The Word Inventory

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
L6
Asked in
Natera

Problem

Given a list of words, return a dict with two keys. 'counts' maps each word to its frequency. 'unique' is the sorted list of words that appear exactly once.

Every word, twice over.

Example 1

Input:

words = ["apple","banana","apple","cherry","banana","date"]

Output:

{"counts":{"date":1,"apple":2,"banana":2,"cherry":1},"unique":["cherry","date"]}

Example 2

Input:

words = ["x","x","x"]

Output:

{"counts":{"x":3},"unique":[]}

Worked solution and explanation

Why this problem exists in real interviews

This tests frequency counting with a secondary filter for singletons. It combines dict accumulation with a filtering pass, a pattern common in data quality and analytics pipelines.

Break down the requirements

Step 1: Count word frequencies

Build a dict mapping each word to its occurrence count.

Step 2: Extract words appearing exactly once

Filter the frequency dict to find singletons and sort them alphabetically.

Step 3: Return both the counts dict and the unique list

Package both results in a dict with 'counts' and 'unique' keys.

The solution

Frequency dict with singleton extraction
def word_inventory(words: list) -> dict:
    counts = {}
    for word in words:
        counts[word] = counts.get(word, 0) + 1
    unique = []
    for word in sorted(counts.keys()):
        if counts[word] == 1:
            unique.append(word)
    result = {"counts": counts, "unique": unique}
    return result

Common follow-up questions