Longest Consecutive Sequence Using Hashing — O(n) Explained
Consecutive sequence problems appear everywhere in the real world — think about detecting unbroken login streaks in a fitness app, finding contiguous ID ranges in a database audit, or identifying runs of sensor readings in IoT pipelines. Whenever you need to find structure inside a pile of unsorted numbers, this problem and its pattern keep showing up. It's also one of the most frequently asked medium-level questions at FAANG-tier interviews because it forces you to think about data access patterns, not just loops.
The naive approach — sort the array and scan for breaks — costs O(n log n) and tempts almost everyone at first glance. But sorting is expensive, and it destroys the original order. The real insight is that you don't need sorted order at all. You just need to answer one question in O(1): 'Does the number X exist in the input?' A hash set is purpose-built for exactly that. With it, you can scan the array once and trace every consecutive chain from its true starting point, bringing the total cost down to O(n).
By the end of this article you'll understand exactly why the hash set approach works and not just mechanically how to code it, you'll be able to trace through any example by hand, spot the two most common implementation bugs before they bite you, and confidently explain your reasoning in an interview when the follow-up questions start flying.
Why Sorting Is the Wrong First Instinct
When most developers see 'find the longest run of consecutive numbers', their brain jumps straight to sorting. Sort the array, walk through it, count runs, track the max. Clean, simple, done. And it works — but it's costing you more than you need to pay.
Sorting is O(n log n). For an array of 10 million elements, that's a meaningful difference from O(n). More importantly, the sorting instinct reveals a deeper misunderstanding: you're paying to impose order on the entire dataset when you only care about local neighbourhood relationships — 'is num+1 in the array?' and 'is num-1 in the array?'.
This is the mental shift that separates intermediate from advanced thinking. Instead of restructuring the data, change how you query it. A HashSet gives you O(1) membership checks. Once every number is in the set, you can ask 'does 5 exist?' or 'does 6 exist?' for free, without touching any sorted structure.
The sorted approach also doesn't handle duplicates gracefully without extra logic — you have to skip them manually. The hash set absorbs duplicates automatically because a set stores only unique values. Two problems solved with one data structure choice.
import java.util.Arrays; import java.util.HashSet; import java.util.Set; public class NaiveSortVsHashSet { // ---- APPROACH 1: Sort-based O(n log n) ---- public static int longestConsecutiveSorting(int[] numbers) { if (numbers.length == 0) return 0; Arrays.sort(numbers); // O(n log n) — reorders entire array int longestRun = 1; int currentRun = 1; for (int i = 1; i < numbers.length; i++) { if (numbers[i] == numbers[i - 1]) { continue; // skip duplicates manually } if (numbers[i] == numbers[i - 1] + 1) { currentRun++; // extend the current run longestRun = Math.max(longestRun, currentRun); } else { currentRun = 1; // break in sequence, reset } } return longestRun; } // ---- APPROACH 2: HashSet-based O(n) ---- public static int longestConsecutiveHashing(int[] numbers) { // Load every number into a set — O(n) time, duplicates disappear Set<Integer> numberSet = new HashSet<>(); for (int num : numbers) { numberSet.add(num); } int longestRun = 0; for (int num : numberSet) { // Only start counting from the BEGINNING of a sequence. // If (num - 1) exists in the set, this number is NOT a sequence start. boolean isSequenceStart = !numberSet.contains(num - 1); if (isSequenceStart) { int currentNumber = num; int currentRun = 1; // Walk forward as long as consecutive numbers exist while (numberSet.contains(currentNumber + 1)) { currentNumber++; currentRun++; } longestRun = Math.max(longestRun, currentRun); } } return longestRun; } public static void main(String[] args) { int[] input = {100, 4, 200, 1, 3, 2, 2, 3}; // note duplicate 2s and 3s System.out.println("Input: " + Arrays.toString(input)); System.out.println("Sorting approach: " + longestConsecutiveSorting(input)); System.out.println("HashSet approach: " + longestConsecutiveHashing(input)); } }
Sorting approach: 4
HashSet approach: 4
The Core Insight — Only Count From Sequence Starts
Here's the question that unlocks everything: what would happen if you started counting a chain from every single number in the set? You'd recount the same sequence multiple times. Starting from 1 you'd count 1→2→3→4 (length 4). Starting from 2 you'd count 2→3→4 (length 3). Starting from 3, length 2. You'd do redundant work proportional to the square of the sequence length in the worst case.
The genius of the algorithm is the guard: only begin a chain-count when the current number has no left-neighbour in the set. If num - 1 is absent, then num must be the true beginning of a consecutive run. You can safely count forward from there, knowing you'll never double-count that sequence.
This single condition — !numberSet.contains(num - 1) — is what keeps the whole algorithm at O(n). Every number is visited at most twice: once when you check it as a potential sequence start, and once when it's stepped over during a forward count from an earlier start. Those two passes across n elements together are still O(n).
Think of it like painting. You only dip your brush at the leftmost edge of each white stripe. You never start mid-stripe. The result: every stripe gets painted exactly once.
import java.util.HashSet; import java.util.Set; public class LongestConsecutiveSequence { public static int findLongestConsecutiveSequence(int[] numbers) { if (numbers == null || numbers.length == 0) { return 0; // guard against empty or null input } // Step 1: Populate the set — O(n) // Using a Set means duplicates are ignored for free. Set<Integer> numberSet = new HashSet<>(); for (int num : numbers) { numberSet.add(num); } int longestSequenceLength = 0; // Step 2: Iterate over unique numbers only (iterating the set, not the array) for (int num : numberSet) { // THE KEY GUARD: skip this number if it's not a sequence start. // If (num - 1) exists, some smaller number already owns this chain. if (numberSet.contains(num - 1)) { continue; // not a start — skip entirely } // We've confirmed: num IS the start of a new consecutive sequence. int sequenceStart = num; int currentLength = 1; // Walk forward one step at a time until the chain breaks while (numberSet.contains(sequenceStart + currentLength)) { currentLength++; // found the next consecutive number } // Update the global maximum longestSequenceLength = Math.max(longestSequenceLength, currentLength); } return longestSequenceLength; } public static void main(String[] args) { // Test Case 1: multiple sequences, clear winner int[] mixed = {10, 5, 6, 100, 7, 101, 8, 102, 9, 103}; System.out.println("Test 1 — Mixed sequences:"); System.out.println(" Input length: " + mixed.length + " numbers"); System.out.println(" Longest consecutive: " + findLongestConsecutiveSequence(mixed)); // Sequences: [5,6,7,8,9,10] length 6 and [100,101,102,103] length 4 // Test Case 2: all elements form one sequence int[] continuous = {3, 2, 1, 0, -1, -2}; System.out.println("\nTest 2 — Continuous descending input:"); System.out.println(" Longest consecutive: " + findLongestConsecutiveSequence(continuous)); // Test Case 3: single element int[] single = {42}; System.out.println("\nTest 3 — Single element:"); System.out.println(" Longest consecutive: " + findLongestConsecutiveSequence(single)); // Test Case 4: all duplicates int[] duplicates = {5, 5, 5, 5}; System.out.println("\nTest 4 — All duplicates:"); System.out.println(" Longest consecutive: " + findLongestConsecutiveSequence(duplicates)); } }
Input length: 10 numbers
Longest consecutive: 6
Test 2 — Continuous descending input:
Longest consecutive: 6
Test 3 — Single element:
Longest consecutive: 1
Test 4 — All duplicates:
Longest consecutive: 1
Proving O(n) — Walking Through the Work Count
When interviewers push back with 'but isn't there a nested while loop inside your for loop — doesn't that make it O(n²)?', this is the moment that separates candidates who understand their own code from those who memorised it.
The answer lies in amortised analysis. Ask: how many total times can any single number be visited by the while loop? Exactly once. The while loop only runs when we're at a confirmed sequence start. Every number that gets consumed by a while-loop forward-walk is never itself a sequence start (because its predecessor exists in the set), so it will hit the continue in the outer for-loop and never trigger its own while-walk.
Total outer loop iterations: at most n (one per unique number). Total inner while-loop steps across ALL iterations: also at most n, because each of the n numbers is stepped over by a while loop at most once. That's 2n operations total — O(n).
This is the same amortised reasoning used to prove that a dynamic array's push operation is O(1) amortised. The expensive step is rare and its cost is spread evenly across all operations. Knowing this explanation cold will visibly impress an interviewer.
import java.util.HashSet; import java.util.Set; /** * This version adds counters to PROVE the O(n) complexity claim. * Run it and watch that totalSteps stays proportional to input size, * not to input size squared. */ public class ComplexityTracer { public static void tracedLongestConsecutive(int[] numbers) { Set<Integer> numberSet = new HashSet<>(); for (int num : numbers) { numberSet.add(num); } int longestSequenceLength = 0; int outerLoopCount = 0; // how many times the for-loop body executes int innerWhileSteps = 0; // total steps across ALL while-loop runs for (int num : numberSet) { outerLoopCount++; if (numberSet.contains(num - 1)) { continue; // not a sequence start, skip } // Confirmed start: count forward int sequenceStart = num; int currentLength = 1; while (numberSet.contains(sequenceStart + currentLength)) { currentLength++; innerWhileSteps++; // track each forward step } longestSequenceLength = Math.max(longestSequenceLength, currentLength); } int uniqueCount = numberSet.size(); System.out.println(" Unique numbers (n): " + uniqueCount); System.out.println(" Outer loop iterations: " + outerLoopCount); System.out.println(" Inner while total steps: " + innerWhileSteps); System.out.println(" Combined total steps: " + (outerLoopCount + innerWhileSteps)); System.out.println(" Longest sequence: " + longestSequenceLength); System.out.println(" Steps / n ratio: " + String.format("%.2f", (double)(outerLoopCount + innerWhileSteps) / uniqueCount)); } public static void main(String[] args) { System.out.println("--- Small input (n=8) ---"); tracedLongestConsecutive(new int[]{1, 2, 3, 4, 10, 11, 50, 51}); System.out.println("\n--- Medium input: one long chain (n=10) ---"); tracedLongestConsecutive(new int[]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}); System.out.println("\n--- Worst-case-looking input: many chains (n=10) ---"); tracedLongestConsecutive(new int[]{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}); } }
Unique numbers (n): 8
Outer loop iterations: 8
Inner while total steps: 6
Combined total steps: 14
Longest sequence: 4
Steps / n ratio: 1.75
--- Medium input: one long chain (n=10) ---
Unique numbers (n): 10
Outer loop iterations: 10
Inner while total steps: 9
Combined total steps: 19
Longest sequence: 10
Steps / n ratio: 1.90
--- Worst-case-looking input: many chains (n=10) ---
Unique numbers (n): 10
Outer loop iterations: 10
Inner while total steps: 0
Combined total steps: 10
Longest sequence: 1
Steps / n ratio: 1.00
Real-World Application — Detecting Login Streak Gaps in a User System
Here's a scenario you'll actually encounter. Your app tracks which calendar days (as day-numbers since epoch) a user logged in. You want to find the user's longest unbroken login streak. The login days come from a database query — unsorted, possibly containing duplicates if the schema logs multiple events per day.
This is the Longest Consecutive Sequence problem in production clothing. The 'array of integers' is your list of day numbers. The 'consecutive sequence' is unbroken daily logins. The hash set approach handles duplicates (multiple logins on one day) without special casing, works in O(n) time even for users with years of history, and doesn't require you to sort a potentially large dataset.
The same pattern applies to: finding contiguous free blocks in a memory allocator, detecting unbroken sensor uptime windows in an IoT dashboard, finding runs of consecutive student attendance days, and identifying sequential order IDs to spot gaps in an audit trail.
Whenever your domain has 'find the longest run of something that can be modelled as consecutive integers', this algorithm is your friend.
import java.util.HashSet; import java.util.List; import java.util.Set; /** * Real-world simulation: find a user's longest consecutive login streak. * Login days are represented as integers (days since epoch). * Data is unsorted and may contain duplicates (multiple logins in one day). */ public class LoginStreakAnalyzer { /** * Returns the length of the longest unbroken daily login streak. * * @param loginDays list of day-numbers when the user logged in (unsorted, may repeat) * @return number of consecutive days in the longest streak */ public static int longestLoginStreak(List<Integer> loginDays) { if (loginDays == null || loginDays.isEmpty()) { return 0; } // Load into a set — duplicates (multiple logins per day) vanish automatically Set<Integer> uniqueLoginDays = new HashSet<>(loginDays); int longestStreak = 0; for (int day : uniqueLoginDays) { // Only start counting from the first day of a streak block. // If the previous day also has a login, this is mid-streak — skip. boolean isStreakStart = !uniqueLoginDays.contains(day - 1); if (isStreakStart) { int streakDay = day; int streakLength = 1; // Advance through consecutive login days while (uniqueLoginDays.contains(streakDay + 1)) { streakDay++; streakLength++; } longestStreak = Math.max(longestStreak, streakLength); System.out.println(" Streak found: days " + day + " to " + streakDay + " (" + streakLength + " days)"); } } return longestStreak; } public static void main(String[] args) { // Simulated login days for a user over several weeks // Days 101-105: 5-day streak // Days 110-112: 3-day streak // Day 120: isolated login // Day 103 appears twice — user logged in twice that day List<Integer> userLoginHistory = List.of( 101, 102, 103, 103, 104, 105, 110, 111, 112, 120 ); System.out.println("=== Login Streak Analysis ==="); System.out.println("Raw login events: " + userLoginHistory.size()); System.out.println("Streaks detected:"); int best = longestLoginStreak(userLoginHistory); System.out.println("\nLongest login streak: " + best + " day(s)"); } }
Raw login events: 10
Streaks detected:
Streak found: days 101 to 105 (5 days)
Streak found: days 110 to 112 (3 days)
Streak found: days 120 to 120 (1 days)
Longest login streak: 5 day(s)
| Aspect | Sort-Based Approach O(n log n) | HashSet Approach O(n) |
|---|---|---|
| Time Complexity | O(n log n) — dominated by sort | O(n) — two linear passes total |
| Space Complexity | O(1) extra if in-place sort | O(n) for the hash set |
| Handles Duplicates | Manual skip logic needed | Set absorbs duplicates automatically |
| Preserves Input Order | No — array is sorted in-place | Yes — original array untouched |
| Handles Negatives | Yes, sorts correctly | Yes, hash set is key-agnostic |
| Code Complexity | Simple to write and read | Slightly more logic, but still concise |
| Best Use Case | When input is nearly sorted already | General case, large unsorted inputs |
| Interview Preference | Acceptable baseline answer | Expected optimal answer at FAANG level |
🎯 Key Takeaways
- The hash set's O(1) lookup is what makes O(n) possible — the algorithm is essentially trading O(n) space for the elimination of sorting.
- The sequence-start guard (
!numberSet.contains(num - 1)) is the entire reason the nested loop doesn't blow up to O(n²) — never implement this algorithm without it. - Iterating over the set (not the original array) in the outer loop silently eliminates duplicate processing — a small detail that matters for correctness on inputs with repeated values.
- This pattern — load into a set, then query cheaply — generalises to dozens of problems: anagram detection, two-sum, first missing positive. Recognising it as a pattern is more valuable than memorising this one problem.
⚠ Common Mistakes to Avoid
- ✕Mistake 1: Starting the count from every number, not just sequence starts — Symptom: the inner while-loop runs from every number, giving correct results but O(n²) runtime on large inputs — Fix: add the guard
if (numberSet.contains(num - 1)) continue;before the while-loop. This single line is what keeps the algorithm linear. - ✕Mistake 2: Iterating over the original array instead of the set in the outer loop — Symptom: duplicates cause the outer loop to re-visit sequence starts, potentially counting the same sequence multiple times, inflating the answer for inputs like [1,1,2,2,3] — Fix: replace
for (int num : numbers)withfor (int num : numberSet). Iterating the set guarantees each unique value is processed exactly once. - ✕Mistake 3: Forgetting to handle null or empty input — Symptom: NullPointerException or ArrayIndexOutOfBoundsException on edge-case inputs, which will cost you the question in an interview even if the main logic is perfect — Fix: add a null-and-length guard at the top:
if (numbers == null || numbers.length == 0) return 0;. Always handle edges before touching the data.
Interview Questions on This Topic
- QCan you walk me through why your solution is O(n) even though there's a while loop nested inside a for loop? (Interviewers want to hear 'amortised analysis' — each element is touched by the while loop at most once across all iterations, so total inner steps ≤ n.)
- QWhat changes if the input can contain negative numbers or Integer.MIN_VALUE? (The hash set handles negatives transparently. The only real edge case is Integer.MIN_VALUE — checking `num - 1` causes integer underflow in Java. A safe fix is to cast to long: `!numberSet.contains((long) num - 1)`.)
- QHow would you modify this to return the actual sequence, not just its length? (Track the start number and length of the best sequence separately. After the loops, reconstruct with a simple for-loop from `bestStart` to `bestStart + longestLength - 1`. This tests whether you understand the algorithm's state, not just its output.)
Frequently Asked Questions
What is the time complexity of the longest consecutive sequence using a hash set?
It's O(n) time and O(n) space. The set is built in one linear pass, and although there's a nested while loop in the algorithm, each number is visited by the while loop at most once across all outer-loop iterations — that's amortised O(n) for the whole traversal, not O(n²).
Why does the longest consecutive sequence problem use a HashSet and not a HashMap?
You only need to answer 'does this number exist?' — a yes/no question. A HashSet is perfectly designed for membership queries and uses less memory than a HashMap. A HashMap would make sense if you also needed to store metadata per number (like index or frequency), but the core algorithm needs none of that.
Does the algorithm work correctly if the input array has duplicate numbers?
Yes, and it handles them more cleanly than the sorting approach. When you load the array into a HashSet, duplicates are silently dropped because sets only store unique values. The algorithm then operates on unique numbers only, so duplicates never inflate the sequence length or cause double-counting.
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