Frequency map + bounded heap
Choose the Popular Stops
01. The question
A shuttle records numeric stop IDs for each visit. Return k stops ordered from most visited to least visited. Break frequency ties by the smaller stop ID, even at the cutoff.
Before revealing the idea, predict the result and describe what information you would need to track.
02. Worked example
Inputvisits = [8, 3, 8, 5, 3, 8, 5, 9], k = 2
Output[8, 3]
- Stop 8 has 3 visits; stops 3 and 5 have 2 each; stop 9 has 1.
- Keep at most two candidates. Use (frequency, -stop) so a larger tied ID is removed first.
- Rank the survivors for the output: stop 8, then stop 3. Stop 3 wins the tie against stop 5.
Try your code
Define solve(data, parameter="") and return your result. Your code runs locally, not on a server.
Edit your test cases
Use 1-10 objects with data and an optional string parameter. Expected outputs are computed from the lesson. An optional expected field must be correct.
Input contract & reference code
Example data: [8,3,8,5,3,8,5,9]. Parameter: "2". Numeric parameters arrive as strings; convert them with int(parameter) when needed.
The example and generated cases use the question’s exact variant. This checks behavior on a small test batch—not performance on every possible input.
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Printed output
Five-second execution limit per batch. Standard-library Python only; additional packages and network access are unavailable. Stop ends execution without blocking the lesson.
03. The idea
Try sketching your own approach first. What would you track, and why would it work?
Reveal the idea
- Count visits for each stop.
- Keep a size-k min-heap ordered by (frequency, negative stop ID).
- Sort the retained pairs in descending priority, then convert the negative IDs back.
04. Watch it happen
Use Step to move slowly, or Play to follow the whole trace. Change a preset or load your own input.
Input & settings
Red = current item / blue = active window or cells / green = completed work. A value docks at its actual destination; the original input stays unchanged. Positions start at 0.
Working state / what changes
Ready to begin
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Python / why it changes
The line behind the step
from collections import Counter
import heapq
def popular_stops(visits, k):
heap = []
for stop, frequency in Counter(visits).items():
heapq.heappush(heap, (frequency, -stop))
if len(heap) > k:
heapq.heappop(heap)
return [-negative_stop for frequency, negative_stop
in sorted(heap, reverse=True)]How to use this walkthrough
Hover or focus a control for its label. Play forms the input and follows values into the working state. Step shows one change; Back and the slider restore exact snapshots. Reduced motion uses instant updates. Saved snapshots and sorted heap displays are teaching overhead, not algorithm complexity.
05. Python solution
from collections import Counter
import heapq
def popular_stops(visits, k):
heap = []
for stop, frequency in Counter(visits).items():
heapq.heappush(heap, (frequency, -stop))
if len(heap) > k:
heapq.heappop(heap)
return [-negative_stop for frequency, negative_stop
in sorted(heap, reverse=True)]06. Cost & edge cases
- Time
- O(n + u log(k + 1)), where u is the number of distinct values
- Space
- O(u + k)
- Require 1 <= k <= the number of distinct stops.
- Stop IDs are nonnegative integers. Equal counts prefer the smaller ID.
- The final sorting step is required: the heap array is not the requested ranked output.