The One-Timers

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
L4

Problem

Given a dict, return the sorted list of values that appear exactly once across all entries.

Values that never repeated.

Example 1

Input:

d = {"a":1,"b":2,"c":1,"d":3}

Output:

[2,3]

Example 2

Input:

d = {"x":5,"y":5}

Output:

[]

Worked solution and explanation

The trap hiding in 'exactly once'

Read this fast and it looks like a dedup: return the distinct values, sorted. So the hand reaches for sorted(set(d.values())) and moves on. That is the wrong answer. 'Appears exactly once' is a frequency filter, not a distinctness filter. On {"x": 5, "y": 5} the set trick returns [5]; the real answer is [], because 5 occurred twice. The real skill here is tallying how often each value shows up and keeping only the ones with a count of one. A set literally cannot express that, since it throws away the counts the moment you build it.

Break down the requirements

Step 1: Count how many times each value appears

Walk the dict and tally each value into a frequency map. The keys are irrelevant to the count; two different keys holding 5 push its tally to 2.

Step 2: Filter values with count exactly 1

Keep only the values whose tally is exactly 1. This is where the set-based approach silently fails: it keeps a repeated value as a single copy instead of dropping it entirely.

Step 3: Sort and return

Sort the survivors ascending and return them. When nothing has a count of 1, that is an empty list, not an error.

The solution

Value frequency counting with uniqueness filter
def the_one_timers(d):
    value_counts = {}
    for key in d:
        val = d[key]
        if val in value_counts:
            value_counts[val] += 1
        else:
            value_counts[val] = 1
    result = []
    for val in value_counts:
        if value_counts[val] == 1:
            result.append(val)
    result.sort()
    return result

Common follow-up questions