What is the time complexity of Top K Frequent Elements?

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

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

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

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

Insight 1: Using a hash map (dictionary) is essential to efficiently count the frequency of each element in the array. Insight 2: The follow-up requirement of better than O(n log n) time complexity suggests avoiding sorting-based approaches. Bucket sort (or frequency array) provides O(n) time complexity.

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Top K Frequent Elements - LeetCode 347 Solution

Q: How to solve Top K Frequent Elements?

The provided solutions use a frequency counting approach combined with bucket sort. First, a hash map is used to count the occurrences of each element in the input array. Then, a 'bucket' array (or frequency array) is created. The index of this array represents the frequency, and the values stored at each index are the elements that occur with that frequency. Finally, the code iterates through the frequency array from the highest frequency to the lowest, adding elements to the result list until the list has `k` elements.

Q: Which companies ask LeetCode 347. Top K Frequent Elements in interviews?

This problem is frequently asked at Arista Networks, Atlassian, Avito, ByteDance, Chewy and 30 more companies.

Top K Frequent Elements - Complete Solution Guide

Top K Frequent Elements is LeetCode problem 347, 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 integer array nums and an integer k , return the k most frequent elements . You may return the answer in any order . Example 1: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] Example 2: Input: nums = [1], k = 1 Output: [1] Example 3: Input: nums = [1,2,1,2,1,2,3,1,3,2], k = 2 Output: [1,2] Constraints: 1 <= nums.length <= 10 5 -10 4 <= nums[i] <= 10 4 k is in the range [1, the number of unique elements in the array] . It is guaranteed that the answer is unique . Follow up: Your algori

Detailed Explanation

The problem asks us to find the `k` most frequent elements in a given integer array `nums`. The input includes the array `nums` and an integer `k`. The output should be a list of the `k` most frequent elements. The order of elements in the output list doesn't matter. The constraints state that `k` is always valid (between 1 and the number of unique elements) and the answer is guaranteed to be unique. The follow-up requires an algorithm with time complexity better than O(n log n).

Solution Approach

The provided solutions use a frequency counting approach combined with bucket sort. First, a hash map is used to count the occurrences of each element in the input array. Then, a 'bucket' array (or frequency array) is created. The index of this array represents the frequency, and the values stored at each index are the elements that occur with that frequency. Finally, the code iterates through the frequency array from the highest frequency to the lowest, adding elements to the result list until the list has `k` elements.

Step-by-Step Algorithm

Step 1: Create a hash map (dictionary) to store the frequency of each element in the input array `nums`. Iterate through `nums` and update the frequency count for each element.
Step 2: Create a frequency array (bucket sort). The size of the frequency array is `len(nums) + 1`, where the index represents the frequency and the value is a list of elements with that frequency.
Step 3: Iterate through the hash map created in Step 1 and populate the frequency array. For each element and its frequency, add the element to the list at the corresponding index in the frequency array.
Step 4: Create an empty result list to store the top `k` frequent elements.
Step 5: Iterate through the frequency array from the highest frequency (last index) down to the lowest frequency (index 1). For each frequency, iterate through the list of elements at that frequency and add them to the result list.
Step 6: Check if the size of the result list has reached `k`. If so, return the result list. If not, continue iterating through the frequency array.
Step 7: If the loop finishes without finding `k` elements (unlikely due to the problem constraints), return the elements found so far.

Key Insights

Insight 1: Using a hash map (dictionary) is essential to efficiently count the frequency of each element in the array.
Insight 2: The follow-up requirement of better than O(n log n) time complexity suggests avoiding sorting-based approaches. Bucket sort (or frequency array) provides O(n) time complexity.
Insight 3: The uniqueness guarantee simplifies the problem as we don't have to deal with ties for frequency at the k-th position.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Divide and Conquer, Sorting, Heap (Priority Queue), Bucket Sort, Counting, Quickselect.

Companies

Asked at: Arista Networks, Atlassian, Avito, ByteDance, Chewy, Cisco, Docusign, DoorDash, Dropbox, EPAM Systems, Goldman Sachs, Hubspot, Intuit, J.P. Morgan, Microstrategy, Netflix, Nutanix, PayPal, Pinterest, Pocket Gems, Robinhood, Roku, Salesforce, Smartsheet, Snap, Snowflake, SoFi, Tesla, Tiger Analytics, Twilio, Visa, Walmart Labs, Yandex, Yelp, eBay.

Top K Frequent Elements | LeetCode 347

Medium

Array Hash Table Divide and Conquer Sorting Heap (Priority Queue)Bucket Sort Counting Quickselect

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> count = new HashMap<>();
        for (int n : nums) {
            count.put(n, count.getOrDefault(n, 0) + 1);
        }

        List<Integer>[] freq = new List[nums.length + 1];
        for (int i = 0; i < freq.length; i++) {
            freq[i] = new ArrayList<>();
        }

        for (int n : count.keySet()) {
            int c = count.get(n);
            freq[c].add(n);
        }

        List<Integer> resList = new ArrayList<>();
        for (int i = freq.length - 1; i > 0; i--) {
            for (int n : freq[i]) {
                resList.add(n);
                if (resList.size() == k) {
                    int[] res = new int[k];
                    for (int j = 0; j < k; j++) {
                        res[j] = resList.get(j);
                    }
                    return res;
                }
            }
        }

        int[] res = new int[resList.size()];
        for (int j = 0; j < resList.size(); j++) {
            res[j] = resList.get(j);
        }
        return res;
    }
}

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Problem Info

Description

Problem 347: Top K Frequent Elements - Detailed explanation and solution approaches.

Topics

Array Hash Table Divide and Conquer Sorting Heap (Priority Queue)Bucket Sort Counting Quickselect

Companies

Arista NetworksAtlassianAvitoByteDanceChewyCiscoDocusignDoorDash

Stats

Difficulty:Medium

Languages:4

Companies:35

Problem #:347

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 (dictionary) to store the frequency of each element in the input array `nums`. Iterate through `nums` and update the frequency count for each element.
2
Step 2: Create a frequency array (bucket sort). The size of the frequency array is `len(nums) + 1`, where the index represents the frequency and the value is a list of elements with that frequency.
3
Step 3: Iterate through the hash map created in Step 1 and populate the frequency array. For each element and its frequency, add the element to the list at the corresponding index in the frequency array.
4
Step 4: Create an empty result list to store the top `k` frequent elements.
5
Step 5: Iterate through the frequency array from the highest frequency (last index) down to the lowest frequency (index 1). For each frequency, iterate through the list of elements at that frequency and add them to the result list.
6
Step 6: Check if the size of the result list has reached `k`. If so, return the result list. If not, continue iterating through the frequency array.
7
Step 7: If the loop finishes without finding `k` elements (unlikely due to the problem constraints), return the elements found so far.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the linear traversals of the input array and the hash map. The space complexity is determined by the hash map and the frequency array, which can both grow linearly with the input size.

📖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

Heap (Priority Queue)

Build a min-heap of size `k` to store the `k` most frequent elements. First, calculate frequencies using a hash map. Then, iterate through the hash map and add elements to the heap. If the heap size exceeds `k`, remove the element with the lowest frequency. Finally, extract elements from the heap to get the result.

Time: O(n log k), where n is the length of `nums` and k is the number of frequent elements to return.Space: O(n), to store the frequencies in a hash map and O(k) for the heap.

Trade-offs: Heap approach has a time complexity of O(n log k), which is better than O(n) bucket sort approach when k is much smaller than n. It might have a slightly higher constant factor due to heap operations. Bucket sort is generally faster if k is close to n.

Sorting

Calculate frequencies using a hash map. Then, sort the entries of the hash map by frequency in descending order. Take the first `k` elements from the sorted entries.

Time: O(n log n) because of the sorting step, where n is the number of unique elements in the input array.Space: O(n) for the hash map.

Trade-offs: Sorting is not an ideal solution as it doesn't satisfy the follow-up requirement of achieving a time complexity better than O(n log n). While straightforward, it is less efficient than bucket sort or heap for larger input sizes.

✨

Key Insights

1Insight 1: Using a hash map (dictionary) is essential to efficiently count the frequency of each element in the array.
2Insight 2: The follow-up requirement of better than O(n log n) time complexity suggests avoiding sorting-based approaches. Bucket sort (or frequency array) provides O(n) time complexity.
3Insight 3: The uniqueness guarantee simplifies the problem as we don't have to deal with ties for frequency at the k-th position.

📚

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!

Sorting

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

Edge Cases

Edge case 1: `nums` is empty. The code handles this gracefully as the loops will simply not execute, returning an empty list, which might be incorrect but acceptable based on the problem definition that assumes the input meets the constraint 1 <= nums.length <= 10^5.
Edge case 2: `nums` contains only one unique element and `k` is 1. The code correctly identifies and returns that single element.
Edge case 3: All elements in `nums` are the same. The code correctly identifies the element and returns it if k is 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle the edge case where the input array is empty or has only one element.

✗Common mistake 2: Using sorting algorithms directly without considering the follow-up requirement of O(n) time complexity, resulting in an inefficient O(n log n) solution.

✗Common mistake 3: Incorrectly implementing the frequency counting step, leading to inaccurate counts and an incorrect output.

✗Common mistake 4: Using bucket sort, but incorrectly allocating the 'buckets', resulting in out-of-bounds array access errors.

✅

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 website traffic to identify the most popular pages or products.

🏦Real-world scenario 2: Identifying trending topics on social media platforms by analyzing the frequency of keywords in posts.

🎮Real-world scenario 3: In data compression algorithms, frequent symbols are assigned shorter codes to reduce the overall size of the compressed data.

📱Real-world scenario 4: In search engines, identifying the most frequent search queries helps optimize indexing and query processing.

💡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: November 15, 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 13

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

@CodeNinjaJul 6

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

@TechLeadJun 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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Top K Frequent Elements"?

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.

🤖 Need help with this problem?