What is the time complexity of Maximum Score Words Formed by Letters?

The optimal solution has a time complexity of O(2^n).

What data structures are used to solve Maximum Score Words Formed by Letters?

This problem uses Array, String, Dynamic Programming, Backtracking, Bit Manipulation, Bitmask. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Score Words Formed by Letters?

Insight 1: Backtracking is suitable because we need to explore all possible combinations of words to find the maximum score. Insight 2: Efficiently checking if a word can be formed with available letters is crucial. Using a frequency counter for letters and words helps in quick validation.

Which companies ask LeetCode 1255. Maximum Score Words Formed by Letters in interviews?

This problem is frequently asked at Infosys.

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Maximum Score Words Formed by Letters - LeetCode 1255 Solution

Maximum Score Words Formed by Letters - Complete Solution Guide

Maximum Score Words Formed by Letters is LeetCode problem 1255, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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","

Detailed Explanation

The problem asks us to find the maximum score achievable by forming words from a given list of words using a limited set of letters, where each letter has a score. We can only use each letter as many times as it appears in the input `letters` list, and each word can be used at most once. The score of a word is the sum of the scores of its letters. The goal is to find the subset of words that yields the highest possible total score while respecting letter availability.

Solution Approach

The solution uses a backtracking algorithm to explore all possible combinations of words. It first creates a frequency counter for the available letters. Then, it iterates through the words and filters those that can be formed using available letters. For each word, it has two choices: either include the word in the current combination or skip it. The backtracking function recursively explores these choices, updating the score and letter counts accordingly. It maintains the maximum score encountered so far and returns it at the end.

Step-by-Step Algorithm

Step 1: Initialize a letter frequency counter (dictionary/map/array) from the input `letters` list.
Step 2: Filter `words` to create `usable_words_data` which stores the words that can be formed with the available letters, along with their letter counts and scores.
Step 3: Define a recursive backtracking function that takes the current index in `usable_words_data` and the current score as input.
Step 4: Base Case: If the index reaches the end of `usable_words_data`, update the maximum score if the current score is greater.
Step 5: Recursive Step: For each word at the current index, explore two choices:
Step 6: Choice 1: Skip the current word and recursively call the backtracking function with the next index and the same current score.
Step 7: Choice 2: Try to include the current word if it can be formed with the remaining letters.
Step 8: Check if the current word can be formed using the current letter counts.
Step 9: If the word can be formed, update the letter counts by subtracting the word's letter counts.
Step 10: Recursively call the backtracking function with the next index and the updated score.
Step 11: After the recursive call, restore the letter counts to their original values (backtracking step).
Step 12: Call the backtracking function starting from index 0 and a score of 0.
Step 13: Return the maximum score obtained during the backtracking process.

Key Insights

Insight 1: Backtracking is suitable because we need to explore all possible combinations of words to find the maximum score.
Insight 2: Efficiently checking if a word can be formed with available letters is crucial. Using a frequency counter for letters and words helps in quick validation.
Insight 3: Representing letter counts using dictionaries or arrays allows for efficient lookups and updates during the backtracking process.
Insight 4: Pre-filtering the list of words by checking if they can be formed using the available letters before the backtracking process to reduce the number of branches explored

Complexity Analysis

Time Complexity: O(2^n)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Dynamic Programming, Backtracking, Bit Manipulation, Bitmask.

Companies

Asked at: Infosys.

Maximum Score Words Formed by Letters | LeetCode 1255

Hard

Array String Dynamic Programming Backtracking Bit Manipulation Bitmask

Time: O(2^n)Space: O(n)

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

class Solution {
    public int maxScoreWords(String[] words, char[] letters, int[] score) {
        Map<Character, Integer> letterCounts = new HashMap<>();
        for (char letter : letters) {
            letterCounts.put(letter, letterCounts.getOrDefault(letter, 0) + 1);
        }

        class WordData {
            Map<Character, Integer> counts;
            int score;

            public WordData(Map<Character, Integer> counts, int score) {
                this.counts = counts;
                this.score = score;
            }
        }

        java.util.List<WordData> usableWordsData = new java.util.ArrayList<>();
        for (String word : words) {
            Map<Character, Integer> wordCount = new HashMap<>();
            for (char c : word.toCharArray()) {
                wordCount.put(c, wordCount.getOrDefault(c, 0) + 1);
            }

            boolean canEverForm = true;
            for (Map.Entry<Character, Integer> entry : wordCount.entrySet()) {
                char character = entry.getKey();
                int count = entry.getValue();
                if (letterCounts.getOrDefault(character, 0) < count) {
                    canEverForm = false;
                    break;
                }
            }

            if (canEverForm) {
                int wordScore = 0;
                for (char c : word.toCharArray()) {
                    wordScore += score[c - 'a'];
                }
                usableWordsData.add(new WordData(wordCount, wordScore));
            }
        }

        int n = usableWordsData.size();
        int maxScore = 0;

        class BacktrackHelper {
            public void backtrack(int index, int currentScore, Map<Character, Integer> currentLetterCounts) {
                if (index == n) {
                    maxScore = Math.max(maxScore, currentScore);
                    return;
                }

                // Choice 1: Skip usableWordsData[index]
                backtrack(index + 1, currentScore, new HashMap<>(currentLetterCounts));

                // Choice 2: Try to include usableWordsData[index]
                WordData wordInfo = usableWordsData.get(index);
                Map<Character, Integer> wCount = wordInfo.counts;

                boolean canForm = true;
                for (Map.Entry<Character, Integer> entry : wCount.entrySet()) {
                    char character = entry.getKey();
                    int count = entry.getValue();
                    if (currentLetterCounts.getOrDefault(character, 0) < count) {
                        canForm = false;
                        break;
                    }
                }

                if (canForm) {
                    Map<Character, Integer> nextLetterCounts = new HashMap<>(currentLetterCounts);
                    for (Map.Entry<Character, Integer> entry : wCount.entrySet()) {
                        char character = entry.getKey();
                        int count = entry.getValue();
                        nextLetterCounts.put(character, nextLetterCounts.get(character) - count);
                    }
                   
                    backtrack(index + 1, currentScore + wordInfo.score, nextLetterCounts);

                }
            }
        }

        BacktrackHelper helper = new BacktrackHelper();
        helper.backtrack(0, 0, new HashMap<>(letterCounts));
        return maxScore;
    }
}

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Problem Info

Description

Problem 1255: Maximum Score Words Formed by Letters - Detailed explanation and solution approaches.

Topics

Array String Dynamic Programming Backtracking Bit Manipulation Bitmask

Companies

Infosys

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1255

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

13 steps

1
Step 1: Initialize a letter frequency counter (dictionary/map/array) from the input `letters` list.
2
Step 2: Filter `words` to create `usable_words_data` which stores the words that can be formed with the available letters, along with their letter counts and scores.
3
Step 3: Define a recursive backtracking function that takes the current index in `usable_words_data` and the current score as input.
4
Step 4: Base Case: If the index reaches the end of `usable_words_data`, update the maximum score if the current score is greater.
5
Step 5: Recursive Step: For each word at the current index, explore two choices:
6
Step 6: Choice 1: Skip the current word and recursively call the backtracking function with the next index and the same current score.
7
Step 7: Choice 2: Try to include the current word if it can be formed with the remaining letters.
8
Step 8: Check if the current word can be formed using the current letter counts.
9
Step 9: If the word can be formed, update the letter counts by subtracting the word's letter counts.
10
Step 10: Recursively call the backtracking function with the next index and the updated score.
11
Step 11: After the recursive call, restore the letter counts to their original values (backtracking step).
12
Step 12: Call the backtracking function starting from index 0 and a score of 0.
13
Step 13: Return the maximum score obtained during the backtracking process.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(2^n)

🐌 Exponential - Very slow! Often needs optimization.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is exponential because the backtracking algorithm explores all possible combinations of words (include or exclude). For each combination, the algorithm needs to verify if the word can be made of available letters which could potentially go through each character of a word, so m is included. The space complexity is dominated by the recursion depth, which is at most the number of words if we choose all words. We also use auxiliary space to store the counts of the letters.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming with Bitmasking

We can use dynamic programming with bitmasking to solve this problem. Each bit in the bitmask represents whether a word has been used or not. We iterate through all possible bitmasks, checking the validity of the word selection and calculating the score.

Time: O(2^n * m), where n is the number of words, and m is the average word length.Space: O(2^n), where n is the number of words.

Trade-offs: DP avoids redundant calculations compared to backtracking, but it has a higher space complexity. DP can be more efficient when memory is not a concern and the number of words is small enough. Backtracking is preferred when space is limited.

✨

Key Insights

1Insight 1: Backtracking is suitable because we need to explore all possible combinations of words to find the maximum score.
2Insight 2: Efficiently checking if a word can be formed with available letters is crucial. Using a frequency counter for letters and words helps in quick validation.
3Insight 3: Representing letter counts using dictionaries or arrays allows for efficient lookups and updates during the backtracking process.
4Insight 4: Pre-filtering the list of words by checking if they can be formed using the available letters before the backtracking process to reduce the number of branches explored

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: `words` is empty. The algorithm correctly handles this by returning 0 since the loop over words won't execute.
Edge case 2: `letters` is empty. If there are no available letters, only words formed from no letters (empty words) or no words will return a non-zero result. The filter logic ensures that no word will ever be chosen if the letters are unavailable.
Edge case 3: `score` is all zeros. All word scores will become zero, so the maximum possible score will be zero, which the solution will correctly return.
Edge case 4: A very large input will lead to a stack overflow issue since this is a recursive solution. Iterative DFS with an explicit stack to manage recursion could potentially mitigate stack overflow issues.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to restore the letter counts after exploring the 'include' choice in the backtracking step. This leads to incorrect results as the available letters are not correctly tracked.

✗Common mistake 2: Incorrectly calculating the score of a word, or not using the given `score` array at all.

✗Common mistake 3: Failing to handle edge cases such as an empty `words` list or an empty `letters` list.

✗Common mistake 4: Not realizing the exponential time complexity of backtracking and thus, not being able to anticipate the performance for large input sizes. Understanding and being able to communicate about exponential algorithms is essential.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing resource allocation. For example, given a set of tasks (words) with different resource requirements (letters) and values (scores), find the combination of tasks that maximizes the total value without exceeding resource constraints.

🏦Real-world scenario 2: Combinatorial optimization in manufacturing. Suppose you want to build different products (words) using available components (letters) where each component has a cost (negative score). Find the combination of products to build that minimizes the overall cost while respecting available component inventories.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Score Words Formed by Letters"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?