Galactic Anomaly Detector — Problem Statement & Solution Guide
Problem Description
You are given a sequence of energy readings collected by a galactic sensor array. Each reading is an integer that may appear multiple times in the sequence. Your task is to produce a new sequence that contains each distinct reading exactly once, preserving the order in which the first occurrence of each reading appears in the original sequence.
Input format: The first line contains a single integer n (1 ≤ n ≤ 10^5), the number of readings. The second line contains n space‑separated integers, each in the range [−10^9, 10^9].
Output format: Print the deduplicated sequence as space‑separated integers on a single line. The sequence must contain each unique reading from the input exactly once, in the order of their first appearance.
The solution should run in linear time and use linear additional memory.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Galactic Anomaly Detector"
WHY DOES IT MATTER?
Deduplication while preserving order is a foundational pattern in data cleaning, log processing, and streaming analytics; failing to handle it efficiently can cause bottlenecks in pipelines that ingest massive event streams.
OPTIMIZATION CHALLENGE
The key insight is to replace repeated linear scans with constant‑time membership checks via a hash table, collapsing an O(n^2) process into a single O(n) pass.
REAL-WORLD CONNECTION
Think of a distributed message queue where each message ID must be processed exactly once; using a hash set of seen IDs mirrors how brokers ensure idempotent consumption across nodes.
During an interview, write the hash‑set solution first, then discuss edge cases (negative numbers, large ranges) and possible in‑place tweaks to demonstrate depth of understanding.
COMPLEXITY AT A GLANCE
O(n)O(k)Core Theory — Why This Approach?
The task of extracting unique elements while preserving their first‑occurrence order is a classic example of a linear‑time filtering problem. A naive solution would scan the array for each element, checking all previous positions to see if it has already appeared, which leads to O(n^2) time on large inputs. The optimal paradigm leverages a hash‑based set (or boolean map) to record which values have been seen, allowing constant‑time membership checks and enabling a single pass over the data.
By maintaining a hash set of encountered readings, each element is processed exactly once: if the reading is not in the set, we append it to the result and insert it into the set; otherwise we skip it. This approach exploits the average O(1) amortized cost of hash table operations, yielding overall O(n) time and O(k) auxiliary space, where k is the number of distinct readings. The algorithm is stable because it respects the original ordering of first appearances, a property essential for many real‑world deduplication pipelines.
Interview Questions on This Problem
Q1How would you modify the solution if the input size exceeds available memory, requiring an external‑memory (disk‑based) approach?
Chunk the input into manageable blocks, write each block's unique elements (preserving order) to temporary files, and also record a global hash set persisted on disk. In a second pass, stream the temporary files while consulting the persisted set to emit the final deduplicated sequence, ensuring only the first occurrence of each value is kept.
Q2Can you achieve O(1) additional space if the input array is mutable and the range of readings is bounded (e.g., 0 ≤ value < N)?
Yes. Use the input array itself as a bitmap or mark visited values by swapping unique elements to the front and using a sentinel (or in‑place flag) for already‑seen values, achieving O(1) extra space while still preserving order through careful index management.
Q3Why is a hash set preferred over a sorted container (like TreeSet) for this problem, and what trade‑offs does each present?
A hash set offers average O(1) insertion and lookup, leading to linear overall time, whereas a TreeSet provides O(log k) operations, increasing total time to O(n log k). The trade‑off is that TreeSet maintains elements in sorted order, which is unnecessary here and adds overhead; however, it guarantees worst‑case logarithmic performance, useful when hash collisions are a concern.
Examples
Input
5 5 3 5 2 3
Output
5 3 2
Explanation: The first reading is 5, so 5 is kept. The next reading is 3, which is new, so 3 is kept. The third reading is 5 again, which has already appeared, so it is discarded. The fourth reading is 2, a new value, so 2 is kept. The final reading is 3, which has already appeared, so it is discarded. The resulting sequence is 5 3 2.
Input
4 1 1 1 1
Output
1
Explanation: All four readings are identical. Only the first occurrence (1) is retained; the rest are removed, leaving a single element sequence.
Input
6 10 -1 10 0 -1 5
Output
10 -1 0 5
Explanation: The first reading 10 is kept. The second reading -1 is new, so it is kept. The third reading 10 repeats, so it is discarded. The fourth reading 0 is new, so it is kept. The fifth reading -1 repeats, so it is discarded. The sixth reading 5 is new, so it is kept. The final sequence is 10 -1 0 5.
Constraints
- 1 <= n <= 10^5
- -10^9 <= reading <= 10^9
- The output sequence contains each distinct reading from the input exactly once
- Time complexity must be O(n)
- Additional memory usage must be O(n)
Optimal Approach & Strategy
Maintain a hash set of seen values while iterating once through the array; output an element only if it's not already in the set. This yields O(n) time and O(k) extra space.
Brute Force Approach
For each element, scan all previous elements to see if it has appeared before; if not, copy it to the result. This requires nested loops and runs in O(n^2) time.
Verified Code Solutions
function solution(nums) { return [...new Set(nums)]; }class Solution { public: std::vector<int> solution(std::vector<int>& nums) { std::set<int> set; for (int num : nums) { set.insert(num); } std::vector<int> result; for (int num : set) { result.push_back(num); } return result; } };import java.util.*; class Solution { public int[] solution(int[] nums) { Set<Integer> set = new HashSet<>(); for (int num : nums) { set.add(num); } int[] result = new int[set.size()]; int i = 0; for (int num : set) { result[i++] = num; } return result; } }def solution(nums): return list(set(nums))function solution(nums) { return [...new Set(nums)]; }Asked in Top Tech Interviews
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