solutions: 2050 - Parallel Courses III (Hard)

solutions/2000-2099/2050-parallel-courses-iii-hard.md

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+---
+description: 'Author: @wingkwong | https://leetcode.com/problems/parallel-courses-iii/'
+tags: [Array, Dynamic Programming, Graph, Topological Sort]
+---
+
+# 2050 - Parallel Courses III (Hard) 
+
+## Problem Link
+
+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. 
+
+<Tabs>
+<TabItem value="py" label="Python">
+<SolutionAuthor name="@wingkwong"/>
+
+```py
+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)])
+```
+
+</TabItem>
+</Tabs>