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Apprenticeship Skill Plan

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A trade academy signs off apprentices on skillCount skills, labelled 0 through skillCount - 1. An apprentice attempts one skill at a time and must respect the academy's gating rules.

You are given requirements, where requirements[i] = [skill, gate] means the assessor will not let anyone attempt skill until gate has already been signed off. A skill that gates itself can therefore never be attempted. Only feasibility matters here — you are not asked which order to use.

Return true when at least one attempt order signs off the whole catalogue, and false when no order does.

Examples

Example 1

Input: skillCount = 4, requirements = [[1, 0], [2, 0], [3, 1], [3, 2]]
Output: true

Skill 0 opens both skill 1 and skill 2, and once those two are signed off skill 3 becomes available, so the whole catalogue clears.

Example 2

Input: skillCount = 2, requirements = [[0, 1], [1, 0]]
Output: false

Each of the two skills gates the other, so neither can ever be the first attempt.

Example 3

Input: skillCount = 6, requirements = [[1, 0], [3, 2], [5, 4]]
Output: true

The catalogue splits into three unrelated pairs, and each pair is cleared by taking the gate before the skill it guards.

Constraints

1 <= skillCount <= 2000
0 <= requirements.length <= 5000
requirements[i].length == 2
0 <= skill_i, gate_i < skillCount
Every pair in requirements is distinct.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def plan_is_feasible(skillCount: int, requirements: list[list[int]]) -> bool:

Java

public boolean planIsFeasible(int skillCount, int[][] requirements)

The intended approach

Time O(V + E)Space O(V + E)

Count, for each skill, how many gates still block it, and keep a pool of skills whose blocker count is zero. Repeatedly take one from the pool, mark it signed off, and decrement the blocker count of every skill it gates, adding any that drop to zero. The catalogue clears exactly when the number of skills signed off this way reaches skillCount; anything left over sits inside a ring of mutual gates.

Pythonentry point: plan_is_feasible

def plan_is_feasible(skillCount, requirements):
    opens = [[] for _ in range(skillCount)]
    blockers = [0] * skillCount
    for skill, gate in requirements:
        opens[gate].append(skill)
        blockers[skill] += 1
    pool = [s for s in range(skillCount) if blockers[s] == 0]
    cleared = 0
    while pool:
        current = pool.pop()
        cleared += 1
        for nxt in opens[current]:
            blockers[nxt] -= 1
            if blockers[nxt] == 0:
                pool.append(nxt)
    return cleared == skillCount

The obvious version

Build the full reachability table by relaxing every intermediate skill in turn, then declare the plan infeasible if any skill turns out to gate itself indirectly. O(V^3).

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_plan_is_feasible(skillCount, requirements):
    reach = [[False] * skillCount for _ in range(skillCount)]
    for skill, gate in requirements:
        reach[gate][skill] = True
    for mid in range(skillCount):
        for src in range(skillCount):
            if reach[src][mid]:
                row = reach[mid]
                for dst in range(skillCount):
                    if row[dst]:
                        reach[src][dst] = True
    for s in range(skillCount):
        if reach[s][s]:
            return False
    return True

How this problem was checked

Reference solution executed against 8 test cases — 8 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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