Indegree elimination

Check Workshop Prerequisites

Original & variantCourse Schedule ↗

Original scenario and examples. The task states the exact input adapters and deterministic output rules; serialization, cloning and operation sequences return observable JSON results.

01. The question

Given {"count":n,"pairs":[[course,prerequisite],...]}, determine whether every workshop can be completed. IDs range from 0 to n-1; dependency cycles make the answer false.

Before revealing the idea, predict the result and describe what information you would need to track.

02. Worked example

Input{"count":4,"pairs":[[1,0],[2,1],[3,1]]}

Outputtrue

  1. Count prerequisites.
  2. Finish zero-indegree workshops and release their dependents.
  3. Compare finished count with total.

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: {"count":4,"pairs":[[1,0],[2,1],[3,1]]}. 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.

Python runtime notices

Python downloads only when you run code, using compressed runtime assets. Code is not saved or uploaded. Enable JavaScript to run it.

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
  1. Count prerequisites.
  2. Finish zero-indegree workshops and release their dependents.
  3. Compare finished count with total.

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

Small verified inputs: at most 15 binary slots, 10 graph vertices, 12 trie operations or a 4×4 board. Binary children use indexed slots 2i+1 and 2i+2. See the task for the object format.

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=""):
    count = data["count"]
    outgoing = [[] for _ in range(count)]
    indegree = [0] * count
    for course, prerequisite in data["pairs"]:
        outgoing[prerequisite].append(course)
        indegree[course] += 1
    ready = [i for i in range(count) if indegree[i] == 0]
    finished = 0
    for current in ready:
        finished += 1
        for neighbor in outgoing[current]:
            indegree[neighbor] -= 1
            if indegree[neighbor] == 0:
                ready.append(neighbor)
    return finished == count
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=""):
    count = data["count"]
    outgoing = [[] for _ in range(count)]
    indegree = [0] * count
    for course, prerequisite in data["pairs"]:
        outgoing[prerequisite].append(course)
        indegree[course] += 1
    ready = [i for i in range(count) if indegree[i] == 0]
    finished = 0
    for current in ready:
        finished += 1
        for neighbor in outgoing[current]:
            indegree[neighbor] -= 1
            if indegree[neighbor] == 0:
                ready.append(neighbor)
    return finished == count

06. Cost & edge cases

Time
O(V + E) with a queue index
Space
O(V + E)
  • Validate the exact input format and bounds before tracing.
  • Equal labels do not collapse vertex identity; missing nodes and invalid pointers are explicit.

Keep going