Two linear DPs for a cycle
Schedule Checks Around a Courtyard
01. The question
Gallery rewards form a ring: the first and last galleries are also neighbors. Select no adjacent galleries and return the largest reward. Original courtyard scenario; one gallery can be checked, and an empty ring earns 0.
Before revealing the idea, predict the result and describe what information you would need to track.
02. Worked example
Input[6, 2, 8, 5]
Output14
- Exclude the last gallery and solve [6,2,8], earning 14.
- Exclude the first and solve [2,8,5], earning 8.
- The better legal case earns 14; first and last are never taken together.
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: [6,2,8,5]. Parameter: "". 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
- Handle empty and one-gallery rings directly.
- Run the linear take-or-skip DP without the last gallery.
- Run it without the first gallery and take the larger result.
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
Enable JavaScript to step through the animation. The worked example and solution below remain readable without it.
Python / why it changes
The line behind the step
def solve(data, parameter=''):
if len(data) <= 1:
return data[0] if data else 0
def linear(values):
best = [0] * (len(values) + 1)
for count, value in enumerate(values, 1):
take = value + (best[count - 2] if count >= 2 else 0)
best[count] = max(best[count - 1], take)
return best[-1]
without_last = linear(data[:-1])
without_first = linear(data[1:])
return max(without_last, without_first)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
def solve(data, parameter=''):
if len(data) <= 1:
return data[0] if data else 0
def linear(values):
best = [0] * (len(values) + 1)
for count, value in enumerate(values, 1):
take = value + (best[count - 2] if count >= 2 else 0)
best[count] = max(best[count - 1], take)
return best[-1]
without_last = linear(data[:-1])
without_first = linear(data[1:])
return max(without_last, without_first)06. Cost & edge cases
- Time
- O(n)
- Space
- O(n) for the two teaching DP rows
- One gallery is not treated as its own conflicting neighbor.
- Two galleries allow at most one selection.
- Use 0-16 integer rewards from 0 to 100.