Problema
Courses are numbered 0..n−1. Each pair [course, prerequisite] means the prerequisite must be taken before the course. Return an order to take all courses; when several courses are available, take the lowest-numbered one first so the answer is unique.
Implement course_order(n, prerequisites).
Input
n: The number of courses. prerequisites: Pairs [course, prerequisite].
Output
Return the course order, or an empty list if a cycle makes it impossible.
Vincoli
1 ≤ n ≤ 100,000; 0 ≤ pairs ≤ 200,000
Esempi
Input:
n = 4
prerequisites = [
[1, 0],
[2, 0],
[3, 1],
[3, 2]
]
Output atteso:
expected = [0, 1, 2, 3]Input:
long long n = 4;
std::vector<std::vector<long long>> prerequisites = {
{1, 0},
{2, 0},
{3, 1},
{3, 2}
};
Output atteso:
std::vector<long long> expected = {0, 1, 2, 3};Input:
long n = 4;
long[][] prerequisites = new long[][] {
new long[] {1, 0},
new long[] {2, 0},
new long[] {3, 1},
new long[] {3, 2}
};
Output atteso:
long[] expected = new long[] {0, 1, 2, 3};Spiegazione
0 first, then 1 and 2 (lowest first), then 3: [0, 1, 2, 3].
Input:
n = 2
prerequisites = [
[0, 1],
[1, 0]
]
Output atteso:
expected = []Input:
long long n = 2;
std::vector<std::vector<long long>> prerequisites = {
{0, 1},
{1, 0}
};
Output atteso:
std::vector<long long> expected = {};Input:
long n = 2;
long[][] prerequisites = new long[][] {
new long[] {0, 1},
new long[] {1, 0}
};
Output atteso:
long[] expected = new long[] {};Spiegazione
Each course requires the other, so no order exists: [].
Come affrontarlo
Prova prima da solo, poi apri un suggerimento alla volta.
Suggerimento 1
This is a dependency ordering problem: courses with no unmet prerequisites must be taken first.
Suggerimento 2
Use topological sorting, and when multiple courses are ready, always choose the smallest-numbered one to make the order deterministic.
Suggerimento 3
If you cannot process all courses, a cycle exists; also, duplicate prerequisite pairs should not break the result.
Complessità attesa
O(n + p log n) time, O(n + p) extra space
Riferimenti
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