Tugas · algoritma sulit Python ·

Course order

def course_order(n: int, prereqs: list[list[int]]) -> list[int]:

course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])  # -> [0, 1, 2, 3]
course_order(2, [[0, 1], [1, 0]])  # -> []
course_order(3, [])  # -> [0, 1, 2]

• #1Z.ai GLM 5.3selesai · 11 dari 11 pengujianrespons dalam 1.3 dtk · waktu eksekusi 555 md · $0.0003

  ✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])diharapkan [0, 1, 2, 3]	4 md
  ✓	course_order(2, [[0, 1], [1, 0]])diharapkan []	<1 md
  ✓	course_order(3, [])diharapkan [0, 1, 2]	<1 md

  ✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md⏱ 546 md
  Kode

  import heapq
  
  def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
      adj = [[] for _ in range(n)]
      indeg = [0] * n
      for a, b in prereqs:
          adj[b].append(a)
          indeg[a] += 1
      heap = [i for i in range(n) if indeg[i] == 0]
      heapq.heapify(heap)
      res = []
      while heap:
          c = heapq.heappop(heap)
          res.append(c)
          for nxt in adj[c]:
              indeg[nxt] -= 1
              if indeg[nxt] == 0:
                  heapq.heappush(heap, nxt)
      return res if len(res) == n else []

• #1OpenAI GPT-6 Lunaselesai · 11 dari 11 pengujianrespons dalam 2.1 dtk · waktu eksekusi 580 md · $0.0001

  ✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])diharapkan [0, 1, 2, 3]	<1 md
  ✓	course_order(2, [[0, 1], [1, 0]])diharapkan []	<1 md
  ✓	course_order(3, [])diharapkan [0, 1, 2]	<1 md

  ✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md⏱ 572 md
  Kode

  import heapq
  
  def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
      graph = [[] for _ in range(n)]
      indegree = [0] * n
      for a, b in prereqs:
          graph[b].append(a)
          indegree[a] += 1
  
      available = [i for i, degree in enumerate(indegree) if degree == 0]
      heapq.heapify(available)
      order = []
  
      while available:
          course = heapq.heappop(available)
          order.append(course)
          for nxt in graph[course]:
              indegree[nxt] -= 1
              if indegree[nxt] == 0:
                  heapq.heappush(available, nxt)
  
      return order if len(order) == n else []

• #3SpaceXAI Grok Build 0.1selesai · 11 dari 11 pengujianrespons dalam 16 dtk · waktu eksekusi 831 md · $0.0052

  ✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])diharapkan [0, 1, 2, 3]	<1 md
  ✓	course_order(2, [[0, 1], [1, 0]])diharapkan []	1 md
  ✓	course_order(3, [])diharapkan [0, 1, 2]	<1 md

  ✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md⏱ 823 md
  Kode

  import heapq
  from collections import defaultdict
  
  def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
      if n == 0:
          return []
      graph = defaultdict(set)
      indegree = [0] * n
      for a, b in prereqs:
          if a not in graph[b]:
              graph[b].add(a)
              indegree[a] += 1
      heap = [i for i in range(n) if indegree[i] == 0]
      heapq.heapify(heap)
      order = []
      while heap:
          node = heapq.heappop(heap)
          order.append(node)
          for nei in graph[node]:
              indegree[nei] -= 1
              if indegree[nei] == 0:
                  heapq.heappush(heap, nei)
      if len(order) == n:
          return order
      else:
          return []

• #3MoonshotAI Kimi K2.7 Codeselesai · 11 dari 11 pengujianrespons dalam 43 dtk · waktu eksekusi 917 md · $0.01

  ✓	course_order(4, [[1, 0], [2, 0], [3, 1], [3, 2]])diharapkan [0, 1, 2, 3]	1 md
  ✓	course_order(2, [[0, 1], [1, 0]])diharapkan []	<1 md
  ✓	course_order(3, [])diharapkan [0, 1, 2]	<1 md

  ✓ 2 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md✓ <1 md⏱ 908 md
  Kode

  import heapq
  
  def course_order(n: int, prereqs: list[list[int]]) -> list[int]:
      adj = [[] for _ in range(n)]
      indeg = [0] * n
      seen = set()
  
      for a, b in prereqs:
          if not (0 <= a < n and 0 <= b < n):
              continue
          key = (b, a)
          if key in seen:
              continue
          seen.add(key)
          adj[b].append(a)
          indeg[a] += 1
  
      heap = [i for i, d in enumerate(indeg) if d == 0]
      heapq.heapify(heap)
  
      order = []
      while heap:
          u = heapq.heappop(heap)
          order.append(u)
          for v in adj[u]:
              indeg[v] -= 1
              if indeg[v] == 0:
                  heapq.heappush(heap, v)
  
      return order if len(order) == n else []

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