207. Course Schedule

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207. Course Schedule

Description

Difficulty: Medium

Related Topics: Depth-First Search, Breadth-First Search, Graph, Topological Sort

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 true if you can finish all courses. Otherwise, return false.

Example 1:

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Input: numCourses = 2, prerequisites = [[1,0]]
Output: true
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:

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Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
Output: false
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
  • 0 <= prerequisites.length <= 5000
  • prerequisites[i].length == 2
  • 0 <= ai, bi < numCourses
  • All the pairs prerequisites[i] are unique.

Hints/Nodes

  • 2023/09/01
  • Traverse the graph and check if there’s a cycle
  • 0x3F’s solution(checked)

Solution

Language: C++

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class Solution {
public:
    vector<vector<int>> graph;
    vector<bool> visited;
    vector<bool> onPath;
    int can = true;

    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        graph = vector<vector<int>>(numCourses, vector<int>());
        visited = vector<bool>(numCourses, false);
        onPath = vector<bool>(numCourses, false);

        for (auto pair : prerequisites) {
            int from = pair[0], to = pair[1];
            graph[from].push_back(to);
        }

        for (int i = 0; i < numCourses; i++) {
            traverse(i);
        }
        return can;
    }

    void traverse(int node) {
        if (onPath[node]) {
            can = false;
        }

        if (!can || visited[node]) {
            return;
        }

        onPath[node] = true;
        visited[node] = true;

        for (int to : graph[node]) {
            traverse(to);
        }

        onPath[node] = false;
    }
};