Introduction

In this blog post, we will explore how to solve the LeetCode problem "1255" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a list of words, list of single letters (might be repeating) and score of every character. Return the maximum score of any valid set of words formed by using the given letters (words[i] cannot be used two or more times). It is not necessary to use all characters in letters and each letter can only be used once. Score of letters 'a', 'b', 'c', ... ,'z' is given by score[0], score[1], ... , score[25] respectively. Example 1: Input: words = ["dog","cat","dad","good"], letters = ["a","a","c","d","d","d","g","o","o"], score = [1,0,9,5,0,0,3,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0] Output: 23 Explanation: Score a=1, c=9, d=5, g=3, o=2 Given letters, we can form the words "dad" (5+1+5) and "good" (3+2+2+5) with a score of 23. Words "dad" and "dog" only get a score of 21. Example 2: Input: words = ["xxxz","ax","bx","cx"], letters = ["z","a","b","c","x","x","x"], score = [4,4,4,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,0,10] Output: 27 Explanation: Score a=4, b=4, c=4, x=5, z=10 Given letters, we can form the words "ax" (4+5), "bx" (4+5) and "cx" (4+5) with a score of 27. Word "xxxz" only get a score of 25. Example 3: Input: words = ["leetcode"], letters = ["l","e","t","c","o","d"], score = [0,0,1,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0] Output: 0 Explanation: Letter "e" can only be used once. Constraints: 1 <= words.length <= 14 1 <= words[i].length <= 15 1 <= letters.length <= 100 letters[i].length == 1 score.length == 26 0 <= score[i] <= 10 words[i], letters[i] contains only lower case English letters.

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• Approach:

  • Use recursion (backtracking) to explore all possible combinations of words.
  • Maintain a count of available letters and update it as words are chosen/un-chosen.
  • Calculate the score of each combination and keep track of the maximum score found so far.

• Complexity:

  • Runtime: O(2ⁿ * m), where n is the number of words and m is the total length of all words. The 2ⁿ comes from the number of possible subsets of words we can choose. For each subset, we need to check if we can form the words in the subset, which takes O(m) time.
  • Storage: O(n + 26), where n is the depth of the recursion (number of words) and 26 is to store letter counts.

Code

    def maxScoreWords(words, letters, score):
    """
    Calculates the maximum score of any valid set of words formed by using the given letters.

    Args:
        words (List[str]): A list of words.
        letters (List[str]): A list of letters.
        score (List[int]): A list of scores for each letter (a-z).

    Returns:
        int: The maximum score.
    """

    letter_counts = {}
    for letter in letters:
        letter_counts[letter] = letter_counts.get(letter, 0) + 1

    def calculate_word_score(word):
        word_score = 0
        for char in word:
            word_score += score[ord(char) - ord('a')]
        return word_score

    def backtrack(index, current_score, available_letters):
        if index == len(words):
            return current_score

        # Option 1: Don't include the current word
        max_score = backtrack(index + 1, current_score, available_letters)

        # Option 2: Include the current word (if possible)
        word = words[index]
        word_counts = {}
        can_form = True
        for char in word:
            word_counts[char] = word_counts.get(char, 0) + 1
            if char not in available_letters or word_counts[char] > available_letters[char]:
                can_form = False
                break

        if can_form:
            new_available_letters = available_letters.copy()
            for char in word:
                new_available_letters[char] -= 1

            word_score = calculate_word_score(word)
            max_score = max(max_score, backtrack(index + 1, current_score + word_score, new_available_letters))

        return max_score

    return backtrack(0, 0, letter_counts)