Course Schedule Topological Sort
There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.
For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.
Return true if you can finish all courses. Otherwise, return false.
#voodoo bfs solution
def canFinish(self, n, prerequisites):
G = [[] for i in range(n)]
degree = [0] * n
for j, i in prerequisites:
G[i].append(j)
degree[j] += 1
bfs = [i for i in range(n) if degree[i] == 0]
for i in bfs:
for j in G[i]:
degree[j] -= 1
if degree[j] == 0:
bfs.append(j)
return len(bfs) == n#bfs solution
class Solution:
def canFinish(self, n: int, prerequisites: List[List[int]]) -> bool:
adj = [[] for _ in range(n)]
indegree = [0] * n
ans = []
for pair in prerequisites:
course = pair[0]
prerequisite = pair[1]
adj[prerequisite].append(course)
indegree[course] += 1
queue = deque()
for i in range(n):
if indegree[i] == 0:
queue.append(i)
while queue:
current = queue.popleft()
ans.append(current)
for next_course in adj[current]:
indegree[next_course] -= 1
if indegree[next_course] == 0:
queue.append(next_course)
return len(ans) == nTopological Sort DFS Solution
class Solution:
def canFinish(self, numCourses, prerequisites):
self.adj_dict = defaultdict(set)
for i, j in prerequisites:
self.adj_dict[i].add(j)
self.Visited = [0] * numCourses
self.FoundCycle = 0
for i in range(numCourses):
if self.FoundCycle == 1: break # early stop if the loop is found
if self.Visited[i] == 0:
self.DFS(i)
return self.FoundCycle == 0
def DFS(self, start):
if self.FoundCycle == 1: return # early stop if the loop is found
if self.Visited[start] == 1:
self.FoundCycle = 1 # loop is found
if self.Visited[start] == 0: # node is not visited yet, visit it
self.Visited[start] = 1 # color current node as gray
for neib in self.adj_dict[start]: # visit all its neibours
self.DFS(neib)
self.Visited[start] = 2 # color current node as blackRelated Problems
A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:
Every adjacent pair of words differs by a single letter.
Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
sk == endWord
Given two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each sequence should be returned as a list of the words [beginWord, s1, s2, ..., sk].
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The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.
Given an array nums of n integers where nums[i] is in the range [1, n], return an array of all the integers in the range [1, n] that do not appear in nums.
Given a string s and an integer k, return the length of the longest substring of s that contains at most k distinct characters.