README.md

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.

Solutions (Rust)

1. Solution

use std::collections::HashMap;

impl Solution {
    pub fn can_finish(num_courses: i32, prerequisites: Vec<Vec<i32>>) -> bool {
        let mut indegrees = vec![0; num_courses as usize];
        let mut follows = HashMap::new();
        let mut courses = vec![];

        for prerequisite in prerequisites {
            indegrees[prerequisite[0] as usize] += 1;
            follows
                .entry(prerequisite[1] as usize)
                .or_insert(vec![])
                .push(prerequisite[0] as usize);
        }

        for b in 0..num_courses as usize {
            if indegrees[b] == 0 {
                courses.push(b);
            }
        }

        while let Some(b) = courses.pop() {
            for a in follows.remove(&b).unwrap_or(vec![]) {
                indegrees[a] -= 1;
                if indegrees[a] == 0 {
                    courses.push(a);
                }
            }
        }

        indegrees.iter().all(|&x| x == 0)
    }
}