Python 3.11+
The code uses standard Python syntax and containers supported by the browser Judge. The Sorting proof does not depend on a minor Python release.
Verify in Python docs(opens in a new tab)Reorder values by a comparison or key to expose structure for subsequent processing. Learn the algorithm's preconditions, state transition, correctness argument, Python cost model, and transfer from one focused drill to a full Interview Problem.
Recognize it when
Consider Sorting when the prompt's constraints and required operations match this shape: Reorder values by a comparison or key to expose structure for subsequent processing.

Sign in to save your code and progress across devices. The lesson and problem statement remain public.
Sorting pays O(n log n) once to expose adjacency, rank, sweep order, or duplicate groups. After ordering, a problem that required repeated global comparison can often be solved with one linear scan.
Encode every priority in one tuple: negative numeric fields for descending order, positive fields for ascending order, and a deterministic final tie-breaker. A key function is easier to audit than a custom comparator.
def rank_key(record):
name, score, attempts = record
return (-score, attempts, name)Implement rank_key(record). record is [name, score, attempts]. Return a tuple key that sorts higher scores first, fewer attempts next, then names alphabetically.
Loading interactive editor.If it does not appear, the reference and starter code remain readable.Reload editor
Python sorting is stable: equal keys retain their original relative order. sorted() returns a new list, while list.sort() mutates and returns None. Choose deliberately so an interview helper does not unexpectedly destroy caller-owned order.
def stable_group(records):
return sorted(records, key=lambda record: record[0])Implement stable_group(records). records contains [group, value] pairs. Return a new list sorted by group while preserving original order within equal groups.
Loading interactive editor.If it does not appear, the reference and starter code remain readable.Reload editor
Comparison sorting is the default for arbitrary values. Counting or buckets can reach O(n + k) when the key domain is small and known, but costs O(k) memory. A heap is preferable when only the top k items matter and k is much smaller than n.
Say: “I sort by this explicit tuple key, then scan adjacent items under this invariant.” State whether stability matters, whether mutation is allowed, and include the linear post-pass in the O(n log n) bound. The Python list reference(opens in a new tab) defines sort, and Car Fleet(opens in a new tab) demonstrates sorting to reveal processing order.
Python 3.11+
The code uses standard Python syntax and containers supported by the browser Judge. The Sorting proof does not depend on a minor Python release.
Verify in Python docs(opens in a new tab)Interview bounds depend on the stated representation, input model, and real Python operations.
| Operation | Average | Worst | Interview note |
|---|---|---|---|
| Sorting complete workflow | O(n log n) | O(n log n) | Sort keys by a compound negative-count and word key. |
O(n) worst-case for the focused Rank Top Words implementation.
These are the mistakes most likely to survive a happy-path example and fail a boundary case.
Verify algorithm preconditions such as sorted input, nonnegative weights, acyclicity, or admissible heuristics before applying it.
Prevent it: State and verify this precondition before coding: The comparison, key domain, stability need, and memory constraints match the selected ordering method.
Avoid
def top_words(counts, limit):
passUse instead
def _rank_words(counts):
return sorted(counts, key=lambda word: (-counts[word], word))
def top_words(counts, limit):
return _rank_words(counts)[:limit]Where you will hit this: Rank Top Words(opens in a new tab)
This implementation violates the stage invariant: Alphabetical sorting alone ignores the primary frequency rank.
Prevent it: Preserve this proof obligation: The resolved prefix, suffix, rank, digit pass, or bucket remains in its documented final relation.
Avoid
def top_words(counts, limit):
return sorted(counts)[:limit]Use instead
def _rank_words(counts):
return sorted(counts, key=lambda word: (-counts[word], word))
def top_words(counts, limit):
return _rank_words(counts)[:limit]Where you will hit this: Rank Top Words(opens in a new tab)
Include visited state, boundary cases, recursion depth, and hidden copying costs in correctness and complexity analysis.
Prevent it: Count every sort, slice, copy, membership check, heap update, and recursive frame before claiming O(n log n).
Avoid
def top_words(counts, limit):
return sorted(counts)[:limit]Use instead
def _rank_words(counts):
return sorted(counts, key=lambda word: (-counts[word], word))
def top_words(counts, limit):
return _rank_words(counts)[:limit]Where you will hit this: Car Fleet(opens in a new tab)
Official Python documentation supports language behavior. Canonical problem pages provide additional practice context.
python-docs · checked 2026-07-27
python-docs · checked 2026-07-12