By What They Answer To

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

Problem

A stream of event-type names comes off the message bus, and you want them filed into families by the letter they start with. Group events by their first character into a dict, keeping each group in the order the names arrived.

Every name reports to its first letter. Sort the roll call as it comes in.

Example 1

Input:

events = ["login","logout","click","checkout","signup"]

Output:

{"c":["click","checkout"],"l":["login","logout"],"s":["signup"]}

Example 2

Input:

events = []

Output:

{}

Worked solution and explanation

What this really is

This is bucket accumulation wearing a grouping costume. The output is a dict of lists, and that shape is the entire point: a list or a set can tell you which first letters showed up, but only a dict of lists can hold, per letter, every name that starts with it in arrival order. Anyone can write the loop. What separates candidates is the first-occurrence case: the moment a key appears for the first time its bucket does not exist yet, and if you index groups[key] before creating the list, or write groups[key] = event instead of appending, every earlier name that shared that letter vanishes. Get that wrong and each letter keeps only its last name, so every multi-item group quietly collapses to one.

How to get there

Step 1: Derive the key

The first character is the partition key. You read it once per name with event[0]; the inputs are real names, so there is always a character to read.

Step 2: Create-then-append, in one pass

Before appending, make sure the bucket exists. setdefault(key, []) returns the existing list or creates an empty one in a single step, then you append. Because you scan left to right and only ever append, within-group order matches input order for free.

The solution

First-character grouping with setdefault
def group_events(events):
    groups = {}
    for event in events:
        groups.setdefault(event[0], []).append(event)
    return groups

One pass, one bucket per first letter, appended in arrival order.

Overwrite (wrong)
groups[key] = event assigns a single name to the key. The second name that shares a letter replaces the first, so 'logout' erases 'login' and the group ends with one element.
Accumulate (right)
groups.setdefault(key, []).append(event) keeps a growing list per key, so every name that shares a letter is retained in order.

Common follow-up questions