What is the time complexity of Top K Frequent Words?

The optimal solution has a time complexity of O(nlogk).

What data structures are used to solve Top K Frequent Words?

This problem uses Array, Hash Table, String, Trie, Sorting, Heap (Priority Queue), Bucket Sort, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Top K Frequent Words?

Insight 1: Counting the frequency of each word is essential for determining the most frequent ones. A hash map (or Counter in Python) is suitable for efficiently storing word counts. Insight 2: A min-heap (priority queue) can be used to keep track of the `k` most frequent words seen so far. By maintaining a heap of size `k`, we ensure that only the top `k` frequent words are retained, leading to an O(n log k) time complexity.

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Top K Frequent Words - LeetCode 692 Solution

Q: Which companies ask LeetCode 692. Top K Frequent Words in interviews?

This problem is frequently asked at Attentive, Box, Intel, Netflix, Palo Alto Networks and 6 more companies.

Top K Frequent Words - Complete Solution Guide

Top K Frequent Words is LeetCode problem 692, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an array of strings words and an integer k , return the k most frequent strings . Return the answer sorted by the frequency from highest to lowest. Sort the words with the same frequency by their lexicographical order . Example 1: Input: words = ["i","love","leetcode","i","love","coding"], k = 2 Output: ["i","love"] Explanation: "i" and "love" are the two most frequent words. Note that "i" comes before "love" due to a lower alphabetical order. Example 2: Input: words = ["the","day","is","s

Detailed Explanation

The problem requires you to find the `k` most frequent words from a given list of words. The output should be a list of these `k` words, sorted first by frequency in descending order (most frequent first) and then, for words with the same frequency, by lexicographical order (alphabetical order). The input consists of a list of strings `words` and an integer `k`, representing the number of most frequent words to return. Constraints include the length of the `words` array (1 to 500), the length of each word (1 to 10), and the range of `k` (1 to the number of unique words).

Solution Approach

The solution uses a combination of a hash map and a min-heap. First, it counts the frequency of each word using a hash map. Then, it iterates through the hash map, pushing each word-frequency pair onto the min-heap. The min-heap is structured to maintain the `k` most frequent words. If the heap size exceeds `k`, the least frequent word (according to the defined comparison logic) is removed from the heap. Finally, the words are extracted from the min-heap and arranged in the correct order (descending frequency, then lexicographical order) to generate the final result.

Step-by-Step Algorithm

Step 1: Create a hash map (or Counter) to store the frequency of each word in the input `words` list.
Step 2: Iterate through the `words` list, updating the frequency count in the hash map for each word.
Step 3: Create a min-heap (priority queue) of size `k`. The heap will store word-frequency pairs.
Step 4: Define a custom comparator for the min-heap. The comparator should prioritize words with lower frequencies and, for words with equal frequencies, prioritize words with higher lexicographical order.
Step 5: Iterate through the hash map, adding each word-frequency pair to the min-heap. If the heap size exceeds `k`, remove the element at the root (the smallest element).
Step 6: Once all words have been processed, extract the elements from the min-heap in reverse order (because it's a min-heap) and add their words to the result list. The order within the heap guarantees that the results are sorted according to the problem's requirements.
Step 7: Return the result list.

Key Insights

Insight 1: Counting the frequency of each word is essential for determining the most frequent ones. A hash map (or Counter in Python) is suitable for efficiently storing word counts.
Insight 2: A min-heap (priority queue) can be used to keep track of the `k` most frequent words seen so far. By maintaining a heap of size `k`, we ensure that only the top `k` frequent words are retained, leading to an O(n log k) time complexity.
Insight 3: The comparison logic within the min-heap is crucial. We need to compare words based on frequency first (lower frequency is smaller) and then lexicographically (higher alphabetical order is smaller) for words with the same frequency, to satisfy the sorting requirements.

Complexity Analysis

Time Complexity: O(nlogk)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, String, Trie, Sorting, Heap (Priority Queue), Bucket Sort, Counting.

Companies

Asked at: Attentive, Box, Intel, Netflix, Palo Alto Networks, Pocket Gems, Robinhood, ServiceNow, Smartsheet, Yandex, Yelp.

Top K Frequent Words | LeetCode 692

Medium

Array Hash Table String Trie Sorting Heap (Priority Queue)Bucket Sort Counting

Time: O(nlogk)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        Map<String, Integer> wordCounts = new HashMap<>();
        for (String word : words) {
            wordCounts.put(word, wordCounts.getOrDefault(word, 0) + 1);
        }

        class WordFreq {
            String word;
            int freq;

            public WordFreq(String word, int freq) {
                this.word = word;
                this.freq = freq;
            }
        }

        PriorityQueue<WordFreq> minHeap = new PriorityQueue<>(
            (a, b) -> {
                if (a.freq != b.freq) {
                    return a.freq - b.freq;
                }
                return b.word.compareTo(a.word);
            }
        );

        for (Map.Entry<String, Integer> entry : wordCounts.entrySet()) {
            minHeap.offer(new WordFreq(entry.getKey(), entry.getValue()));
            if (minHeap.size() > k) {
                minHeap.poll();
            }
        }

        List<String> result = new ArrayList<>(k);
        for (int i = k - 1; i >= 0; i--) {
            result.add(0, minHeap.poll().word);
        }

        return result;
    }
}

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Problem Info

Description

Problem 692: Top K Frequent Words - Detailed explanation and solution approaches.

Topics

Array Hash Table String Trie Sorting Heap (Priority Queue)Bucket Sort Counting

Companies

AttentiveBoxIntelNetflixPalo Alto NetworksPocket GemsRobinhoodServiceNow

Stats

Difficulty:Medium

Languages:4

Companies:11

Problem #:692

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: Create a hash map (or Counter) to store the frequency of each word in the input `words` list.
2
Step 2: Iterate through the `words` list, updating the frequency count in the hash map for each word.
3
Step 3: Create a min-heap (priority queue) of size `k`. The heap will store word-frequency pairs.
4
Step 4: Define a custom comparator for the min-heap. The comparator should prioritize words with lower frequencies and, for words with equal frequencies, prioritize words with higher lexicographical order.
5
Step 5: Iterate through the hash map, adding each word-frequency pair to the min-heap. If the heap size exceeds `k`, remove the element at the root (the smallest element).
6
Step 6: Once all words have been processed, extract the elements from the min-heap in reverse order (because it's a min-heap) and add their words to the result list. The order within the heap guarantees that the results are sorted according to the problem's requirements.
7
Step 7: Return the result 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(nlogk)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The hash map stores the frequency of each unique word, taking O(n) space in the worst case. The min-heap holds at most 'k' word-frequency pairs. The loop iterates through all 'n' words and might perform heap operations of O(log k) complexity for each word, hence O(n log k) time complexity. The extraction of results is bounded by O(k), and in the worst case is less than O(n log k).

📖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

Bucket Sort

Instead of a heap, we can use bucket sort. Create buckets where the index represents the frequency. Place words into the appropriate buckets based on their frequency. Then, iterate through the buckets in reverse order of frequency, and sort words within each bucket lexicographically. Extract the top 'k' words.

Time: O(n) on average, but can be O(n^2) in worst case if many words have the same frequency.Space: O(n)

Trade-offs: Bucket sort can be faster in some cases, especially if the frequency distribution is relatively uniform. However, the heap-based approach guarantees O(n log k) time complexity, making it more predictable.

Sorting

Count the frequency of each word using a hash map. Then, convert the hash map to a list of word-frequency pairs and sort this list based on frequency (descending) and lexicographical order (ascending). Finally, take the first 'k' elements from the sorted list.

Time: O(n log n), where n is the number of unique words.Space: O(n)

Trade-offs: This approach is simpler to implement but has a higher time complexity than the heap-based solution, especially when 'k' is much smaller than 'n'. It's less efficient for large datasets and small 'k' values.

✨

Key Insights

1Insight 1: Counting the frequency of each word is essential for determining the most frequent ones. A hash map (or Counter in Python) is suitable for efficiently storing word counts.
2Insight 2: A min-heap (priority queue) can be used to keep track of the `k` most frequent words seen so far. By maintaining a heap of size `k`, we ensure that only the top `k` frequent words are retained, leading to an O(n log k) time complexity.
3Insight 3: The comparison logic within the min-heap is crucial. We need to compare words based on frequency first (lower frequency is smaller) and then lexicographically (higher alphabetical order is smaller) for words with the same frequency, to satisfy the sorting requirements.

📚

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!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: When the input list is empty. The solution will correctly return an empty list.
Edge case 2: When k is equal to the number of unique words. The solution will return all unique words sorted by frequency and then lexicographically.
Edge case 3: When k is 1. The solution will return the single most frequent word.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the comparator for the min-heap, leading to incorrect sorting of words with the same frequency.

✗Common mistake 2: Not handling the edge case where the input list is empty or 'k' is invalid.

✗Common mistake 3: Using a max-heap instead of a min-heap, requiring extra processing to get the correct order.

✅

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: Analyzing search query logs to find the most popular search terms.

🏦Real-world scenario 2: Identifying trending topics on social media platforms by counting hashtag frequencies.

🎮Real-world scenario 3: Building auto-complete suggestions based on the most frequently typed words in a text editor.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: October 31, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

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

@CodeNinjaJul 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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Top K Frequent Words"?

For this medium-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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