From ca73d3235c11c4700946a315980641eff912545d Mon Sep 17 00:00:00 2001 From: Prad Nukala Date: Tue, 22 Sep 2026 14:24:38 -0400 Subject: [PATCH] feat(work): add Course Schedule II solution and clean up Course Schedule I implementation --- .../topological-sort/207-course-schedule.mdx | 22 +++++++ .../210-course-schedule-ii.mdx | 39 +++++++++++ .../Depth-First Search/207.course-schedule.py | 42 +++++++++--- .../210.course-schedule-ii.py | 64 +++++++++++++++++++ 4 files changed, 157 insertions(+), 10 deletions(-) create mode 100644 apps/docs/content/topological-sort/210-course-schedule-ii.mdx create mode 100644 work/1/Medium/Depth-First Search/210.course-schedule-ii.py diff --git a/apps/docs/content/topological-sort/207-course-schedule.mdx b/apps/docs/content/topological-sort/207-course-schedule.mdx index 1c51120..ec82721 100644 --- a/apps/docs/content/topological-sort/207-course-schedule.mdx +++ b/apps/docs/content/topological-sort/207-course-schedule.mdx @@ -31,4 +31,26 @@ sidebar: ```py class Solution: def canFinish(self, numCourses: int, prerequisites: list[list[int]]) -> bool: + indegree = [0] * numCourses + adj = [[] for x in range(numCourses)] + for prereq in prerequisites: + adj[prereq[1]].append(prereq[0]) + indegree[prereq[0]] += 1 + + queue = [] + + for i in range(numCourses): + if indegree[i] == 0: + queue.append(i) + + visited = 0 + while queue: + node = queue.pop(0) + visited += 1 + for neighbor in adj[node]: + indegree[neighbor] -= 1 + if indegree[neighbor] == 0: + queue.append(neighbor) + + return numCourses == visited ``` diff --git a/apps/docs/content/topological-sort/210-course-schedule-ii.mdx b/apps/docs/content/topological-sort/210-course-schedule-ii.mdx new file mode 100644 index 0000000..c2a99f1 --- /dev/null +++ b/apps/docs/content/topological-sort/210-course-schedule-ii.mdx @@ -0,0 +1,39 @@ +--- +title: '210. Course Schedule II' +description: There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai +sidebar: + label: 'Course Schedule II' + badge: 'Medium' +--- + +Depth-First Search + +### Example 1: +- Input: `numCourses = 2, prerequisites = [[1,0]]` +- Output: `[0,1]` +- 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]`. + +### Example 2: +- Input: `numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]` +- Output: `[0,2,1,3]` +- 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`. So one correct course order is `[0,1,2,3]`. Another correct ordering is `[0,2,1,3]`. + +### Example 3: +- Input: `numCourses = 1, prerequisites = []` +- Output: `[0]` + +### Constraints: + +- `1 <= numCourses <= 2000` +- `0 <= prerequisites.length <= numCourses * (numCourses - 1)` +- `prerequisites[i].length == 2` +- `0 <= ai, bi < numCourses` +- `ai != bi` +- All the pairs [ai, bi] are distinct. + +## Solution + +```py +class Solution: + def findOrder(self, numCourses: int, prerequisites: list[list[int]]) -> list[int]: +``` diff --git a/work/1/Medium/Depth-First Search/207.course-schedule.py b/work/1/Medium/Depth-First Search/207.course-schedule.py index c617af0..0ff5f32 100644 --- a/work/1/Medium/Depth-First Search/207.course-schedule.py +++ b/work/1/Medium/Depth-First Search/207.course-schedule.py @@ -15,38 +15,60 @@ have to first take course 1. Return true if you can finish all courses. Otherwise, return false. - + Example 1: Input: numCourses = 2, prerequisites = [[1,0]] Output: true -Explanation: There are a total of 2 courses to take. +Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible. Example 2: Input: numCourses = 2, prerequisites = [[1,0],[0,1]] Output: false -Explanation: There are a total of 2 courses to take. +Explanation: There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible. - + Constraints: - • 1 <= numCourses <= 2000 + • 1 <= numCourses <= 2000 - • 0 <= prerequisites.length <= 5000 + • 0 <= prerequisites.length <= 5000 - • prerequisites[i].length == 2 + • prerequisites[i].length == 2 - • 0 <= ai, bi < numCourses + • 0 <= ai, bi < numCourses - • All the pairs prerequisites[i] are unique. + • All the pairs prerequisites[i] are unique. """ + class Solution: def canFinish(self, numCourses: int, prerequisites: list[list[int]]) -> bool: - + indegree = [0] * numCourses + adj = [[] for x in range(numCourses)] + for prereq in prerequisites: + adj[prereq[1]].append(prereq[0]) + indegree[prereq[0]] += 1 + + queue = [] + + for i in range(numCourses): + if indegree[i] == 0: + queue.append(i) + + visited = 0 + while queue: + node = queue.pop(0) + visited += 1 + for neighbor in adj[node]: + indegree[neighbor] -= 1 + if indegree[neighbor] == 0: + queue.append(neighbor) + + return numCourses == visited diff --git a/work/1/Medium/Depth-First Search/210.course-schedule-ii.py b/work/1/Medium/Depth-First Search/210.course-schedule-ii.py new file mode 100644 index 0000000..fde7a6b --- /dev/null +++ b/work/1/Medium/Depth-First Search/210.course-schedule-ii.py @@ -0,0 +1,64 @@ +""" +210. Course Schedule II +Difficulty: Medium +https://leetcode.com/problems/course-schedule-ii/ + +────────────────────────────────────────────────── + +There are a total of numCourses courses you have to take, labeled +from 0 to numCourses - 1. You are given an array prerequisites where +prerequisites[i] = [ai, bi] indicates that you must take course bi +first if you want to take course ai. + +• For example, the pair [0, 1], indicates that to take course 0 you +have to first take course 1. + +Return the ordering of courses you should take to finish all courses. +If there are many valid answers, return any of them. If it is +impossible to finish all courses, return an empty array. + + + +Example 1: + +Input: numCourses = 2, prerequisites = [[1,0]] +Output: [0,1] +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]. + +Example 2: + +Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]] +Output: [0,2,1,3] +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. +So one correct course order is [0,1,2,3]. Another correct ordering is +[0,2,1,3]. + +Example 3: + +Input: numCourses = 1, prerequisites = [] +Output: [0] + + + +Constraints: + + • 1 <= numCourses <= 2000 + + • 0 <= prerequisites.length <= numCourses * (numCourses - 1) + + • prerequisites[i].length == 2 + + • 0 <= ai, bi < numCourses + + • ai != bi + + • All the pairs [ai, bi] are distinct. +""" + +class Solution: + def findOrder(self, numCourses: int, prerequisites: list[list[int]]) -> list[int]: +