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2050-parallel-courses-iii/2050-parallel-courses-iii.py

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+class Solution:
+    def minimumTime(self, n: int, relations: List[List[int]], time: List[int]) -> int:
+        """
+        Complexities:
+        Time:  O(n^2)
+        Space: O(r)
+
+        where n = n, r = len(relations)
+        """
+
+        # courses been taken array
+        # - O(r) time / O(r) space
+        hasPrereq = [False] * n
+        prereqMap = [set() for i in range(n)]
+        leadsToMap = [set() for i in range(n)]
+        for prevCourse, nextCourse in relations:
+            hasPrereq[nextCourse - 1] = True
+            prereqMap[nextCourse - 1].add(prevCourse)
+            leadsToMap[prevCourse - 1].add(nextCourse)
+
+        # O(n) space
+        timeUntilComplete = [-1] * n
+        coursesCalculated = 0
+        maxOverallTimeFound = 0
+        toResolve = set()
+
+        # calculate timeUntilComplete for no prereq courses
+        # - O(n) time
+        for i in range(n):
+            if not hasPrereq[i]:
+                timeUntilComplete[i] = time[i]
+                maxOverallTimeFound = max(maxOverallTimeFound, time[i])
+                coursesCalculated += 1
+                toResolve.update(leadsToMap[i])
+
+        # find all timeUntilComplete
+        # - if all prereq times are known, then calc
+        # - O(n^2) time
+        while len(toResolve) > 0:
+            nextCourse = toResolve.pop()
+            maxPrereqTime = -1
+            for prereq in prereqMap[nextCourse - 1]:
+                if timeUntilComplete[prereq - 1] == -1:
+                    maxPrereqTime = -1
+                    break
+                maxPrereqTime = max(maxPrereqTime, timeUntilComplete[prereq - 1])
+
+            if maxPrereqTime > -1:
+                timeUntilComplete[nextCourse - 1] = maxPrereqTime + time[nextCourse - 1]
+                maxOverallTimeFound = max(maxOverallTimeFound, timeUntilComplete[nextCourse - 1])
+                coursesCalculated += 1
+                toResolve.update(leadsToMap[nextCourse - 1])
+        
+
+
+        return maxOverallTimeFound

2050-parallel-courses-iii/README.md

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+<h2><a href="https://leetcode.com/problems/parallel-courses-iii">2050. Parallel Courses III</a></h2><h3>Hard</h3><hr><p>You are given an integer <code>n</code>, which indicates that there are <code>n</code> courses labeled from <code>1</code> to <code>n</code>. You are also given a 2D integer array <code>relations</code> where <code>relations[j] = [prevCourse<sub>j</sub>, nextCourse<sub>j</sub>]</code> denotes that course <code>prevCourse<sub>j</sub></code> has to be completed <strong>before</strong> course <code>nextCourse<sub>j</sub></code> (prerequisite relationship). Furthermore, you are given a <strong>0-indexed</strong> integer array <code>time</code> where <code>time[i]</code> denotes how many <strong>months</strong> it takes to complete the <code>(i+1)<sup>th</sup></code> course.</p>
+
+<p>You must find the <strong>minimum</strong> number of months needed to complete all the courses following these rules:</p>
+
+<ul>
+	<li>You may start taking a course at <strong>any time</strong> if the prerequisites are met.</li>
+	<li><strong>Any number of courses</strong> can be taken at the <strong>same time</strong>.</li>
+</ul>
+
+<p>Return <em>the <strong>minimum</strong> number of months needed to complete all the courses</em>.</p>
+
+<p><strong>Note:</strong> The test cases are generated such that it is possible to complete every course (i.e., the graph is a directed acyclic graph).</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<strong><img alt="" src="https://assets.leetcode.com/uploads/2021/10/07/ex1.png" style="width: 392px; height: 232px;" /></strong>
+
+<pre>
+<strong>Input:</strong> n = 3, relations = [[1,3],[2,3]], time = [3,2,5]
+<strong>Output:</strong> 8
+<strong>Explanation:</strong> 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.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<strong><img alt="" src="https://assets.leetcode.com/uploads/2021/10/07/ex2.png" style="width: 500px; height: 365px;" /></strong>
+
+<pre>
+<strong>Input:</strong> n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], time = [1,2,3,4,5]
+<strong>Output:</strong> 12
+<strong>Explanation:</strong> 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.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= relations.length &lt;= min(n * (n - 1) / 2, 5 * 10<sup>4</sup>)</code></li>
+	<li><code>relations[j].length == 2</code></li>
+	<li><code>1 &lt;= prevCourse<sub>j</sub>, nextCourse<sub>j</sub> &lt;= n</code></li>
+	<li><code>prevCourse<sub>j</sub> != nextCourse<sub>j</sub></code></li>
+	<li>All the pairs <code>[prevCourse<sub>j</sub>, nextCourse<sub>j</sub>]</code> are <strong>unique</strong>.</li>
+	<li><code>time.length == n</code></li>
+	<li><code>1 &lt;= time[i] &lt;= 10<sup>4</sup></code></li>
+	<li>The given graph is a directed acyclic graph.</li>
+</ul>