How do you solve Number of Valid Words for Each Puzzle on LeetCode?

Number of Valid Words for Each Puzzle (LeetCode #1178) 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: Using bit manipulation can significantly speed up checks for character presence.

What is the brute force solution for Number of Valid Words for Each Puzzle?

The brute force approach for Number of Valid Words for Each Puzzle is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each word against each puzzle to see if it meets the validity criteria. This is straightforward but inefficient, as it requires checking every character in the word against the characters in the puzzle. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Number of Valid Words for Each Puzzle?

Number of Valid Words for Each Puzzle uses the following concepts: Array, Hash Table, String, Bit Manipulation, Trie. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Valid Words for Each Puzzle?

Clearly explain your thought process while coding. Consider edge cases, such as empty words or puzzles. Practice bit manipulation problems to become comfortable with this technique.

What are common mistakes when solving Number of Valid Words for Each Puzzle?

Not checking if the first letter of the puzzle is present in the word. Overlooking the need to ensure all characters in the word are within the puzzle.

#1178

Number of Valid Words for Each Puzzle

Hard

ArrayHash TableStringBit ManipulationTrieHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n)

💡

Intuition

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

The optimal solution uses bit manipulation to represent words and puzzles, allowing for quick checks of character presence. This drastically reduces the time complexity by leveraging the small size of the puzzles.

⚙️

Algorithm

3 steps

1Step 1: Create a bitmask for each word and puzzle, where each bit represents whether a character is present.
2Step 2: For each puzzle, check the first letter's bit and then use bitwise operations to validate the words.
3Step 3: Count valid words for each puzzle based on the bitmask comparisons.

solution.py18 lines

1def findNumOfValidWords(words, puzzles):
2    from collections import defaultdict
3    def bitmask(word):
4        mask = 0
5        for char in word:
6            mask |= 1 << (ord(char) - ord('a'))
7        return mask
8    word_masks = [bitmask(word) for word in words]
9    result = []
10    for puzzle in puzzles:
11        first_letter_mask = 1 << (ord(puzzle[0]) - ord('a'))
12        puzzle_mask = bitmask(puzzle)
13        count = 0
14        for word_mask in word_masks:
15            if (word_mask & first_letter_mask) and (word_mask & puzzle_mask) == word_mask:
16                count += 1
17        result.append(count)
18    return result

ℹ

Complexity note: The time complexity is O(n * m) where n is the number of words and m is the number of puzzles. Each word is processed once to create its bitmask, and each puzzle checks against all word masks. The space complexity is O(n) for storing the word masks.

1Using bit manipulation can significantly speed up checks for character presence.
2The first letter of the puzzle is crucial for determining valid words.

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