mn), where 'm' is the length of the words array, and 'n' is the maximum length of a word in words. This is because we iterate through each word and count its letter frequencies which is a linear process based on the word's length. The initial processing of the licensePlate is O(k), where k is the length of licensePlate, which is dominated by O(mn).
• Space Complexity: O(1). We use arrays of size 26 (to store letter counts) which takes constant space irrespective of input size.

Code (Python)

from collections import Counter
 
def shortestCompletingWord(licensePlate, words):
    plate_counts = Counter(c.lower() for c in licensePlate if c.isalpha())  # Count license plate letter frequencies
    shortest_word = ""
    for word in words:
        word_counts = Counter(word)
        if all(word_counts[char] >= plate_counts[char] for char in plate_counts): # Check if word completes
            if not shortest_word or len(word) < len(shortest_word):
                shortest_word = word
    return shortest_word

The code in other languages (Java, C++, Go, TypeScript, Rust) follows a similar structure, adapting the syntax and data structures accordingly. The core algorithm remains consistent across all implementations.