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Commit 72d7625

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added Course Schedule II and Reconstruct Itinerary (Medium)
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‎Medium/CourseSchedule2/README.md

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# Course Schedule II
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[Leetcode Link](https://leetcode.com/problems/course-schedule-ii/)
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## Problem:
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There are a total of `n` courses you have to take labelled from `0` to `n - 1`.
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Some courses may have `prerequisites`, for example, if `prerequisites[i] = [ai, bi]` this means you must take the course bi before the course ai.
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Given the total number of courses `numCourses` and a list of the `prerequisite` pairs, return the ordering of courses you should take to finish all courses.
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If there are many valid answers, return _any_ of them. If it is impossible to finish all courses, return _an empty array_.
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## Example:
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```
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Input: numCourses = 2, prerequisites = [[1,0]]
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Output: [0,1]
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Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So the correct course order is [0,1].
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```
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```
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Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
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Output: [0,2,1,3]
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Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
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So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].
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```
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```
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Input: numCourses = 1, prerequisites = []
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Output: [0]
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```
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## Note:
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- 1 <= numCourses <= 2000
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- 0 <= prerequisites.length <= numCourses \* (numCourses - 1)
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- prerequisites[i].length == 2
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- 0 <= ai, bi < numCourses
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- ai != bi
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- All the pairs [ai, bi] are distinct.

‎Medium/CourseSchedule2/solution.py

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from typing import List
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from collections import defaultdict, deque
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# this method does not work if there's cyclic dependency (ie. a course A with prereq B that has prereq A)
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class Solution:
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def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
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availableCourses = defaultdict(list)
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for course, prereq in prerequisites:
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availableCourses[course].append(prereq)
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def takeCourse(course):
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if not availableCourses[course]:
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if course not in courses:
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courses.append(course)
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return
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for c in availableCourses[course]:
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takeCourse(c)
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if course not in courses:
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courses.append(course)
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courses = list()
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for i in range(0, numCourses):
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takeCourse(i)
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return courses
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sol = Solution()
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numCourses = 6
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prereq = [[1, 0], [2, 0], [3, 1], [3, 2], [3, 5]]
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print(sol.findOrder(numCourses, prereq))
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from typing import List
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from collections import defaultdict
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class Solution:
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def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]:
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indegree = [0]*numCourses
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topological_list = list()
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queue = list()
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# associate indegree and prereq for each courses
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graph = defaultdict(list)
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for course, prereq in prerequisites:
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graph[prereq].append(course)
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indegree[course] += 1
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print(graph)
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# start by adding all courses from 0 that have indegree of 0
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for i in range(numCourses):
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# add all the courses with indegree of 0
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if indegree[i] == 0:
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queue.append(i)
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while queue:
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print("Indegree:", indegree)
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print("Queue:", queue)
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currentCourse = queue.pop(0)
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topological_list.append(currentCourse)
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# for each course that needs currentCourse as prereq, decrement indegree by 1 (since currentCourse is "completed"), when indegree of course == 0, then add to queue to be added in topological order
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for course in graph[currentCourse]:
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indegree[course] -= 1
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if indegree[course] == 0:
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queue.append(course)
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return topological_list if len(topological_list) == numCourses else []
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sol = Solution()
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numCourses = 3
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prereq = [[1, 0], [1, 2], [0, 1]]
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print(sol.findOrder(numCourses, prereq))

‎Medium/ReconstructItinerary/README.md

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# Reconstruct Itinerary
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[Leetcode Link](https://leetcode.com/problems/reconstruct-itinerary/)
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## Problem:
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Given a list of airline tickets represented by pairs of departure and arrival airports `[from, to]`, reconstruct the itinerary in order. All of the tickets belong to a man who departs from `JFK`. Thus, the itinerary must begin with `JFK`.
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## Example:
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```
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Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
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Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]
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```
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```
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Input: [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
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Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
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Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].
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But it is larger in lexical order.
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```
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## Note:
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1. If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string. For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].
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2. All airports are represented by three capital letters (IATA code).
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3. You may assume all tickets form at least one valid itinerary.
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4. One must use all the tickets once and only once.

‎Medium/ReconstructItinerary/solution.py

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from typing import List
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from collections import defaultdict
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class Solution:
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def findItinerary(self, tickets: List[List[str]]) -> List[str]:
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destinations = defaultdict(list)
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for departure, destination in sorted(tickets, reverse=True):
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destinations[departure] += destination,
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print(destinations)
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route = []
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def visit(airport):
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while destinations[airport]:
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visit(destinations[airport].pop())
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route.append(airport)
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visit('JFK')
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route.reverse()
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return route
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sol = Solution()
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tickets = [["JFK", "KUL"], ["JFK", "NRT"], ["NRT", "JFK"]]
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print(sol.findItinerary(tickets))

‎README.md

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- [Invert Binary Tree](Easy/InvertBinaryTree)
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- Medium
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- [Minimum Add to Make Parentheses Valid](Medium/MinimumAddtoMakeParenthesesValid)
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- [Distribute Coins in Binary Tree](Medium/DistributionCoinsInBinaryTree)
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- [Distribute Coins in Binary Tree](Medium/DistributeCoinsInBinaryTree)
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- [Find Minimum Number of Fibonacci Numbers Whose Sum is K](Medium/FindMinNumFibNumSumK)
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- [Find the Duplicate Number](Medium/FindDuplicateNumber)
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- [Unique Paths](Medium/UniquePaths)
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- [Number of Islands](Medium/NumberOfIslands)
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- [Reverse Linked List II](Medium/ReverseLinkedList2)
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- [Binary Tree Postorder Traversal](Medium/BinaryPostorder)
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- [Reconstruct Itinerary](Medium/ReconstructItinerary)
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- [Course Schedule II](Medium/CourseSchedule2)
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- Hard
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- [Maximum Score Words Formed by Letters](Hard/MaximumScoreWords)

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