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Given a paragraph and a list of banned words, identify the most frequent word in the paragraph that is not on the banned list. The input words in the banned list are in lowercase with no punctuation, while the paragraph words are case insensitive. It is guaranteed that at least one non-banned word exists, and the answer will be unique and in lowercase

February 18, 20252 min read
MediumCodingText AnalysisProblem-SolvingCritical Thinkingdata analystsoftware engineer
Given a paragraph and a list of banned words, identify the most frequent word in the paragraph that is not on the banned list. The input words in the banned list are in lowercase with no punctuation, while the paragraph words are case insensitive. It is guaranteed that at least one non-banned word exists, and the answer will be unique and in lowercase

Approach Understand the Requirement : Determine the task of identifying the most frequent non-banned word in the paragraph. Input Parsing : Read the paragraph and banned words list, ensuring case insensitivity for the paragraph. Word Frequency Calculation :…

Approach

  1. Understand the Requirement: Determine the task of identifying the most frequent non-banned word in the paragraph.
  2. Input Parsing: Read the paragraph and banned words list, ensuring case insensitivity for the paragraph.
  3. Word Frequency Calculation: Count occurrences of each word in the paragraph that is not on the banned list.
  4. Determine the Most Frequent Word: Identify the word with the highest frequency and ensure it is unique.

Key Points

  • Clarity: Ensure you understand the difference between banned and allowed words.
  • Frequency Count: Use a methodical approach to count word occurrences accurately.
  • Uniqueness: The solution guarantees that the most frequent non-banned word will be unique.

Standard Response

  • Input: A paragraph and a list of banned words.
  • Output: The most frequent non-banned word in lowercase.
def most_frequent_word(paragraph, banned):
 # Convert banned list to a set for faster lookup
 banned_set = set(banned)
 
 # Normalize the paragraph and split into words
 words = paragraph.lower().split()
 
 # Create a frequency dictionary for non-banned words
 frequency = {}
 
 for word in words:
 if word not in banned_set:
 if word in frequency:
 frequency[word] += 1
 else:
 frequency[word] = 1
 
 # Find the word with the maximum frequency
 most_frequent = max(frequency, key=frequency.get)
 
 return most_frequent

Tips & Variations

Common Mistakes to Avoid

  • Ignoring Case Sensitivity: Ensure the comparison is case insensitive.
  • Overlooking Punctuation: Clean the input to avoid punctuation affecting word counts.

Alternative Ways to Answer

  • For longer texts, consider using advanced data structures like Counter from the collections module for efficiency.

Role-Specific Variations

  • Technical Roles: Emphasize the importance of efficiency in counting algorithms.
  • Creative Roles: Focus on language variety and the richness of vocabulary used.

Follow-Up Questions

  • What strategies did you use to ensure accuracy in your word count?
  • How would you handle ties in word frequency, if applicable?

Formatting & SEO Guidelines

  • Keywords: Identify, frequency count, non-banned words, unique word, case insensitivity.
  • Readability: Use bullet points, short paragraphs, and clear headings.

This structured approach ensures a comprehensive understanding of the task at hand, providing clarity and direction for job seekers or anyone tackling similar problems

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