2050-parallel-courses-iii.rb

# frozen_string_literal: true

# 2050. Parallel Courses III
# Hard
# https://leetcode.com/problems/parallel-courses-iii

=begin
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.
=end

# @param {Integer} n
# @param {Integer[][]} relations
# @param {Integer[]} time
# @return {Integer}
def minimum_time(n, relations, times)
  prevs = Array.new(n + 1) { [] }
  nexts = Array.new(n + 1) { [] }

  times.unshift(0)

  relations.each do |a, b|
    prevs[b] << a
    nexts[a] << b
  end

  taken = Set[]
  time = 0

  min_heap = []
  starts = (1..n).to_a.select { |val| prevs[val].length == 0 }
  starts.each { |start| min_heap << [times[start], start] }
  min_heap.sort!

  until min_heap.empty?
    val, num = min_heap.shift
    time = val
    taken.add(num)

    nexts[num].each do |adj|
      if prevs[adj].all? { |node| taken.include?(node) }
        idx = min_heap.bsearch_index do |subarr|
            subarr[0] >= time + times[adj]
          end || min_heap.length

        min_heap.insert(idx, [times[adj] + time, adj])
      end
    end
  end

  time
end

# **************** #
#       TEST       #
# **************** #

require "test/unit"
class Test_minimum_time < Test::Unit::TestCase
  def test_
    assert_equal 8, minimum_time(3, [[1, 3], [2, 3]], [3, 2, 5])
    assert_equal 12, minimum_time(5, [[1, 5], [2, 5], [3, 5], [3, 4], [4, 5]], [1, 2, 3, 4, 5])
  end
end