diff --git a/2050-parallel-courses-iii/README.md b/2050-parallel-courses-iii/README.md new file mode 100644 index 0000000..7f310f0 --- /dev/null +++ b/2050-parallel-courses-iii/README.md @@ -0,0 +1,54 @@ +
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:
+ +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:
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+
++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 * 1040 <= relations.length <= min(n * (n - 1) / 2, 5 * 104)relations[j].length == 21 <= prevCoursej, nextCoursej <= nprevCoursej != nextCoursej[prevCoursej, nextCoursej] are unique.time.length == n1 <= time[i] <= 104