Take-or-skip prefix DP
Schedule Nonadjacent Gallery Checks
01. The question
A line of galleries offers nonnegative inspection rewards. Adjacent galleries share a guard shift, so select no two neighbors. Return the greatest total reward; checking none is allowed. Original inspection scenario; the nonadjacent-maximization task is unchanged.
Before revealing the idea, predict the result and describe what information you would need to track.
02. Worked example
Input[3, 8, 4, 9, 2]
Output17
- For the first two galleries, choose reward 8 rather than 3.
- Taking gallery 3 adds 9 to the best reward before its neighbor, 8.
- Skipping the last gallery preserves the best reward 17.
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: [3,8,4,9]. 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
- Start the empty prefix at reward 0.
- Either skip the current gallery or take it plus the prefix ending before its neighbor.
- Store the better total for every prefix.
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=''):
best = [0] * (len(data) + 1)
for count, value in enumerate(data, 1):
take = value + (best[count - 2] if count >= 2 else 0)
best[count] = max(best[count - 1], take)
return best[-1]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=''):
best = [0] * (len(data) + 1)
for count, value in enumerate(data, 1):
take = value + (best[count - 2] if count >= 2 else 0)
best[count] = max(best[count - 1], take)
return best[-1]06. Cost & edge cases
- Time
- O(n)
- Space
- O(n) for the teaching DP table
- Empty or all-zero rewards return 0.
- One gallery may be selected.
- Use 0-16 rewards, each an integer from 0 to 100.