What is the time complexity of Most Common Word?

The optimal solution has a time complexity of O(n+b).

What is the space complexity of Most Common Word?

The space complexity is O(n+b).

What data structures are used to solve Most Common Word?

This problem uses Array, Hash Table, String, Counting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Most Common Word?

Insight 1: Need to handle case-insensitivity by converting the paragraph to lowercase. Insight 2: Need to remove punctuation and split the paragraph into words.

Which companies ask LeetCode 819. Most Common Word in interviews?

This problem is frequently asked at Datadog.

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Most Common Word - LeetCode 819 Solution

Most Common Word - Complete Solution Guide

Most Common Word is LeetCode problem 819, a Easy 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 string paragraph and a string array of the banned words banned , return the most frequent word that is not banned . It is guaranteed there is at least one word that is not banned, and that the answer is unique . The words in paragraph are case-insensitive and the answer should be returned in lowercase . Note that words can not contain punctuation symbols. Example 1: Input: paragraph = "Bob hit a ball, the hit BALL flew far after it was hit.", banned = ["hit"] Output: "ball" Explanation:

Detailed Explanation

The problem requires finding the most frequent word in a given paragraph that is not in a provided list of banned words. The search should be case-insensitive, and punctuation should be ignored. The output must be the most frequent *non-banned* word in lowercase. It's guaranteed there is at least one valid word and that the answer is unique. The input is a string `paragraph` and a string array `banned`. The output is a string: the most frequent non-banned word.

Solution Approach

The solution involves first preprocessing the paragraph by converting it to lowercase and splitting it into words, removing any punctuation marks. Then, it creates a set of banned words for efficient lookup. It iterates through the extracted words, counts their occurrences, and ignores banned words. Finally, it identifies the most frequent word among the non-banned words.

Step-by-Step Algorithm

Step 1: Convert the paragraph to lowercase.
Step 2: Remove punctuation from the paragraph and split it into words.
Step 3: Create a set from the banned words array for O(1) lookup.
Step 4: Iterate through the extracted words, and for each word, check if it's in the banned set.
Step 5: If the word is not banned, increment its count in a word count map (or counter).
Step 6: After iterating through all the words, find the word with the highest count in the word count map (or counter).
Step 7: Return the word with the highest count.

Key Insights

Insight 1: Need to handle case-insensitivity by converting the paragraph to lowercase.
Insight 2: Need to remove punctuation and split the paragraph into words.
Insight 3: Need to efficiently check if a word is banned using a set data structure for O(1) lookup.

Complexity Analysis

Time Complexity: O(n+b)

Space Complexity: O(n+b)

Topics

This problem involves: Array, Hash Table, String, Counting.

Companies

Asked at: Datadog.

Most Common Word | LeetCode 819

Easy

Array Hash Table String Counting

Time: O(n+b)Space: O(n+b)

Solution Code

Available in:

class Solution {
    public String mostCommonWord(String paragraph, String[] banned) {
        Set<String> bannedSet = new HashSet<>(Arrays.asList(banned));
        Map<String, Integer> wordCounts = new HashMap<>();
        String[] words = paragraph.toLowerCase().split("[^a-zA-Z]+");
        for (String word : words) {
            if (!bannedSet.contains(word)) {
                wordCounts.put(word, wordCounts.getOrDefault(word, 0) + 1);
            }
        }
        String mostCommonWord = "";
        int maxCount = 0;
        for (Map.Entry<String, Integer> entry : wordCounts.entrySet()) {
            if (entry.getValue() > maxCount) {
                maxCount = entry.getValue();
                mostCommonWord = entry.getKey();
            }
        }
        return mostCommonWord;
    }
}

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Problem Info

Description

Problem 819: Most Common Word - Detailed explanation and solution approaches.

Topics

Array Hash Table String Counting

Companies

Datadog

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:819

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

7 steps

1
Step 1: Convert the paragraph to lowercase.
2
Step 2: Remove punctuation from the paragraph and split it into words.
3
Step 3: Create a set from the banned words array for O(1) lookup.
4
Step 4: Iterate through the extracted words, and for each word, check if it's in the banned set.
5
Step 5: If the word is not banned, increment its count in a word count map (or counter).
6
Step 6: After iterating through all the words, find the word with the highest count in the word count map (or counter).
7
Step 7: Return the word with the highest count.

💡

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(n+b)

💾

Space Complexity

O(n+b)

📝

Why?

The time complexity is determined by the dominant operations, which are processing the paragraph and creating the banned word set. The space complexity is determined by the data structures used to store the word counts and banned 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

Using regular expressions for word extraction

Instead of manual parsing using delimiters, regular expressions can be used to extract words more concisely, as done in the Python solution.

Time: O(n + b)Space: O(n + b)

Trade-offs: Regular expressions can be more concise but might have a slightly higher constant factor in time complexity. Also might be harder to understand.

✨

Key Insights

1Insight 1: Need to handle case-insensitivity by converting the paragraph to lowercase.
2Insight 2: Need to remove punctuation and split the paragraph into words.
3Insight 3: Need to efficiently check if a word is banned using a set data structure for O(1) lookup.

📚

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 paragraph. The solution correctly handles this by returning an empty string.
Edge case 2: Paragraph with only banned words. The problem statement guarantees there is at least one non-banned word, so this scenario doesn't need special handling. However, if this constraint wasn't given, we would need a check.
Edge case 3: Paragraph with punctuation immediately next to words (e.g., 'ball,'). The solution needs to correctly parse these words.
Edge case 4: Empty banned list. The solution correctly handles this as no words are banned.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling case-insensitivity. Forgetting to convert the paragraph to lowercase will lead to incorrect counts.

✗Common mistake 2: Not correctly handling punctuation. Failing to remove punctuation will result in incorrect word recognition.

✗Common mistake 3: Using a list instead of a set for banned words. This results in O(b) lookup time for banned words, increasing the overall time complexity to O(n*b).

✗Common mistake 4: Incorrectly splitting the paragraph into words. Missing spaces or incorrect regex leads to incorrect results.

✅

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: Sentiment analysis. Identifying the most frequent positive or negative words in a text while excluding common stop words (similar to banned words).

🏦Real-world scenario 2: Keyword extraction from a document. Finding the most relevant keywords while excluding generic words like 'the', 'a', 'is', etc. (which are similar to banned words).

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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 8, 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

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

@TechLeadJul 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 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.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Most Common Word"?

For this easy-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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