Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Parallel Courses in Python
Suppose there are N courses, and these are labelled from 1 to N. We also gave a relation array, where relations[i] = [X, Y], is representing a prerequisite relationship between course X and course Y. So, this means course X has to be studied before course Y.
In one semester we can study any number of courses as long as we have studied all the prerequisites for the course we are studying. We have to find the minimum number of semesters needed to study all courses. And if there is no way to study all the courses, then return -1.
So, if the input is like N = 3, relations = [[1,3],[2,3]], then the output will be 2 as In the first semester, courses 1 and 2 are studied. In the second semester, course 3 is studied.
To solve this, we will follow these steps −
courses := n
visited := an array of size n+1, and fill this with false
queue := a new list
graph := a list of n+1 sublists
in_degree := an array of size n+1, and fill this with 0
for each i in relations, do
insert i[1] at the end of graph[i[0]]
in_degree[i[1]] := in_degree[i[1]] + 1
semeseter := 1
for i in range 1 to n+1, do
if in_degree[i] is zero, then
insert i at the end of queue
visited[i] := True
semester := 1
courses := courses - size of queue
while queue is not empty and courses is non-zero, do
current_size := size of queue
while current_size is non-zero, do
current_course := queue[0]
delete first element from queue
current_size := current_size - 1
for each i in graph[current_course], do
in_degree[i] := in_degree[i] - 1
if i is not visited and in_degree[i] is zero, then
courses := courses - 1
insert i at the end of queue
visited[i]:= True
semester := semester + 1
return semester if courses is 0 otherwise -1
Let us see the following implementation to get better understanding −
Example
class Solution(object): def minimumSemesters(self, n, relations): courses = n visited = [False for i in range(n+1)] queue = [] graph = [[] for i in range(n+1)] in_degree = [0 for i in range(n+1)] for i in relations: graph[i[0]].append(i[1]) in_degree[i[1]]+=1 semeseter = 1 for i in range(1,n+1): if not in_degree[i]: queue.append(i) visited[i] = True semester = 1 courses -= len(queue) while queue and courses: current_size = len(queue) while current_size: current_course = queue[0] queue.pop(0) current_size-=1 for i in graph[current_course]: in_degree[i]-=1 if not visited[i] and not in_degree[i]: courses-=1 queue.append(i) visited[i]=True semester+=1 return semester if not courses else -1 ob = Solution() print(ob.minimumSemesters(3,[[1,3],[2,3]]))
Input
3, [[1,3],[2,3]]
Output
-1
145 Views
