What is the time complexity of Number of Valid Words for Each Puzzle?

The optimal solution has a time complexity of O(p * 2^6 + n).

What data structures are used to solve Number of Valid Words for Each Puzzle?

This problem uses Array, Hash Table, String, Bit Manipulation, Trie. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Valid Words for Each Puzzle?

Insight 1: The limited size of puzzles (length 7) and words being lowercase letters suggests using bit manipulation to represent words and puzzles as integers where each bit corresponds to a letter. Insight 2: The core idea is to precompute a bitmask for each word and then, for each puzzle, efficiently check which words are valid by generating submasks of the puzzle's characters (excluding the first character) and combining them with the first character's mask.

Which companies ask LeetCode 1178. Number of Valid Words for Each Puzzle in interviews?

This problem is frequently asked at Dropbox.

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Number of Valid Words for Each Puzzle - LeetCode 1178 Solution

Number of Valid Words for Each Puzzle - Complete Solution Guide

Number of Valid Words for Each Puzzle is LeetCode problem 1178, 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

With respect to a given puzzle string, a word is valid if both the following conditions are satisfied: word contains the first letter of puzzle . For each letter in word , that letter is in puzzle . For example, if the puzzle is "abcdefg" , then valid words are "faced" , "cabbage" , and "baggage" , while invalid words are "beefed" (does not include 'a' ) and "based" (includes 's' which is not in the puzzle). Return an array answer , where answer[i] is the number of words in the given word list w

Detailed Explanation

The problem asks us to determine, for each puzzle in a given list of puzzles, how many words from a given word list are 'valid' with respect to that puzzle. A word is considered valid if it satisfies two conditions: 1) It contains the first letter of the puzzle, and 2) Every letter in the word is also present in the puzzle. Both words and puzzles consist of lowercase English letters. We need to return an array where each element at index 'i' represents the count of valid words for the puzzle at index 'i' in the puzzles list. Crucially, each puzzle has a length of 7 and contains no repeated characters.

Solution Approach

The solution employs a bit manipulation approach. First, each word in the `words` array is converted into a bitmask, where the i-th bit is set if the i-th letter of the alphabet ('a' + i) is present in the word. Then, a hash map (or equivalent data structure) is used to store the count of each unique word mask. For each puzzle, a similar bitmask is created, and all possible submasks of the puzzle's characters (excluding the first character) are generated. For each submask, the first character's bit is added to it to form a potential 'word mask'. If this 'word mask' exists in the hash map, the count of valid words for that puzzle is incremented. Finally, the counts for all puzzles are returned as a list.

Step-by-Step Algorithm

Step 1: Create a hash map `word_counts` to store the count of each unique word mask.
Step 2: Iterate through each word in `words`. For each word, create a bitmask representing the presence of each character in the word. Update `word_counts` with this mask and its count.
Step 3: Initialize an empty list `answer` to store the counts of valid words for each puzzle.
Step 4: Iterate through each puzzle in `puzzles`. For each puzzle:
Step 5: Extract the bitmask corresponding to the first character of the puzzle.
Step 6: Extract the bitmask corresponding to the remaining 6 characters of the puzzle.
Step 7: Initialize a count variable `count` to 0.
Step 8: Generate all possible submasks of the 6-character mask. This is done by iterating from the 6-character mask down to 0, and using the bitwise AND operator to ensure each submask only contains bits present in the original 6-character mask.
Step 9: For each submask, combine it with the first character's mask to create a potential word mask.
Step 10: If this potential word mask exists in `word_counts`, increment the `count` by the number of times the word mask occurs.
Step 11: Add the `count` to the `answer` list.
Step 12: Return the `answer` list.

Key Insights

Insight 1: The limited size of puzzles (length 7) and words being lowercase letters suggests using bit manipulation to represent words and puzzles as integers where each bit corresponds to a letter.
Insight 2: The core idea is to precompute a bitmask for each word and then, for each puzzle, efficiently check which words are valid by generating submasks of the puzzle's characters (excluding the first character) and combining them with the first character's mask.
Insight 3: Generating all possible submasks of a 6-character string (excluding the first letter) has a complexity of O(2^6), which is acceptable given the constraints.

Complexity Analysis

Time Complexity: O(p * 2^6 + n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, String, Bit Manipulation, Trie.

Companies

Asked at: Dropbox.

Number of Valid Words for Each Puzzle | LeetCode 1178

Hard

Array Hash Table String Bit Manipulation Trie

Time: O(p * 2^6 + n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public List<Integer> findNumOfValidWords(String[] words, String[] puzzles) {
        Map<Integer, Integer> wordCounts = new HashMap<>();
        for (String word : words) {
            int mask = 0;
            for (char c : word.toCharArray()) {
                mask |= (1 << (c - 'a'));
            }

            if (Integer.bitCount(mask) <= 7) {
                wordCounts.put(mask, wordCounts.getOrDefault(mask, 0) + 1);
            }
        }

        List<Integer> answer = new ArrayList<>();
        for (String puzzle : puzzles) {
            int firstCharMask = 1 << (puzzle.charAt(0) - 'a');

            int restMask = 0;
            for (int i = 1; i < 7; i++) {
                restMask |= (1 << (puzzle.charAt(i) - 'a'));
            }

            int count = 0;
            int submask = restMask;
            while (true) {
                int wordMask = submask | firstCharMask;
                if (wordCounts.containsKey(wordMask)) {
                    count += wordCounts.get(wordMask);
                }

                if (submask == 0) {
                    break;
                }
                submask = (submask - 1) & restMask;
            }

            answer.add(count);
        }

        return answer;
    }
}

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Problem Info

Description

Problem 1178: Number of Valid Words for Each Puzzle - Detailed explanation and solution approaches.

Topics

Array Hash Table String Bit Manipulation Trie

Companies

Dropbox

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1178

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

12 steps

1
Step 1: Create a hash map `word_counts` to store the count of each unique word mask.
2
Step 2: Iterate through each word in `words`. For each word, create a bitmask representing the presence of each character in the word. Update `word_counts` with this mask and its count.
3
Step 3: Initialize an empty list `answer` to store the counts of valid words for each puzzle.
4
Step 4: Iterate through each puzzle in `puzzles`. For each puzzle:
5
Step 5: Extract the bitmask corresponding to the first character of the puzzle.
6
Step 6: Extract the bitmask corresponding to the remaining 6 characters of the puzzle.
7
Step 7: Initialize a count variable `count` to 0.
8
Step 8: Generate all possible submasks of the 6-character mask. This is done by iterating from the 6-character mask down to 0, and using the bitwise AND operator to ensure each submask only contains bits present in the original 6-character mask.
9
Step 9: For each submask, combine it with the first character's mask to create a potential word mask.
10
Step 10: If this potential word mask exists in `word_counts`, increment the `count` by the number of times the word mask occurs.
11
Step 11: Add the `count` to the `answer` list.
12
Step 12: Return the `answer` list.

💡

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(p * 2^6 + n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from the fact that we iterate through all words once and then iterate through all puzzles. For each puzzle, we generate all possible submasks of its characters (excluding the first letter). Since puzzles have 7 characters, there are at most 2^6 submasks to generate for each puzzle. The space complexity is derived from the size of the hash map which will, in the worst case, store a count for each of the words.

📖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

Trie-based approach

Store words in a Trie data structure. For each puzzle, traverse the Trie and check if the words formed by the puzzle's characters are present. The first character of the puzzle must be included in the traversal.

Time: O(p * m), where p is the number of puzzles and m is the average length of words in the trie.Space: O(N * K), where N is the number of words and K is the average length of a word.

Trade-offs: The Trie approach may be faster for very long word lists with relatively short words if the Trie is well balanced. However, the bit manipulation approach has better constant factors and is likely faster in practice for the given constraints.

✨

Key Insights

1Insight 1: The limited size of puzzles (length 7) and words being lowercase letters suggests using bit manipulation to represent words and puzzles as integers where each bit corresponds to a letter.
2Insight 2: The core idea is to precompute a bitmask for each word and then, for each puzzle, efficiently check which words are valid by generating submasks of the puzzle's characters (excluding the first character) and combining them with the first character's mask.
3Insight 3: Generating all possible submasks of a 6-character string (excluding the first letter) has a complexity of O(2^6), which is acceptable given the constraints.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

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

Edge Cases

Edge case 1: Empty words or puzzles. The algorithm still works correctly because empty strings would result in a bitmask of 0, and the logic would handle it appropriately. However, the problem statement indicates words have a length between 4 and 50, and puzzles have length 7.
Edge case 2: Duplicate words. The `word_counts` counter handles duplicate words correctly, so they are counted multiple times if they are valid for a given puzzle.
Edge case 3: Words containing characters not in puzzles. The algorithm inherently filters words containing characters not in the puzzle, as the bitwise AND operation ensures that only characters present in both the word and the puzzle contribute to the validity.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to include the first character of the puzzle when checking for valid words. The word must contain the first character of the puzzle to be valid.

✗Common mistake 2: Not correctly generating submasks. The bitwise AND operation `(submask - 1) & rest_mask` is crucial to ensure that only characters present in the puzzle are included in the submasks.

✗Common mistake 3: Not handling duplicate words properly. The `Counter` object in Python and `HashMap` in Java correctly count the number of occurrences of each word mask. If a word appears multiple times, it should be counted multiple times if it's valid.

✅

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: Search suggestions or auto-completion features where you want to filter suggestions based on a limited set of characters (e.g., characters available on a broken keyboard).

🏦Real-world scenario 2: Games involving anagrams or word puzzles where you need to find valid words from a set of available letters. This problem is a simplified version of such scenarios.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

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: September 7, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Number of Valid Words for Each Puzzle"?

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 hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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