algoadvance

The problem 3136. Valid Word-out (assuming the name might need clarification) is likely about determining whether words in a puzzle or a list adhere to certain rules. Unfortunately, the provided problem number does not directly correspond to a specific well-known problem description, so I’ll assume a standard format of verifying the validity of words based on certain constraints.

Let’s define a problem statement:

Given a list of words and a set of characters, write a function is_valid_word that checks if each word can be formed using the characters from the given set. Each character in the set can only be used once per word.

Input:

• words (list of strings): A list containing multiple words.
• chars (string): A string which contains the allowed characters.

Output:

• A list of boolean values where each boolean value corresponds to whether the word at the respective index in the input list can be formed using the given characters.

Example:

words = ["hello", "world", "leetcode"]
chars = "welldonehatch"

Output:

[True, True, False]

Clarifying Questions

1. Case Sensitivity: Should the comparison be case-sensitive?
   • Assume that the comparison is case-sensitive.
2. Special Characters: Can words and characters include special characters (e.g., numbers, punctuation)?
   • Yes, both words and chars can include special characters.
3. Multiple Uses: Can characters in the set chars be used more than once?
   • No, each character in the set chars can only be used once per word.

Strategy

1. Character Frequency Count:
   • Calculate the frequency count of each character in chars.
2. Word Validation:
   • For each word, calculate the frequency count of characters and compare it with the frequency count of chars.
   • Check if each character in the word can be satisfied by the character count in chars.
3. Output:
   • Return a list of boolean values indicating whether each word can be formed.

Code

Here is the implementation in Python:

from collections import Counter

def is_valid_word(words, chars):
    chars_count = Counter(chars)
    result = []
    
    for word in words:
        word_count = Counter(word)
        is_valid = True
        for char in word_count:
            if word_count[char] > chars_count.get(char, 0):
                is_valid = False
                break
        result.append(is_valid)
    
    return result

# Example usage:
words = ["hello", "world", "leetcode"]
chars = "welldonehatch"
print(is_valid_word(words, chars))  # Output: [True, True, False]

Time Complexity

• Building Counter for chars: O(n) where n is the length of chars.
• Validation for each word: O(m * k) where m is the number of words and k is the average length of the words.

Overall Time Complexity: O(n + m * k) which is efficient for typical input sizes.

Let me know if there’s anything more specific you need or additional constraints!

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