framework_note/leetcode/note/medium/128_longest_consecutive_sequence.md

id, title, difficulty, tags, status, date_solved, leetcode_url, review_needed

id	title	difficulty	tags	status	date_solved	leetcode_url	review_needed
128	Longest Consecutive Sequence	Medium	array hash-table union-find	Solved	2026-06-05	https://leetcode.com/problems/longest-consecutive-sequence/	false

128. Longest Consecutive Sequence

[!info] Problem Link: LeetCode - Longest Consecutive Sequence

📝 Problem Description

Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence.

You must write an algorithm that runs in O(n) time.

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📥 Example 1

Input: nums = [100,4,200,1,3,2] Output: 4 Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.

📥 Example 2

Input: nums = [0,3,7,2,5,8,4,6,0,1] Output: 9

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💡 Approaches & Explanations

Approach 1: Hash Set — Optimal

To achieve O(n) time complexity, we use a hash set for O(1) lookups. The core idea is to identify the start of each possible sequence. A number n is the start of a sequence if n - 1 is not present in the set.

1. Insert all numbers from nums into a hash set.
2. Iterate through each number n in the set:
   • Check if n - 1 is in the set.
   • If n - 1 is NOT in the set, n is the start of a sequence.
   • From n, keep checking for n + 1, n + 2, ... and increment the current sequence length.
   • Update the maximum length found so far.

📊 Complexity Analysis

• Time Complexity: O(N) where N is the number of elements. Although there is a nested while loop, each element is visited at most twice (once by the main loop and once by the while loop), resulting in linear time.
• Space Complexity: O(N) to store the elements in the hash set.

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💻 Code Implementations

Python3

class Solution:
    def longestConsecutive(self, nums: List[int]) -> int:
        num_set = set(nums)
        longest = 0
        
        for n in num_set:
            # Check if n is the start of a sequence
            if (n - 1) not in num_set:
                length = 1
                while (n + length) in num_set:
                    length += 1
                longest = max(length, longest)
                
        return longest

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🧠 Key Takeaways & Lessons

• Identifying Sequence Starts: By checking for the absence of n - 1, we ensure that we only start counting from the beginning of a sequence, avoiding redundant work.
• Hash Set for Efficiency: Trading space for time by using a Hash Set allows us to reduce what would be an O(n^2) or O(n \log n) problem into O(n).