Leetcode 207: Course Schedule

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:

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:

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.

Problem Analysis:

After read the problem, we can know if we can finish courses, there must be no prerequisites loop cycle. As show in example 2.

So our task is to check whether some loop cycle happens.

It is a classic dfs problem.

Solution

class Solution {
public:
    bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
        for (const auto& item : prerequisites)
            data_[item.front()].insert(item.back());
        vector<bool> visited(numCourses, false);
        vector<bool> on_stack(numCourses, false);
        for (int i = 0; i < numCourses; ++i) {
            if (!visited[i]) {
                if (isLoop(i, visited, on_stack)) return false;
            }
        }
        
        return true;
    }
    
    bool isLoop(int index, vector<bool>& visited, vector<bool>& on_stack) {
        visited[index] = true;
        on_stack[index] = true;
        
        auto it = data_.find(index);
        if (it != data_.end()) {
            for (const auto& item : it->second) {
                if (!visited[item]) {
                    if (isLoop(item, visited, on_stack)) return true;
                } else if (on_stack[item]) {
                    return true;
                }
            }
        }
        
        on_stack[index] = false;
        
        return false;
    }
private:
    
    unordered_map<int, unordered_set<int>> data_;
};

time complexity is O(n)

Space complexity is O(n)