What is the best approach to solve Longest Consecutive Sequence?

The optimal approach for Longest Consecutive Sequence is the "Optimal — Hash Set (Start of sequence detection)" technique. It runs in O(n) time and uses O(n) space. Add all numbers to a hash set. For each number, only start counting if it's the beginning of a sequence (i.e., num-1 is NOT in the set). Then count how long the sequence extends using the set for O(1)

What is the time complexity of Longest Consecutive Sequence?

The optimal solution for Longest Consecutive Sequence has a time complexity of O(n) and a space complexity of O(n), achieved using the Hash Set pattern.

Is Longest Consecutive Sequence a Medium LeetCode problem?

Yes, LeetCode #128 (Longest Consecutive Sequence) is rated Medium. It uses the Hash Set pattern and is commonly asked in interviews at Amazon, Google, Facebook.

How to solve Longest Consecutive Sequence in Python, Java, C++, and JavaScript?

LumiChats provides complete solutions for Longest Consecutive Sequence in Python, Java, C++, and JavaScript, along with step-by-step visual animations and time/space complexity analysis. All solutions are free to view.

#128MediumHash Set

Longest Consecutive Sequence

Use a hash set to start counting only from sequence beginnings.

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Problem Statement

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.

Examples

Example 1

Input:nums = [100,4,200,1,3,2]

Output:4

Explanation: The longest consecutive sequence is [1,2,3,4], length 4.

Example 2

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

Output:9

Constraints

0 ≤ nums.length ≤ 10⁵
-10⁹ ≤ nums[i] ≤ 10⁹

Solutions

Sort then scan

TimeO(n log n)SpaceO(1)

Sort the array. Iterate and track the current streak length. O(n log n) — violates the O(n) requirement.

Visual Animation

def longestConsecutive(nums: list[int]) -> int:
    if not nums: return 0
    nums = sorted(set(nums))
    best = cur = 1
    for i in range(1, len(nums)):
        if nums[i] == nums[i-1]+1: cur += 1
        else: cur = 1
        best = max(best, cur)
    return best

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