The Response Aggregator

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
Viking

Problem

Given pages (a list of pages, each a list of dicts with 'category' and 'amount'), sum amounts per category across all pages. Return a list of [category, total] pairs sorted by total descending, tie-break alphabetically by category.

Multiple result pages. One clean summary.

Example

Input:

pages = [[{"amount":50,"category":"food"},{"amount":200,"category":"travel"}],[{"amount":30,"category":"food"},{"amount":75,"category":"office"}]]

Output:

[["travel",200],["food",80],["office",75]]

Worked solution and explanation

Why this problem exists in real interviews

This tests flattening paginated data and aggregating by category, a common ETL pattern when consuming paginated API responses. It checks whether candidates can handle two levels of iteration and produce a correctly sorted list of results.

Break down the requirements

Step 1: Flatten all pages into a single stream of records

Iterate through each page, then each record within the page.

Step 2: Accumulate totals by category

Use a dictionary to sum amounts per category across all pages.

Step 3: Sort and return a list of pairs

Convert the dict to a list of [category, total] pairs, then sort by total descending with an alphabetical tie-break on category.

The solution

Page flattening with category aggregation
def aggregate_pages(pages):
    totals = {}
    for page in pages:
        for record in page:
            cat = record['category']
            amt = record['amount']
            if cat in totals:
                totals[cat] += amt
            else:
                totals[cat] = amt
    pairs = [[cat, total] for cat, total in totals.items()]
    pairs.sort(key=lambda p: (-p[1], p[0]))
    return pairs

Common follow-up questions