How do you solve Maximum Score Words Formed by Letters on LeetCode?

Maximum Score Words Formed by Letters (LeetCode #1255) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n) space complexity. Key insight: Use backtracking to explore combinations efficiently.

What is the brute force solution for Maximum Score Words Formed by Letters?

The brute force approach for Maximum Score Words Formed by Letters is Brute Force, with O(2^n * m) time complexity and O(n) space complexity. We can generate all possible subsets of words and calculate their scores based on the available letters. This approach is straightforward but inefficient as it checks every combination. The optimal Optimal Solution approach improves this to O(n * m).

What are the interview tips for Maximum Score Words Formed by Letters?

Explain your thought process clearly. Discuss edge cases, such as empty letters or words. Practice writing recursive functions to build confidence.

What are common mistakes when solving Maximum Score Words Formed by Letters?

Not checking letter availability before including a word. Forgetting to reset letter counts after backtracking.

#1255

Maximum Score Words Formed by Letters

Hard

ArrayHash TableStringDynamic ProgrammingBacktrackingBit ManipulationCountingBitmaskHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^n * m)	O(n * m)
Space	O(n)	O(n)

💡

Intuition

Time O(n * m)Space O(n)

We can use a backtracking approach with pruning to explore valid combinations of words while keeping track of the letter counts. This avoids unnecessary calculations and improves efficiency.

⚙️

Algorithm

4 steps

1Step 1: Create a frequency map of the available letters.
2Step 2: Use a recursive function to explore each word and decide whether to include it or not.
3Step 3: If including a word exceeds the available letters, skip that branch.
4Step 4: Keep track of the maximum score during the recursion.

solution.py24 lines

1# Full working Python code
2from collections import Counter
3
4def maxScoreWords(words, letters, score):
5    letter_count = Counter(letters)
6    max_score = [0]
7
8    def backtrack(index, current_score):
9        if index == len(words):
10            max_score[0] = max(max_score[0], current_score)
11            return
12        # Skip the current word
13        backtrack(index + 1, current_score)
14        # Include the current word
15        word_count = Counter(words[index])
16        if all(word_count[ch] <= letter_count[ch] for ch in word_count):
17            for ch in word_count:
18                letter_count[ch] -= word_count[ch]
19            backtrack(index + 1, current_score + sum(score[ord(ch) - ord('a')] * word_count[ch] for ch in word_count))
20            for ch in word_count:
21                letter_count[ch] += word_count[ch]
22
23    backtrack(0, 0)
24    return max_score[0]

ℹ

Complexity note: The time complexity is O(n * m) where n is the number of words and m is the average length of words. The space complexity is O(n) for storing letter counts.

1Use backtracking to explore combinations efficiently.
2Prune branches that exceed letter limits early.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.