1255. Maximum Score Words Formed by Letters

Problem Description

The given problem is an optimization challenge where we have to find the maximum score achievable by forming valid sets of words from a given list. Each word has an associated score based on the characters it contains and a score table provided for the alphabet. The words come from a list called words, and we are given a list of available single letters which may contain duplicates. Each letter in letters can be used only once. The score for each letter is given by an array score, where score[0] corresponds to the score of 'a', score[1] to the score of 'b', and so on unto score[25], which corresponds to the score of 'z'.

The goal is to select words from the words list in such a way that we can form them from the letters in the letters list without reusing any single letter more than its available count. We cannot use any word more than once. We should return the highest possible score that can be obtained from any combination of the words formed.

Intuition

To solve the problem, we think of all possible subsets of the provided words array. For each subset, we calculate the score of its constituent words if it's possible to create the subset with the given letters. We only consider a subset valid if each letter required to form this subset of words is available the required number of times in the letters list. We use bitwise operations to generate possible subsets.

Generating subsets: We use a binary representation of numbers to iterate over all possible subsets. For a list of n words, there are 2^n possible subsets. We can represent the subsets using a bitmask of length n, where the j-th bit indicates whether or not to include the j-th word in the subset.

Counting letters in words: We use a data structure, Counter, to count the occurrences of each letter in both the currently considered subset of words and the available letters.

Checking validity and score calculation: For a chosen subset, we check if it can be created with the given letters by ensuring the count of each letter in the subset does not exceed the count of available letters. If valid, we calculate the subset's score by summing the individual character scores. This is done by referring to the supplied score array using the character's ASCII value adjusted with the ASCII value of 'a' to correctly index into the score list.

Maximizing the score: We compare each valid subset's score to the current known maximum score and update the maximum score if a higher one is found.

Optimization: Although the proposed solution methodically generates all potential subsets and assesses each one, it is not efficient for large lists of words due to its exponential time complexity. However, this brute-force approach provides a direct and complete search of the solution space, which is acceptable for smaller inputs.

Learn more about Dynamic Programming, Backtracking and Bitmask patterns.

Solution Approach

The implementation of the solution can be explained step by step as follows:

1. Initialization: A Counter object named cnt is created to store the count of each available letter in the letters list. Next, the length of the words list is stored in variable n, and a variable ans is initialized to zero to keep track of the maximum score.

2. Iterating over subsets: A for loop is used to iterate through numbers 0 to 2^n - 1 representing all possible subsets where n is the number of words. Inside the loop, each number represents a subset where each bit in the binary representation of the number indicates the presence (1) or absence (0) of a word at that index in the subset.

3. Building a subset: For each number i in the loop, we use a list comprehension to collect words corresponding to set bits in the binary representation of the number. This is done with [words[j] for j in range(n) if i >> j & 1]. Then, a Counter object named cur is created to store the count of each letter in this subset of words.

4. Checking if the subset is valid: The all function is used to verify that for all characters c and their count v in the subset cur, there is a sufficient number of letters in cnt (v <= cnt[c]). This ensures that the current subset can be formed from the available letters without reusing any letter more than available.

5. Calculating the score of the subset: If the subset is valid, the score is computed using the sum function. For each character c and its count v in cur, the expression v * score[ord(c) - ord('a')] calculates the total score of the character in the subset by multiplying its count with its respective score in score. The ord function gets the ASCII value of the character, and subtracting ord('a') gives the index of the score for that character in the score array.

6. Updating the maximum score: The score of the current valid subset t is compared with the current ans, and if t is greater, ans is updated to t.

7. Returning the result: After the loop finishes executing, the final value of ans is returned, which contains the maximum score obtained from any valid subset of words.

The solution uses a bitmasking technique to generate all possible subsets of words and a Counter data structure to efficiently count the occurrences of letters in both the available letters and each subset of words. The brute-force approach checks each subset for validity and calculates its score independently. Despite its simplicity, the time complexity of this solution is O(2^n * (m + k)), where n is the number of words, m is the length of the longest word, and k is the number of unique characters in letters, which may not be performant for large datasets.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach:

Assume we have the following inputs:

• words = ["dog", "cat", "dad", "good"]
• letters = ["a", "a", "c", "d", "d", "g", "o", "o"]
• score = [1, 0, ..., 2, ..., 0, ..., 1, ..., 0, 2] // score for "a" is 1, "c" is 2, "d" is 1, "g" is 2, "o" is 1

The score array only shows the scores for the relevant characters for brevity.

1. Initialization: We calculate the count of each letter in the letters list. The counter would look like cnt = {'a': 2, 'c': 1, 'd': 2, 'g': 1, 'o': 2}. The number of words n = 4, and the starting maximum score ans = 0.

2. Iterating over subsets: We generate numbers from 0 to 2^n - 1 which is from 0 to 15 for n = 4. Each representation will be from 0000 to 1111 in binary, corresponding to taking none or all the words, respectively.

3. Building a subset: When the loop is at i = 3 (binary 0011), it indicates including words[0] and words[1] which are "dog" and "cat". We determine the count of letters in the subset, cur = Counter("dogcat") = {'d': 1, 'o': 1, 'g': 1, 'c': 1, 'a': 1}.

4. Checking if the subset is valid: We compare cur against cnt and find that all letters in cur are less than or equal to those in cnt ({'d': 1, 'o': 1, 'g': 1, 'c': 1, 'a': 1} vs. {'a': 2, 'c': 1, 'd': 2, 'g': 1, 'o': 2}). Therefore, the subset is valid.

5. Calculating the score of the subset: The score for the subset "dogcat" is calculated as follows:

   • "dog" contributes 1*score[3] + 1*score[14] + 1*score[6] = 1*1 + 1*1 + 1*2 = 4 because score[3] for 'd', score[14] for 'o', and score[6] for 'g'.
   • "cat" contributes 1*score[2] + 1*score[0] + 1*score[19] = 1*2 + 1*1 + 1*0 = 3 because score[2] for 'c', score[0] for 'a'.
   • Total subset score is 4 + 3 = 7.

6. Updating the maximum score: Since 7 is greater than 0, we update ans to 7.

7. Continuing: The loop continues, testing the validity of other subsets and updating ans as necessary. After testing all possible subsets, the highest score found is returned. For instance, if the algorithm later encounters the subset "good" with a higher score than 7, ans will be updated accordingly.

8. Returning the result: After completion, suppose the highest score was achieved by the subset "good" and it was 8. Then, ans will be 8, which is the result returned by the function.

This walkthrough exemplifies the steps and considerations of the solution approach on a smaller scale of inputs, demonstrating how the brute-force method selects and sums up word scores to find the maximum possible score given the constraints.

Solution Implementation

Time and Space Complexity

The given code snippet is a solution to the problem of finding the maximum score from forming words using given letters with each letter having a score.

Time Complexity:

To analyze the time complexity, let's consider n to be the number of words and l to be the total number of letters across all words.

1. The code uses a bit mask to generate all possible combinations of words, which results in 2^n combinations as it iterates over a range of size 2^n.
2. For each combination, it attempts to create a Counter object for the selected words. This operation is linear with respect to the length of the concatenated string of chosen words. In the worst case, this could be O(l) when all words are chosen.
3. The all function inside the loop checks if all letters in the current combination are less than or equal to the count of available letters. This runs in O(26) since there are 26 lowercase English letters, in the worst case where every letter is used.
4. The sum function, which calculates the total score of current combination of words, will iterate over each character in the concatenated string. This is O(l) in the worst case for the same reason as step 2.
5. Updating the maximum answer ans is O(1) for each of the 2^n combinations.

Combining these steps, the worst-case time complexity is O(2^n * (l + 26 + l + 1)), which simplifies to O(2^n * l) as l dominates the constant 26 and 1.

Space Complexity:

Now, let's look at the space complexity:

1. The counter cnt stores the frequency of each letter available. This is at most O(26), which is a constant O(1) because we only have 26 letters.
2. The cur counter inside the loop could store at most O(26) entries due to the lowercase letters, so this is also O(1).
3. The list [words[j] for j in range(n) if i >> j & 1] is ephemeral and its space complexity in terms of the number of characters it can hold is up to O(l), when all words are selected.
4. The combination bitmask takes up to O(n) space to store the indices of words.

Considering all the above, the space complexity of the code is dominated by the space needed for the ephemeral list of selected words and the bitmask, and hence it is O(l + n).

To summarize, the time complexity is O(2^n * l), and the space complexity is O(l + n).

Learn more about how to find time and space complexity quickly using problem constraints.