Remove Duplicate Identifiers — Problem Statement & Solution Guide

HashingMediumMixed

TimeO(n)

SpaceO(n)

Problem Description

You are given an array of strings, identifiers, representing a log of unique system tags. Due to a synchronization error, the same tag may appear multiple times in the sequence. Your task is to process this sequence and return a new array containing only the distinct tags, preserving the order of their first occurrence.

Specifically, iterate through the input array from left to right. If a tag has not been seen before, append it to the result array. If the tag has already been encountered, skip it. Continue this process until all elements have been processed.

Return the resulting array of unique identifiers.

DSA Pattern Breakdown

"Remove Duplicate Identifiers"

medium

WHY DOES IT MATTER?

The "first‑unique" pattern appears in many deduplication tasks—log aggregation, event streaming, and caching—where order matters. Mastering it demonstrates a candidate’s ability to choose the right data structure for constant‑time lookups.

OPTIMIZATION CHALLENGE

The key insight is to avoid repeated scans by storing a hash of already‑seen identifiers, turning a potentially O(n²) problem into a linear one with O(1) average‑case lookups.

REAL-WORLD CONNECTION

Think of a distributed system that receives idempotent messages; each message carries a unique tag, but network glitches may cause repeats. The system must process each tag once, in the order received, to maintain consistency across replicas.

During an interview, write the set‑check logic first, then immediately discuss edge cases (empty input, null strings) and the trade‑off of extra space versus speed; this shows holistic problem‑solving.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(n)

Core Theory — Why This Approach?

The problem of removing duplicate identifiers while preserving the order of first occurrence is a classic example of the "first‑unique" pattern, which can be solved efficiently using a hash‑based set. The naive solution—scanning the array for each element to see if it has appeared before—has a quadratic time complexity because each lookup may require traversing an ever‑growing prefix of the list. This quickly becomes infeasible for large inputs (e.g., millions of tags) due to the O(n²) time blow‑up and the repeated scanning overhead.

The optimal paradigm leverages constant‑time membership checks offered by hash tables (or language‑provided Set structures). By iterating the array once, we maintain a set of identifiers we have already encountered. For each tag, we query the set: if it is absent, we append it to the result list and insert it into the set; otherwise we skip it. This single pass guarantees O(n) time, where n is the length of the input, and O(n) auxiliary space for the set and result array. The approach scales linearly and is cache‑friendly, making it suitable for production‑grade log processing pipelines.

Interview Questions on This Problem

Q1How would you remove duplicates from a list of strings while preserving the original order, and what is the time and space complexity?

Iterate once, keep a HashSet of seen strings, and build a new list adding a string only if it is not in the set. This runs in O(n) time and O(n) extra space.

Q2Can you modify the algorithm to work in‑place without using extra space beyond O(1) auxiliary variables?

If the input array can be mutated, you can use two pointers: one for the write position and one for reading, while maintaining a hash set for seen values. The set still requires O(n) space, but the output can be written back into the original array, achieving O(1) extra array space.

Q3Why might a naïve double‑loop solution pass small test cases but fail on large datasets in a production environment?

A double‑loop solution has O(n²) time complexity, which is acceptable for tiny inputs but becomes prohibitively slow as n grows (e.g., n = 10⁶ leads to ~10¹² operations). In production, such latency can cause timeouts, resource exhaustion, and degraded user experience.

Examples

Example 1

Input

identifiers = ["tag_1", "tag_2", "tag_1", "tag_3", "tag_2"]

Output

["tag_1", "tag_2", "tag_3"]

Explanation: Start with an empty result and a set of seen tags. 1. "tag_1" is not seen. Add to result, mark as seen. Result: ["tag_1"]. 2. "tag_2" is not seen. Add to result, mark as seen. Result: ["tag_1", "tag_2"]. 3. "tag_1" is already seen. Skip. 4. "tag_3" is not seen. Add to result, mark as seen. Result: ["tag_1", "tag_2", "tag_3"]. 5. "tag_2" is already seen. Skip. Final output: ["tag_1", "tag_2", "tag_3"].

Example 2

Input

identifiers = ["alpha", "beta", "gamma", "alpha", "beta", "gamma"]

Output

["alpha", "beta", "gamma"]

Explanation: Start with an empty result and a set of seen tags. 1. "alpha" is not seen. Add to result, mark as seen. Result: ["alpha"]. 2. "beta" is not seen. Add to result, mark as seen. Result: ["alpha", "beta"]. 3. "gamma" is not seen. Add to result, mark as seen. Result: ["alpha", "beta", "gamma"]. 4. "alpha" is already seen. Skip. 5. "beta" is already seen. Skip. 6. "gamma" is already seen. Skip. Final output: ["alpha", "beta", "gamma"].

Example 3

Input

identifiers = ["x", "y", "z", "x", "y", "z", "x"]

Output

["x", "y", "z"]

Explanation: Start with an empty result and a set of seen tags. 1. "x" is not seen. Add to result, mark as seen. Result: ["x"]. 2. "y" is not seen. Add to result, mark as seen. Result: ["x", "y"]. 3. "z" is not seen. Add to result, mark as seen. Result: ["x", "y", "z"]. 4. "x" is already seen. Skip. 5. "y" is already seen. Skip. 6. "z" is already seen. Skip. 7. "x" is already seen. Skip. Final output: ["x", "y", "z"].

Example 4

Input

identifiers = ["unique_1", "unique_2", "unique_3", "unique_4"]

Output

["unique_1", "unique_2", "unique_3", "unique_4"]

Explanation: Start with an empty result and a set of seen tags. 1. "unique_1" is not seen. Add to result, mark as seen. Result: ["unique_1"]. 2. "unique_2" is not seen. Add to result, mark as seen. Result: ["unique_1", "unique_2"]. 3. "unique_3" is not seen. Add to result, mark as seen. Result: ["unique_1", "unique_2", "unique_3"]. 4. "unique_4" is not seen. Add to result, mark as seen. Result: ["unique_1", "unique_2", "unique_3", "unique_4"]. Final output: ["unique_1", "unique_2", "unique_3", "unique_4"].

Constraints

1 <= identifiers.length <= 10^5
1 <= identifiers[i].length <= 100
identifiers[i] consists of lowercase English letters, digits, and underscores.
The total number of characters in identifiers is at most 10^6.

Optimal Approach & Strategy

Traverse once, using a hash set to record seen identifiers and build the result list on the fly.

Brute Force Approach

For each identifier, scan all previous elements to see if it already appeared; if not, copy it to the result.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function removeDuplicateIdentifiers(identifiers) {
    const seen = new Set();
    const result = [];
    for (const id of identifiers) {
        if (!seen.has(id)) {
            seen.add(id);
            result.push(id);
        }
    }
    return result;
}

#include <vector>
#include <string>
#include <unordered_set>

std::vector<std::string> removeDuplicateIdentifiers(const std::vector<std::string>& identifiers) {
    std::vector<std::string> result;
    std::unordered_set<std::string> seen;
    for (const auto& id : identifiers) {
        if (seen.find(id) == seen.end()) {
            seen.insert(id);
            result.push_back(id);
        }
    }
    return result;
}

import java.util.*;

public class Solution {
    public List<String> removeDuplicateIdentifiers(List<String> identifiers) {
        Set<String> seen = new HashSet<>();
        List<String> result = new ArrayList<>();
        for (String id : identifiers) {
            if (seen.add(id)) {
                result.add(id);
            }
        }
        return result;
    }
}

def remove_duplicate_identifiers(identifiers):
    """Remove duplicate identifiers while preserving order."""
    seen = set()
    result = []
    for id in identifiers:
        if id not in seen:
            seen.add(id)
            result.append(id)
    return result

function removeDuplicateIdentifiers(identifiers) {
    const seen = new Set();
    const result = [];
    for (const id of identifiers) {
        if (!seen.has(id)) {
            seen.add(id);
            result.push(id);
        }
    }
    return result;
}

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Remove Duplicate Identifiers Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Hashing

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Remove Duplicate Identifiers — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor