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Question
Formatted question description: https://leetcode.ca/all/210.html
210 Course Schedule II
There are a total of n courses you have to take, labeled from 0 to n-1.
Some courses may have prerequisites, for example to take course 0 you have to first take course 1,
which is expressed as a pair: [0,1]
Given the total number of courses and a list of prerequisite pairs,
return the ordering of courses you should take to finish all courses.
There may be multiple correct orders, you just need to return one of them.
If it is impossible to finish all courses, return an empty array.
Example 1:
Input: 2, [[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: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [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] .
Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices.
Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.
@tag-graph
Algorithm
Topological Sort
of directed graph, in this way, the difficulty increases,
Due to the basis of the previous one, with a slight modification, every time an array is removed from the queue, it is stored in the result.
In the end, if there is a cycle in the directed graph, the number of elements in the result is not equal to the total number of courses, then we clear the result.
Code
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import java.util.*; public class Course_Schedule_II { public static void main(String[] args) { Course_Schedule_II out = new Course_Schedule_II(); Solution s = out.new Solution(); System.out.println(Arrays.toString(s.findOrder(2, new int[][]{ {1,0} }))); } class Solution { public int[] findOrder(int numCourses, int[][] prerequisites) { if(numCourses <= 0) { return new int[0]; } if(prerequisites == null) { return new int[numCourses]; } int[] inDegree = new int[numCourses]; for (int[] each: prerequisites) { inDegree[each[0]]++; } Queue<Integer> q = new LinkedList<>(); for (int i = 0; i < inDegree.length; i++) { if (inDegree[i] == 0) { q.offer(i); } } List<Integer> result = new ArrayList<>(); while (!q.isEmpty()) { int current = q.poll(); result.add(current); for (int[] each: prerequisites) { if (current == each[1]) { inDegree[each[0]]--; if (inDegree[each[0]] == 0) { q.offer(each[0]); } } } } if(!(result.size() == numCourses)) { return new int[0]; } return result.stream().mapToInt(i -> i).toArray(); } } class Solution2 { public int[] findOrder(int numCourses, int[][] prerequisites) { if (numCourses <= 0 || prerequisites == null) { return new int[0]; } // construct graph int[] incomingCount = new int[numCourses]; List<List<Integer>> g = new ArrayList<>(numCourses); constructGraph(numCourses, prerequisites, g, incomingCount); // bfs or dfs return bfs(numCourses, g, incomingCount); } private int[] bfs(int numCourses, List<List<Integer>> g, int[] incomingCount) { int[] result = new int[numCourses]; Queue<Integer> q = new LinkedList<>(); // search for starting point for (int i = 0; i < numCourses; i++) { if (incomingCount[i] == 0) { q.offer(i); // could be multiple node without incoming } } int resultPointer = 0; while (!q.isEmpty()) { int node = q.poll(); result[resultPointer++] = node; for (int each: g.get(node)) { incomingCount[each]--; if (incomingCount[each] == 0) { q.offer(each); } } } // @note: how to check valid result return resultPointer == numCourses ? result : new int[0]; } private void constructGraph(int numCourses, int[][] prerequisites, List<List<Integer>> g, int[] incomingCount) { // empty graph for (int i = 0; i < numCourses; i++) { g.add(new ArrayList<Integer>()); } for (int[] pair: prerequisites) { incomingCount[pair[0]]++; g.get(pair[1]).add(pair[0]); } } } } ############ class Solution { public int[] findOrder(int numCourses, int[][] prerequisites) { List<Integer>[] g = new List[numCourses]; Arrays.setAll(g, k -> new ArrayList<>()); int[] indeg = new int[numCourses]; for (var p : prerequisites) { int a = p[0], b = p[1]; g[b].add(a); ++indeg[a]; } Deque<Integer> q = new ArrayDeque<>(); for (int i = 0; i < numCourses; ++i) { if (indeg[i] == 0) { q.offer(i); } } int[] ans = new int[numCourses]; int cnt = 0; while (!q.isEmpty()) { int i = q.poll(); ans[cnt++] = i; for (int j : g[i]) { if (--indeg[j] == 0) { q.offer(j); } } } return cnt == numCourses ? ans : new int[0]; } }
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// OJ: https://leetcode.com/problems/course-schedule-ii/ // Time: O(V + E) // Space: O(V + E) class Solution { public: vector<int> findOrder(int n, vector<vector<int>>& E) { vector<vector<int>> G(n); vector<int> indegree(n), ans; for (auto &e : E) { G[e[1]].push_back(e[0]); indegree[e[0]]++; } queue<int> q; for (int i = 0; i < n; ++i) { if (indegree[i] == 0) q.push(i); } while (q.size()) { int u = q.front(); q.pop(); ans.push_back(u); for (int v : G[u]) { if (--indegree[v] == 0) q.push(v); } } return ans.size() == n ? ans : vector<int>{}; } };
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class Solution: def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]: g = defaultdict(list) indeg = [0] * numCourses for a, b in prerequisites: g[b].append(a) indeg[a] += 1 q = deque([i for i, v in enumerate(indeg) if v == 0]) ans = [] while q: i = q.popleft() ans.append(i) # assumption is only one path, as in question 'You may assume that there are no duplicate edges in the input prerequisites.' for j in g[i]: indeg[j] -= 1 if indeg[j] == 0: q.append(j) return ans if len(ans) == numCourses else [] ############ class Solution(object): def findOrder(self, numCourses, prerequisites): """ :type numCourses: int :type prerequisites: List[List[int]] :rtype: List[int] """ def dfs(start, visited, graph, ans): visited[start] = 1 for nbr in graph[start]: if visited[nbr] == 1: return False if visited[nbr] != 0: continue if dfs(nbr, visited, graph, ans) == False: return False ans.append(start) visited[start] = 2 return True graph = [[] for _ in range(0, numCourses)] ans = [] for pre in prerequisites: start, end = pre graph[start].append(end) visited = [0 for _ in range(0, numCourses)] for pre in prerequisites: start, end = pre if visited[start] != 0: continue if dfs(start, visited, graph, ans) == False: return [] for i in range(0, numCourses): if visited[i] == 0: ans.append(i) return ans
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func findOrder(numCourses int, prerequisites [][]int) []int { g := make([][]int, numCourses) indeg := make([]int, numCourses) for _, p := range prerequisites { a, b := p[0], p[1] g[b] = append(g[b], a) indeg[a]++ } q := []int{} for i, v := range indeg { if v == 0 { q = append(q, i) } } ans := []int{} for len(q) > 0 { i := q[0] q = q[1:] ans = append(ans, i) for _, j := range g[i] { indeg[j]-- if indeg[j] == 0 { q = append(q, j) } } } if len(ans) == numCourses { return ans } return []int{} }
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function findOrder(numCourses: number, prerequisites: number[][]): number[] { let g = Array.from({ length: numCourses }, () => []); let indeg = new Array(numCourses).fill(0); for (let [a, b] of prerequisites) { g[b].push(a); ++indeg[a]; } let q = []; for (let i = 0; i < numCourses; ++i) { if (!indeg[i]) { q.push(i); } } let ans = []; while (q.length) { const i = q.shift(); ans.push(i); for (let j of g[i]) { if (--indeg[j] == 0) { q.push(j); } } } return ans.length == numCourses ? ans : []; }
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public class Solution { public int[] FindOrder(int numCourses, int[][] prerequisites) { var g = new List<int>[numCourses]; for (int i = 0; i < numCourses; ++i) { g[i] = new List<int>(); } var indeg = new int[numCourses]; foreach (var p in prerequisites) { int a = p[0], b = p[1]; g[b].Add(a); ++indeg[a]; } var q = new Queue<int>(); for (int i = 0; i < numCourses; ++i) { if (indeg[i] == 0) q.Enqueue(i); } var ans = new int[numCourses]; var cnt = 0; while (q.Count > 0) { int i = q.Dequeue(); ans[cnt++] = i; foreach (int j in g[i]) { if (--indeg[j] == 0) q.Enqueue(j); } } return cnt == numCourses ? ans : new int[0]; } }