The Ones That Count

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

Domain
Python
Difficulty
easy
Seniority
L5
Asked in
W. W. Grainger

Problem

A workout tracker logs each session as a record carrying the activity name and its duration in minutes. Given the full log and a short list of activity names you actually care about, return the total minutes spent on just those activities.

Not every minute counts. Add up the ones that do.

Example 1

Input:

targets = ["run","code"], activities = [{"name":"run","minutes":30},{"name":"read","minutes":45},{"name":"code","minutes":60},{"name":"run","minutes":20}]

Output:

110

Example 2

Input:

targets = ["run"], activities = [{"name":"swim","minutes":40}]

Output:

0

Worked solution and explanation

What this really is

Underneath the ledger costume this is a filter-then-sum: keep only the records whose name is one you care about, then add up their minutes. Everyone gets the arithmetic right. What separates people on a warm-up like this is whether you reach for a set. Check membership against the raw targets list and every one of your 10,000 records pays an O(m) scan; the answer is identical but the runtime quietly goes quadratic. The other way to lose points is subtler: forget the filter and sum everything, which passes the tiny example in your head and fails the moment a non-target record slips in.

Break down the requirements

Step 1: Convert targets to a set

List membership checks are O(n) per lookup. set(targets) gives O(1) average-case, which is what keeps filtering 10K activities against 100 targets linear instead of quadratic.

Step 2: Filter and sum in one pass

With the targets in a set, sum the minutes of every activity whose name is in it. A generator expression fed to sum() filters and accumulates in one readable pass, no manual running total to manage.

Step 3: Let the base case fall out

Empty inputs produce 0 naturally: sum() over an empty generator is 0, so no special-case branch is needed. Same story when targets is empty and nothing matches.

The solution

Set lookup with a generator sum
def activity_time_ledger(activities: list[dict], targets: list[str]) -> int:
    target_set = set(targets)
    return sum(
        activity['minutes']
        for activity in activities
        if activity['name'] in target_set
    )

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