L2 · Working engineerPython internals~3 min · 3 tests#70

Sort a List of Dictionaries by Key

Sort a list of employee dicts by salary using sorted() with a key function, ascending and descending. Practical Python sorting with live tests.

The problem

employees is a list of dicts with "name" and "salary". Return it sorted by salary, ascending.

Examples

Example 1

Input

sort_by_salary([{'name': 'John', 'salary': 50000}, {'name': 'Alice', 'salary': 70000}, {'name': 'Bob', 'salary': 60000}])

Expected output

[{'name': 'John', 'salary': 50000}, {'name': 'Bob', 'salary': 60000}, {'name': 'Alice', 'salary': 70000}]

+ 2 hidden tests on Submit.

Edge cases to ask about

Empty list
Ties
Missing key (KeyError)

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · sort_by_salary

def sort_by_salary(employees):
    """Return employees sorted by salary ascending."""
    # Write your solution here
    pass

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Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and measure your code against the optimal one at growing input sizes.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

Measure it

Runs the function on inputs of size 250 up to 16,000 and records the time and peak memory. Slow solutions stop early — a short curve is itself the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
sorted(key=…)	O(n log n)	O(n)	operator.itemgetter('salary') also works.

Walkthrough of the optimal approach (try it yourself first)

Same pattern as the previous question with a dict lookup in the key: sorted(employees, key=lambda e: e["salary"]). For ties broken by name, return a tuple: key=lambda e: (e["salary"], e["name"]).

Complexity: O(n log n) time, O(n) space. A comparison sort of n records.

Reveal the reference solution

def sort_by_salary(employees):
    return sorted(employees, key=lambda e: e["salary"])

Follow-ups interviewers ask

Descending salary, then name.
Some records have no salary key.

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of Sort a List of Dictionaries by Key in Python?

The optimal solution runs in O(n log n) time and O(n) auxiliary space. A comparison sort of n records.

What follow-up questions do interviewers ask about Sort a List of Dictionaries by Key?

Descending salary, then name. Some records have no salary key.