LeetCode #128 — MEDIUM

Longest Consecutive Sequence

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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.

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

Example 3:

Input: nums = [1,0,1,2]
Output: 3

Constraints:

0 <= nums.length <= 10⁵
-10⁹ <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Union-Find

Example 1

[100,4,200,1,3,2]

Example 2

[0,3,7,2,5,8,4,6,0,1]

Example 3

[1,0,1,2]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #128: Longest Consecutive Sequence
class Solution {
    public int longestConsecutive(int[] nums) {
        Set<Integer> s = new HashSet<>();
        for (int x : nums) {
            s.add(x);
        }
        int ans = 0;
        Map<Integer, Integer> d = new HashMap<>();
        for (int x : nums) {
            int y = x;
            while (s.contains(y)) {
                s.remove(y++);
            }
            d.put(x, d.getOrDefault(y, 0) + y - x);
            ans = Math.max(ans, d.get(x));
        }
        return ans;
    }
}

// Accepted solution for LeetCode #128: Longest Consecutive Sequence
func longestConsecutive(nums []int) (ans int) {
	s := map[int]bool{}
	for _, x := range nums {
		s[x] = true
	}
	d := map[int]int{}
	for _, x := range nums {
		y := x
		for s[y] {
			delete(s, y)
			y++
		}
		d[x] = d[y] + y - x
		ans = max(ans, d[x])
	}
	return
}

# Accepted solution for LeetCode #128: Longest Consecutive Sequence
class Solution:
    def longestConsecutive(self, nums: List[int]) -> int:
        s = set(nums)
        ans = 0
        d = defaultdict(int)
        for x in nums:
            y = x
            while y in s:
                s.remove(y)
                y += 1
            d[x] = d[y] + y - x
            ans = max(ans, d[x])
        return ans

// Accepted solution for LeetCode #128: Longest Consecutive Sequence
use std::collections::{HashMap, HashSet};

impl Solution {
    pub fn longest_consecutive(nums: Vec<i32>) -> i32 {
        let mut s: HashSet<i32> = nums.iter().cloned().collect();
        let mut ans = 0;
        let mut d: HashMap<i32, i32> = HashMap::new();
        for &x in &nums {
            let mut y = x;
            while s.contains(&y) {
                s.remove(&y);
                y += 1;
            }
            let length = d.get(&(y)).unwrap_or(&0) + y - x;
            d.insert(x, length);
            ans = ans.max(length);
        }
        ans
    }
}

// Accepted solution for LeetCode #128: Longest Consecutive Sequence
function longestConsecutive(nums: number[]): number {
    const s = new Set(nums);
    let ans = 0;
    const d = new Map<number, number>();
    for (const x of nums) {
        let y = x;
        while (s.has(y)) {
            s.delete(y++);
        }
        d.set(x, (d.get(y) || 0) + (y - x));
        ans = Math.max(ans, d.get(x)!);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.