What is the time complexity of Longest Harmonious Subsequence?

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

What data structures are used to solve Longest Harmonious Subsequence?

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

What are the key insights for solving Longest Harmonious Subsequence?

Insight 1: The longest harmonious subsequence must contain elements 'x' and 'x+1' for some value 'x'. We can iterate through all possible values 'x' and find the counts of 'x' and 'x+1' in the array. Insight 2: Using a hash map (or dictionary) is an efficient way to store and retrieve the frequency of each number in the input array.

Which companies ask LeetCode 594. Longest Harmonious Subsequence in interviews?

This problem is frequently asked at LiveRamp, ZS Associates.

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Longest Harmonious Subsequence - LeetCode 594 Solution

Longest Harmonious Subsequence - Complete Solution Guide

Longest Harmonious Subsequence is LeetCode problem 594, 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

We define a harmonious array as an array where the difference between its maximum value and its minimum value is exactly 1 . Given an integer array nums , return the length of its longest harmonious subsequence among all its possible subsequences. Example 1: Input: nums = [1,3,2,2,5,2,3,7] Output: 5 Explanation: The longest harmonious subsequence is [3,2,2,2,3] . Example 2: Input: nums = [1,2,3,4] Output: 2 Explanation: The longest harmonious subsequences are [1,2] , [2,3] , and [3,4] , all of w

Detailed Explanation

The problem requires us to find the length of the longest harmonious subsequence within a given integer array `nums`. A harmonious array is defined as an array where the difference between its maximum and minimum values is exactly 1. A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. We need to consider all possible subsequences and find the longest one that satisfies the harmonious property. If no harmonious subsequence exists, we should return 0.

Solution Approach

The solution uses a hash map to count the occurrences of each number in the input array. Then, it iterates through the hash map and for each number `num`, it checks if `num + 1` also exists in the hash map. If it does, it calculates the sum of their counts (which represents the length of a harmonious subsequence formed by `num` and `num + 1`) and updates the `max_len` if the current harmonious subsequence length is greater than the current `max_len`. Finally, the algorithm returns `max_len`.

Step-by-Step Algorithm

Step 1: Create an empty hash map (dictionary) `count` to store the frequency of each number in the input array `nums`.
Step 2: Iterate through the `nums` array and update the `count` hash map. For each `num` in `nums`, increment its count in the `count` hash map.
Step 3: Initialize `max_len` to 0. This variable will store the length of the longest harmonious subsequence found so far.
Step 4: Iterate through the `count` hash map. For each `num` in `count`, check if `num + 1` also exists as a key in the `count` hash map.
Step 5: If `num + 1` exists in the `count` hash map, calculate the sum of the counts of `num` and `num + 1` in the `count` hash map.
Step 6: Update `max_len` with the maximum value between the current `max_len` and the sum calculated in step 5.
Step 7: After iterating through all numbers in the `count` hash map, return the final `max_len`.

Key Insights

Insight 1: The longest harmonious subsequence must contain elements 'x' and 'x+1' for some value 'x'. We can iterate through all possible values 'x' and find the counts of 'x' and 'x+1' in the array.
Insight 2: Using a hash map (or dictionary) is an efficient way to store and retrieve the frequency of each number in the input array.
Insight 3: We can find the maximum length by iterating through the keys of the hash map and checking if key+1 exists. The length of the harmonious subsequence will be the sum of counts of the key and key+1.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Sliding Window, Sorting, Counting.

Companies

Asked at: LiveRamp, ZS Associates.

Longest Harmonious Subsequence | LeetCode 594

Easy

Array Hash Table Sliding Window Sorting Counting

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

Solution Code

Available in:

class Solution {
    public int findLHS(int[] nums) {
        Map<Integer, Integer> countMap = new HashMap<>();
        for (int num : nums) {
            countMap.put(num, countMap.getOrDefault(num, 0) + 1);
        }
        int maxLength = 0;
        for (int num : countMap.keySet()) {
            if (countMap.containsKey(num + 1)) {
                maxLength = Math.max(maxLength, countMap.get(num) + countMap.get(num + 1));
            }
        }
        return maxLength;
    }
}

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Problem Info

Description

Problem 594: Longest Harmonious Subsequence - Detailed explanation and solution approaches.

Topics

Array Hash Table Sliding Window Sorting Counting

Companies

LiveRampZS Associates

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:594

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 an empty hash map (dictionary) `count` to store the frequency of each number in the input array `nums`.
2
Step 2: Iterate through the `nums` array and update the `count` hash map. For each `num` in `nums`, increment its count in the `count` hash map.
3
Step 3: Initialize `max_len` to 0. This variable will store the length of the longest harmonious subsequence found so far.
4
Step 4: Iterate through the `count` hash map. For each `num` in `count`, check if `num + 1` also exists as a key in the `count` hash map.
5
Step 5: If `num + 1` exists in the `count` hash map, calculate the sum of the counts of `num` and `num + 1` in the `count` hash map.
6
Step 6: Update `max_len` with the maximum value between the current `max_len` and the sum calculated in step 5.
7
Step 7: After iterating through all numbers in the `count` hash map, return the final `max_len`.

💡

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 O(n) because we iterate through the array once to create the hash map, and then iterate through the keys in the hash map. Since the number of unique keys is at most n, the overall time complexity is O(n). The space complexity is O(n) because, in the worst case where all elements are unique, the hash map will store n key-value pairs.

📖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

Sorting

Sort the array and then iterate through it, keeping track of the count of each element and its adjacent elements (num+1). Find the maximum count among harmonious subsequences.

Time: O(n log n) due to sorting.Space: O(1) if sorting is done in-place, otherwise O(n).

Trade-offs: Sorting approach is generally slower because of O(n log n) time complexity, but can have a better space complexity if in-place sorting is used. The Hashmap is generally faster and uses more space.

✨

Key Insights

1Insight 1: The longest harmonious subsequence must contain elements 'x' and 'x+1' for some value 'x'. We can iterate through all possible values 'x' and find the counts of 'x' and 'x+1' in the array.
2Insight 2: Using a hash map (or dictionary) is an efficient way to store and retrieve the frequency of each number in the input array.
3Insight 3: We can find the maximum length by iterating through the keys of the hash map and checking if key+1 exists. The length of the harmonious subsequence will be the sum of counts of the key and key+1.

📚

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!

Sliding Window

A moving "window" over data to process subsets efficiently.

Sorting

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

Edge Cases

Edge case 1: Empty input array `nums`. The algorithm correctly returns 0 because the `count` hash map will be empty, and the loop will not execute.
Edge case 2: An array containing only one unique number, e.g., `[1, 1, 1, 1]`. The algorithm returns 0 because there's no `num + 1` present in the `count` map.
Edge case 3: An array with negative numbers, e.g., `[-1, 0, 1]`. The hashmap approach works correctly with negative numbers as the keys can be any integer as long as it satisfies constraints.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly assuming the harmonious subsequence needs to be contiguous (a subarray). The problem specifies 'subsequence,' meaning elements don't need to be adjacent.

✗Common mistake 2: Not handling the case where the input array contains only one unique number. The algorithm must correctly return 0 in this scenario.

✗Common mistake 3: Using an inefficient algorithm with a high time complexity (e.g., O(n^2)) that would result in TLE (Time Limit Exceeded) for larger input arrays.

✅

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 customer purchase data to find pairs of products that are frequently bought together, where the price difference is minimal (e.g., $1). This helps in product placement and bundling strategies.

🏦Real-world scenario 2: Identifying pairs of temperature readings from sensors that are close to each other (within a degree Celsius). This is useful in environmental monitoring and climate studies.

💡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: July 25, 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

❓

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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Longest Harmonious Subsequence"?

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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