Skip to main content

2050 - Parallel Courses III (Hard)

https://leetcode.com/problems/parallel-courses-iii/

Problem Statement​

You are given an integer n, which indicates that there are n courses labeled from 1 to n. You are also given a 2D integer array relations where relations[j] = [prevCoursej, nextCoursej] denotes that course prevCoursej has to be completed before course nextCoursej (prerequisite relationship). Furthermore, you are given a 0-indexed integer array time where time[i] denotes how many months it takes to complete the (i+1)th course.

You must find the minimum number of months needed to complete all the courses following these rules:

  • You may start taking a course at any time if the prerequisites are met.
  • Any number of courses can be taken at the same time.

Return the minimum number of months needed to complete all the courses.

Note: The test cases are generated such that it is possible to complete every course (i.e., the graph is a directed acyclic graph).

Example 1:

Input: n = 3, relations = [[1,3],[2,3]], time = [3,2,5]
Output: 8
Explanation: The figure above represents the given graph and the time required to complete each course. 
We start course 1 and course 2 simultaneously at month 0.
Course 1 takes 3 months and course 2 takes 2 months to complete respectively.
Thus, the earliest time we can start course 3 is at month 3, and the total time required is 3 + 5 = 8 months.

Example 2:

Input: n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], time = [1,2,3,4,5]
Output: 12
Explanation: The figure above represents the given graph and the time required to complete each course.
You can start courses 1, 2, and 3 at month 0.
You can complete them after 1, 2, and 3 months respectively.
Course 4 can be taken only after course 3 is completed, i.e., after 3 months. It is completed after 3 + 4 = 7 months.
Course 5 can be taken only after courses 1, 2, 3, and 4 have been completed, i.e., after max(1,2,3,7) = 7 months.
Thus, the minimum time needed to complete all the courses is 7 + 5 = 12 months.

Constraints:

  • 1 <= n <= 5 * 10 ^ 4
  • 0 <= relations.length <= min(n * (n - 1) / 2, 5 * 10 ^ 4)
  • relations[j].length == 2
  • 1 <= prevCoursej, nextCoursej <= n
  • prevCoursej != nextCoursej
  • All the pairs [prevCoursej, nextCoursej] are unique.
  • time.length == n
  • 1 <= time[i] <= 10 ^ 4
  • The given graph is a directed acyclic graph.

Approach 1: DFS​

We can use dfs to find the maximum time of all the paths starting from u. If there is no prerequisite relationship for u, we can simply return time[u]. Otherwise, we check the same for all of the neighbors. We can the maximum of all the results.

Written by @wingkwong
class Solution:
    def minimumTime(self, n: int, relations: List[List[int]], time: List[int]) -> int:
        g = defaultdict(list)
        # build the prerequisite graph 
        # minus one to make it zero indexing
        for x in relations: g[x[0] - 1].append(x[1] - 1)
        # memoize it to improve performance
        @cache
        def dfs(u):
            # no prerequisite relationship -> return its time
            if not g[u]: return time[u]
            # the current time + the max of the dfs result from the neighbors
            return max([dfs(v) for v in g[u]]) + time[u]
        # try all the nodes 
        return max([dfs(i) for i in range(n)])