The Resume Sifter

A medium Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
medium
Seniority
L3
Asked in
Meta

Problem

Given a list of URLs in the format 'https://resumes.io/firstname_lastname_id' (or 'firstname_id' for single-name variants) and a set existing_ids of known ID strings, return a list of [name_prefix, id] pairs for URLs whose id is NOT in existing_ids, preserving input order. The name_prefix is everything before the LAST underscore; the id is what follows it. Each result pair must be a list of two strings, e.g. ['john_doe', '123'].

Pull what's useful. Skip what you know.

Example

Input:

urls = ["https://resumes.io/john_doe_123","https://resumes.io/jane_smith_456","https://resumes.io/bob_789"], existing_ids = ["456"]

Output:

[["john_doe","123"],["bob","789"]]

Worked solution and explanation

Why this problem exists in real interviews

This tests URL parsing, string extraction, and set-based filtering, a practical combination in ETL workflows. Interviewers check whether candidates can extract structured data from URLs and filter against a known set.

Break down the requirements

Step 1: Parse the URL to extract the path segment

Split on / to get the last path segment containing firstname_lastname_id.

Step 2: Split the segment into name_prefix and ID

The id is the token after the LAST underscore. Use segment.rsplit('_', 1) so a name containing underscores (e.g. john_doe) stays intact as the name_prefix while the trailing token becomes the id.

Step 3: Filter out known IDs

Check each extracted ID against the provided set of known IDs. Keep only new candidates, appending each as a [name_prefix, id] list and preserving input order.

The solution

URL parsing with set-based filtering
def filter_new_resumes(urls, existing_ids):
    existing = set(existing_ids)
    result = []
    for url in urls:
        segment = url.rstrip('/').split('/')[-1]
        name_prefix, candidate_id = segment.rsplit('_', 1)
        if candidate_id not in existing:
            result.append([name_prefix, candidate_id])
    return result

Common follow-up questions