2050. Parallel Courses III

Description

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 * 104
• 0 <= relations.length <= min(n * (n - 1) / 2, 5 * 104)
• relations[j].length == 2
• 1 <= prevCoursej, nextCoursej <= n
• prevCoursej != nextCoursej
• All the pairs [prevCoursej, nextCoursej] are unique.
• time.length == n
• 1 <= time[i] <= 104
• The given graph is a directed acyclic graph.

 

Solutions

Solution: Topological Sort + Dynamic Programming

• Time complexity: O(m+n)
• Space complexity: O(m+n)

 

JavaScript

/**
 * @param {number} n
 * @param {number[][]} relations
 * @param {number[]} time
 * @return {number}
 */
const minimumTime = function (n, relations, time) {
  const courses = Array.from({ length: n + 1 }, () => []);
  const indegree = Array.from({ length: n + 1 }, () => 0);
  const dp = Array.from({ length: n + 1 }, () => 0);
  let queue = [];

  for (const [prevCourse, nextCourse] of relations) {
    courses[prevCourse].push(nextCourse);
    indegree[nextCourse] += 1;
  }

  for (let course = 1; course <= n; course++) {
    if (indegree[course]) continue;

    queue.push(course);
    dp[course] = time[course - 1];
  }

  while (queue.length) {
    const nextQueue = [];

    for (const course of queue) {
      for (const nextCourse of courses[course]) {
        const months = dp[course] + time[nextCourse - 1];

        dp[nextCourse] = Math.max(months, dp[nextCourse]);
        indegree[nextCourse] -= 1;

        if (indegree[nextCourse] === 0) {
          nextQueue.push(nextCourse);
        }
      }
    }

    queue = nextQueue;
  }

  return Math.max(...dp);
};