Course Schedule Problem


Description

LeetCode Problem 207.

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] = [a_i, b_i] indicates that you must take course b_i first if you want to take course a_i.

  • 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 <= 10^5
  • 0 <= prerequisites.length <= 5000
  • prerequisites[i].length == 2
  • 0 <= a_i, b_i < numCourses
  • All the pairs prerequisites[i] are unique.


Sample C++ Code

Breadth-First Search

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class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        graph g = buildGraph(numCourses, prerequisites);
        vector<int> degrees = computeIndegrees(g);
        for (int i = 0; i < numCourses; i++) {
            int j = 0;
            for (; j < numCourses; j++) {
                if (!degrees[j]) {
                    break;
                }
            }
            if (j == numCourses) {
                return false;
            }
            degrees[j]--;
            for (int v : g[j]) {
                degrees[v]--;
            }
        }
        return true;
    }
private:
    typedef vector<vector<int>> graph;
    
    graph buildGraph(int numCourses, vector<pair<int, int>>& prerequisites) {
        graph g(numCourses);
        for (auto p : prerequisites) {
            g[p.second].push_back(p.first);
        }
        return g;
    }
    
    vector<int> computeIndegrees(graph& g) {
        vector<int> degrees(g.size(), 0);
        for (auto adj : g) {
            for (int v : adj) {
                degrees[v]++;
            }
        }
        return degrees;
    }
};

Depth-First Search

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class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        graph g = buildGraph(numCourses, prerequisites);
        vector<bool> todo(numCourses, false), done(numCourses, false);
        for (int i = 0; i < numCourses; i++) {
            if (!done[i] && !acyclic(g, todo, done, i)) {
                return false;
            }
        }
        return true;
    }
private:
    typedef vector<vector<int>> graph;
    
    graph buildGraph(int numCourses, vector<pair<int, int>>& prerequisites) {
        graph g(numCourses);
        for (auto p : prerequisites) {
            g[p.second].push_back(p.first);
        }
        return g;
    }
    
    bool acyclic(graph& g, vector<bool>& todo, vector<bool>& done, int node) {
        if (todo[node]) {
            return false;
        }
        if (done[node]) {
            return true;
        }
        todo[node] = done[node] = true;
        for (int v : g[node]) {
            if (!acyclic(g, todo, done, v)) {
                return false;
            }
        }
        todo[node] = false;
        return true;
    }
};

Topological Sort

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bool canFinish(int n, vector<pair<int, int>>& pre) {
    vector<vector<int>> adj(n, vector<int>());
    vector<int> degree(n, 0);
    for (auto &p: pre) {
        adj[p.second].push_back(p.first);
        degree[p.first]++;
    }
    queue<int> q;
    for (int i = 0; i < n; i++)
        if (degree[i] == 0) q.push(i);
    while (!q.empty()) {
        int curr = q.front(); q.pop(); n--;
        for (auto next: adj[curr])
            if (--degree[next] == 0) q.push(next);
    }
    return n == 0;
}




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