What is the time complexity of Wiggle Subsequence?

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

What data structures are used to solve Wiggle Subsequence?

This problem uses Array, Dynamic Programming, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Wiggle Subsequence?

Insight 1: We don't need to find the actual subsequence, only its length. This allows for a dynamic programming or greedy approach that tracks the length of the wiggle sequence ending with an 'upward' or 'downward' wiggle. Insight 2: The core idea is to maintain two states: 'up' which represents the length of the longest wiggle subsequence ending with an increasing difference, and 'down' which represents the length of the longest wiggle subsequence ending with a decreasing difference.

Which companies ask LeetCode 376. Wiggle Subsequence in interviews?

This problem is frequently asked at Electronic Arts.

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Wiggle Subsequence - LeetCode 376 Solution

Wiggle Subsequence - Complete Solution Guide

Wiggle Subsequence is LeetCode problem 376, 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

A wiggle sequence is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences. For example, [1, 7, 4, 9, 2, 5] is a wiggle sequence because the differences (6, -3, 5, -7, 3) alternate between positive and negative. In contrast, [1, 4, 7, 2, 5] and [1, 7, 4, 5, 5] are not

Detailed Explanation

The problem asks us to find the length of the longest 'wiggle subsequence' within a given integer array. A wiggle sequence is one where the differences between consecutive elements strictly alternate between positive and negative. A single-element sequence or a two-element sequence with distinct values is considered a wiggle sequence. We need to return the length of the longest such subsequence that can be formed by deleting elements from the original array while maintaining the original order.

Solution Approach

The provided solution uses a greedy approach with constant space. It iterates through the array, maintaining two variables, 'up' and 'down'. 'up' represents the length of the longest wiggle subsequence ending with an increasing wiggle, and 'down' represents the length of the longest wiggle subsequence ending with a decreasing wiggle. For each element, it checks if it's greater than or less than the previous element. If it's greater, it means we can extend the 'down' sequence by one. If it's less, we can extend the 'up' sequence by one. If they are equal, we don't update 'up' or 'down' because the current element doesn't contribute to a wiggle.

Step-by-Step Algorithm

Step 1: Initialize 'up' and 'down' to 1. A single element is considered a wiggle sequence of length 1.
Step 2: Iterate through the array starting from the second element (index 1).
Step 3: For each element, compare it with the previous element.
Step 4: If the current element is greater than the previous element, it means we have an increasing wiggle. Update 'up' to 'down + 1'.
Step 5: If the current element is less than the previous element, it means we have a decreasing wiggle. Update 'down' to 'up + 1'.
Step 6: If the current element is equal to the previous element, no update is needed because it doesn't contribute to a wiggle.
Step 7: After iterating through the entire array, return the maximum of 'up' and 'down', which represents the length of the longest wiggle subsequence.

Key Insights

Insight 1: We don't need to find the actual subsequence, only its length. This allows for a dynamic programming or greedy approach that tracks the length of the wiggle sequence ending with an 'upward' or 'downward' wiggle.
Insight 2: The core idea is to maintain two states: 'up' which represents the length of the longest wiggle subsequence ending with an increasing difference, and 'down' which represents the length of the longest wiggle subsequence ending with a decreasing difference.
Insight 3: Constant space optimization is possible because at each step, we only need the 'up' and 'down' values from the previous step. We don't need to store the entire sequence.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Greedy.

Companies

Asked at: Electronic Arts.

Wiggle Subsequence | LeetCode 376

Medium

Array Dynamic Programming Greedy

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int wiggleMaxLength(int[] nums) {
        int n = nums.length;
        if (n < 2) {
            return n;
        }
        
        int up = 1;
        int down = 1;
        
        for (int i = 1; i < n; i++) {
            if (nums[i] > nums[i-1]) {
                up = down + 1;
            } else if (nums[i] < nums[i-1]) {
                down = up + 1;
            }
        }
        
        return Math.max(up, down);
    }
}

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Problem Info

Description

Problem 376: Wiggle Subsequence - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy

Companies

Electronic Arts

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:376

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: Initialize 'up' and 'down' to 1. A single element is considered a wiggle sequence of length 1.
2
Step 2: Iterate through the array starting from the second element (index 1).
3
Step 3: For each element, compare it with the previous element.
4
Step 4: If the current element is greater than the previous element, it means we have an increasing wiggle. Update 'up' to 'down + 1'.
5
Step 5: If the current element is less than the previous element, it means we have a decreasing wiggle. Update 'down' to 'up + 1'.
6
Step 6: If the current element is equal to the previous element, no update is needed because it doesn't contribute to a wiggle.
7
Step 7: After iterating through the entire array, return the maximum of 'up' and 'down', which represents the length of the longest wiggle subsequence.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the input array 'nums' once. Therefore, the time complexity is directly proportional to the size of the input, which is O(n). The algorithm only uses a few constant space variables ('up', 'down', 'n', 'i'), regardless of the size of the input. Therefore, the space complexity is O(1).

📖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

Dynamic Programming

A dynamic programming approach can also be used, where dp[i][0] stores the length of the longest wiggle subsequence ending at index i with the last difference being positive and dp[i][1] stores the length of the longest wiggle subsequence ending at index i with the last difference being negative.

Time: O(n^2)Space: O(n)

Trade-offs: The DP approach is less efficient in terms of both time and space compared to the greedy approach. However, it can be useful if you need to reconstruct the actual wiggle subsequence, which the greedy approach doesn't provide directly.

✨

Key Insights

1Insight 1: We don't need to find the actual subsequence, only its length. This allows for a dynamic programming or greedy approach that tracks the length of the wiggle sequence ending with an 'upward' or 'downward' wiggle.
2Insight 2: The core idea is to maintain two states: 'up' which represents the length of the longest wiggle subsequence ending with an increasing difference, and 'down' which represents the length of the longest wiggle subsequence ending with a decreasing difference.
3Insight 3: Constant space optimization is possible because at each step, we only need the 'up' and 'down' values from the previous step. We don't need to store the entire sequence.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: Empty array or array with a single element. The code handles this by returning the length of the array directly since an empty or single-element array is considered a wiggle sequence (of length 0 or 1 respectively).
Edge case 2: Array with all elements being equal. In this case, 'up' and 'down' remain at 1, and the function correctly returns 1 because the longest wiggle subsequence is just a single element.
Edge case 3: Array with only two different numbers that alternate. For example, [1,2,1,2,1]. The code will correctly find the length of the wiggle subsequence which will be equal to the length of the array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where consecutive elements are equal. If consecutive elements are equal, they should not contribute to the wiggle sequence length.

✗Common mistake 2: Not initializing 'up' and 'down' to 1. A single element can be considered a wiggle sequence of length 1.

✗Common mistake 3: Incorrectly updating 'up' and 'down' based on the comparison. It's essential to update 'up' using the previous 'down' value and vice-versa.

✅

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 stock market data to identify patterns of price fluctuations (upward and downward trends). A wiggle sequence could represent a sequence of alternating price increases and decreases.

🏦Real-world scenario 2: Signal processing where you might want to extract a sequence of alternating positive and negative changes in a signal to identify significant events or patterns.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: June 17, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 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.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 "Wiggle Subsequence"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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