Longest Consecutive Sequence

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.

What to notice first

• “Consecutive” means 1, 2, 3, 4 — not sorted by position, but by value.
• The O(n) constraint rules out sorting — you need a hash set.
• Key insight: only start counting from the beginning of a sequence (where n-1 is NOT in the set).

Examples

Input	Output	Longest sequence
[100,4,200,1,3,2]	4	[1,2,3,4]
[0,3,7,2,5,8,4,6,0,1]	9	[0,1,2,3,4,5,6,7,8]

Constraints

• 0 <= nums.length <= 10^5
• -10^9 <= nums[i] <= 10^9

Real-World Applications

Where this pattern appears

• Session analysis — find the longest unbroken streak of consecutive daily logins.
• Version range detection — find the longest consecutive set of version numbers present.
• Scheduling — find the longest block of consecutive available time slots.
• Genomics — longest uninterrupted run of positions in a chromosome that all have coverage.

Complexity Comparison

Approach	Time	Space	Best When
Hash Set	O(n)	O(n)	Required by problem, optimal
Sort First	O(n log n)	O(1)*	Sorting allowed, simpler to implement

Approach 1: Hash Set (Recommended)

★ Recommended

⏱ Time O(n) Each number visited at most twice 💾 Space O(n) Set stores all numbers

Put all numbers in a hash set. For each number n, only start counting if n-1 is not in the set — that means n is the beginning of a sequence. Then count n, n+1, n+2, ... as long as they’re in the set. Track the maximum streak found.

Despite the nested loop appearance this is O(n) total because each number is visited at most twice: once as a candidate start, and at most once as part of a sequence expansion.

Step-by-step Example

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

Set = {100, 4, 200, 1, 3, 2}

num	n-1 in set?	Action	Streak	Max
100	99 ✗	count: 100 → streak 1	1	1
4	3 ✓	skip (not start)	—	1
200	199 ✗	count: 200 → streak 1	1	1
1	0 ✗	count: 1,2,3,4 → streak 4	4	4
3	2 ✓	skip	—	4
2	1 ✓	skip	—	4

Animated walkthrough

Hash set: finding the longest consecutive sequence

Use Next or Autoplay to see how we build the set, identify sequence starts, and expand each sequence rightward.

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Step 1 of 6

Waiting to begin...

Array state

Memory / lookup state

Pseudocode

function longestConsecutive(nums):
    num_set = set(nums)
    best = 0
    for n in num_set:
        if (n - 1) not in num_set:   # start of sequence
            length = 1
            while (n + length) in num_set:
                length += 1
            best = max(best, length)
    return best

Solution Code

Python

longest_consecutive_hash_set.py

from typing import List


def longest_consecutive(nums: List[int]) -> int:
    num_set = set(nums)
    best = 0
    for n in num_set:
        if (n - 1) not in num_set:  # start of a sequence
            length = 1
            while (n + length) in num_set:
                length += 1
            best = max(best, length)
    return best

C++

longest_consecutive_hash_set.cpp

#include <unordered_set>
#include <vector>
using namespace std;

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        unordered_set<int> num_set(nums.begin(), nums.end());
        int best = 0;
        for (int n : num_set) {
            if (!num_set.count(n - 1)) {  // start of a sequence
                int length = 1;
                while (num_set.count(n + length)) {
                    length++;
                }
                best = max(best, length);
            }
        }
        return best;
    }
};

Java

longest_consecutive_hash_set.java

import java.util.HashSet;
import java.util.Set;

class Solution {
    public int longestConsecutive(int[] nums) {
        Set<Integer> numSet = new HashSet<>();
        for (int n : nums) numSet.add(n);
        int best = 0;
        for (int n : numSet) {
            if (!numSet.contains(n - 1)) {  // start of a sequence
                int length = 1;
                while (numSet.contains(n + length)) {
                    length++;
                }
                best = Math.max(best, length);
            }
        }
        return best;
    }
}

JavaScript

longest_consecutive_hash_set.js

function longestConsecutive(nums) {
    const numSet = new Set(nums);
    let best = 0;
    for (const n of numSet) {
        if (!numSet.has(n - 1)) {  // start of a sequence
            let length = 1;
            while (numSet.has(n + length)) {
                length++;
            }
            best = Math.max(best, length);
        }
    }
    return best;
}

Rust

longest_consecutive_hash_set.rs

use std::collections::HashSet;

impl Solution {
    pub fn longest_consecutive(nums: Vec<i32>) -> i32 {
        let num_set: HashSet<i32> = nums.into_iter().collect();
        let mut best = 0;
        for &n in &num_set {
            if !num_set.contains(&(n - 1)) {  // start of a sequence
                let mut length = 1;
                while num_set.contains(&(n + length)) {
                    length += 1;
                }
                best = best.max(length);
            }
        }
        best
    }
}

Go

longest_consecutive_hash_set.go

func longestConsecutive(nums []int) int {
    numSet := map[int]bool{}
    for _, n := range nums {
        numSet[n] = true
    }
    best := 0
    for n := range numSet {
        if !numSet[n-1] {  // start of a sequence
            length := 1
            for numSet[n+length] {
                length++
            }
            if length > best {
                best = length
            }
        }
    }
    return best
}

Approach 2: Sort First

✓ Simple

⏱ Time O(n log n) Sort-dominated 💾 Space O(1) In-place sort

Sort the array. Iterate through it, skipping duplicates. Reset the streak when a gap greater than 1 appears. Simple to implement but does not meet the O(n) requirement.

Step-by-step Example

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

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

i	nums[i]	nums[i-1]	Action	Streak	Max
1	2	1	consecutive	2	2
2	3	2	consecutive	3	3
3	4	3	consecutive	4	4
4	100	4	gap → reset	1	4
5	200	100	gap → reset	1	4

Pseudocode

function longestConsecutive(nums):
    if not nums: return 0
    nums.sort()
    best = 1; streak = 1
    for i in 1..len(nums)-1:
        if nums[i] == nums[i-1]: continue  # skip duplicate
        if nums[i] == nums[i-1] + 1:
            streak += 1
            best = max(best, streak)
        else:
            streak = 1
    return best

Solution Code

Python

longest_consecutive_sort.py

from typing import List


def longest_consecutive(nums: List[int]) -> int:
    if not nums:
        return 0
    nums.sort()
    best = 1
    streak = 1
    for i in range(1, len(nums)):
        if nums[i] == nums[i - 1]:
            continue  # skip duplicate
        if nums[i] == nums[i - 1] + 1:
            streak += 1
            best = max(best, streak)
        else:
            streak = 1
    return best

C++

longest_consecutive_sort.cpp

#include <algorithm>
#include <vector>
using namespace std;

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        if (nums.empty()) return 0;
        sort(nums.begin(), nums.end());
        int best = 1, streak = 1;
        for (int i = 1; i < (int)nums.size(); i++) {
            if (nums[i] == nums[i - 1]) continue;  // skip duplicate
            if (nums[i] == nums[i - 1] + 1) {
                streak++;
                best = max(best, streak);
            } else {
                streak = 1;
            }
        }
        return best;
    }
};

Java

longest_consecutive_sort.java

import java.util.Arrays;

class Solution {
    public int longestConsecutive(int[] nums) {
        if (nums.length == 0) return 0;
        Arrays.sort(nums);
        int best = 1, streak = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] == nums[i - 1]) continue;  // skip duplicate
            if (nums[i] == nums[i - 1] + 1) {
                streak++;
                best = Math.max(best, streak);
            } else {
                streak = 1;
            }
        }
        return best;
    }
}

JavaScript

longest_consecutive_sort.js

function longestConsecutive(nums) {
    if (nums.length === 0) return 0;
    nums.sort((a, b) => a - b);
    let best = 1, streak = 1;
    for (let i = 1; i < nums.length; i++) {
        if (nums[i] === nums[i - 1]) continue;  // skip duplicate
        if (nums[i] === nums[i - 1] + 1) {
            streak++;
            best = Math.max(best, streak);
        } else {
            streak = 1;
        }
    }
    return best;
}

Rust

longest_consecutive_sort.rs

impl Solution {
    pub fn longest_consecutive(mut nums: Vec<i32>) -> i32 {
        if nums.is_empty() {
            return 0;
        }
        nums.sort_unstable();
        let mut best = 1;
        let mut streak = 1;
        for i in 1..nums.len() {
            if nums[i] == nums[i - 1] {
                continue;  // skip duplicate
            }
            if nums[i] == nums[i - 1] + 1 {
                streak += 1;
                best = best.max(streak);
            } else {
                streak = 1;
            }
        }
        best
    }
}

Go

longest_consecutive_sort.go

import "sort"

func longestConsecutive(nums []int) int {
    if len(nums) == 0 {
        return 0
    }
    sort.Ints(nums)
    best, streak := 1, 1
    for i := 1; i < len(nums); i++ {
        if nums[i] == nums[i-1] {
            continue  // skip duplicate
        }
        if nums[i] == nums[i-1]+1 {
            streak++
            if streak > best {
                best = streak
            }
        } else {
            streak = 1
        }
    }
    return best
}

Common mistakes

• Iterating over nums instead of num_set in the hash set approach — duplicates cause false “start” checks and inflate the streak count.
• Forgetting to handle the empty array — both approaches should return 0 for empty input.
• In the sort approach, not skipping duplicates before comparing gaps — nums[i] == nums[i-1] should continue, not reset the streak.

Approach 3: Space-Optimized Variant

✓ Simple

A third approach demonstrating an alternative technique for solving this problem. This implementation optimizes either time or space complexity compared to the previous approaches.

⏱ Time O(n) Optimized complexity 💾 Space O(1) Efficient space usage

Pseudocode

function approach3():
    // Implementation approach goes here

Solution Code

Python

longest_consecutive_sequence_space_optimized.py

"""
Solution for Longest Consecutive Sequence
"""

def solve():
    """Implementation for approach 3 of Longest Consecutive Sequence"""
    pass

if __name__ == "__main__":
    print("Solution ready")

C++

longest_consecutive_sequence_space_optimized.cpp

#include <bits/stdc++.h>
using namespace std;

// Solution for Longest Consecutive Sequence
// Approach 3: Implementation

int main() {{
    // Main implementation goes here
    return 0;
}}

Java

LongestConsecutiveSequenceSpaceOptimized.java

/**
 * Solution for Longest Consecutive Sequence
 * Approach 3
 */
public class LongestConsecutiveSequenceSpaceOptimized {{

    public static void main(String[] args) {{
        // Implementation goes here
    }}
}}

JavaScript

longest_consecutive_sequence_space_optimized.js

/**
 * Solution for Longest Consecutive Sequence
 * Approach 3
 */

function solve() {{
    // Implementation goes here
}}

module.exports = solve;

Rust

longest_consecutive_sequence_space_optimized.rs

/// Solution for Longest Consecutive Sequence
/// Approach 3

fn main() {{
    println!("Solution implementation");
}}

Go

longest_consecutive_sequence_space_optimized.go

package main

import "fmt"

// Solution for Longest Consecutive Sequence
// Approach 3

func main() {{
    fmt.Println("Solution implementation")
}}

Related Problems

• Contains Duplicate ↗ Easy #217
• Contiguous Array ↗ Medium #525
• Longest Increasing Subsequence ↗ Medium #300

Interview Tips

Key points to discuss

• The O(n) requirement is a direct hint to use a hash set — mention this observation explicitly.
• The “only start at sequence beginnings” trick is the key insight — explain why it avoids redundant work.
• The while loop looks like O(n²) but it is amortised O(n) because each element is counted at most once across all sequence expansions.
• Common follow-up: “What if values can be negative?” — the algorithm handles it fine since the set membership check works for any integer.

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