Question

207. Course Schedule

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Description

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

Solution 1: Topological Sorting

For this problem, we can consider the courses as nodes in a graph, and prerequisites as edges in the graph. Thus, we can transform this problem into determining whether there is a cycle in the directed graph.

Specifically, we can use the idea of topological sorting. For each node with an in-degree of $0$, we reduce the in-degree of its out-degree nodes by $1$, until all nodes have been traversed.

If all nodes have been traversed, it means there is no cycle in the graph, and we can complete all courses; otherwise, we cannot complete all courses.

The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the number of courses and prerequisites respectively.

class Solution:
  def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool:
    g = defaultdict(list)
    indeg = [0] * numCourses
    for a, b in prerequisites:
      g[b].append(a)
      indeg[a] += 1
    cnt = 0
    q = deque(i for i, x in enumerate(indeg) if x == 0)
    while q:
      i = q.popleft()
      cnt += 1
      for j in g[i]:
        indeg[j] -= 1
        if indeg[j] == 0:
          q.append(j)
    return cnt == numCourses

class Solution {
  public boolean canFinish(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 cnt = 0;
    while (!q.isEmpty()) {
      int i = q.poll();
      ++cnt;
      for (int j : g[i]) {
        if (--indeg[j] == 0) {
          q.offer(j);
        }
      }
    }
    return cnt == numCourses;
  }
}

class Solution {
public:
  bool canFinish(int numCourses, vector<vector<int>>& prerequisites) {
    vector<vector<int>> g(numCourses);
    vector<int> indeg(numCourses);
    for (auto& p : prerequisites) {
      int a = p[0], b = p[1];
      g[b].push_back(a);
      ++indeg[a];
    }
    queue<int> q;
    for (int i = 0; i < numCourses; ++i) {
      if (indeg[i] == 0) {
        q.push(i);
      }
    }
    int cnt = 0;
    while (!q.empty()) {
      int i = q.front();
      q.pop();
      ++cnt;
      for (int j : g[i]) {
        if (--indeg[j] == 0) {
          q.push(j);
        }
      }
    }
    return cnt == numCourses;
  }
};

func canFinish(numCourses int, prerequisites [][]int) bool {
  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, x := range indeg {
    if x == 0 {
      q = append(q, i)
    }
  }
  cnt := 0
  for len(q) > 0 {
    i := q[0]
    q = q[1:]
    cnt++
    for _, j := range g[i] {
      indeg[j]--
      if indeg[j] == 0 {
        q = append(q, j)
      }
    }
  }
  return cnt == numCourses
}

function canFinish(numCourses: number, prerequisites: number[][]): boolean {
  const g: number[][] = new Array(numCourses).fill(0).map(() => []);
  const indeg: number[] = new Array(numCourses).fill(0);
  for (const [a, b] of prerequisites) {
    g[b].push(a);
    indeg[a]++;
  }
  const q: number[] = [];
  for (let i = 0; i < numCourses; ++i) {
    if (indeg[i] == 0) {
      q.push(i);
    }
  }
  let cnt = 0;
  while (q.length) {
    const i = q.shift()!;
    cnt++;
    for (const j of g[i]) {
      if (--indeg[j] == 0) {
        q.push(j);
      }
    }
  }
  return cnt == numCourses;
}

use std::collections::VecDeque;

impl Solution {
  #[allow(dead_code)]
  pub fn can_finish(num_course: i32, prerequisites: Vec<Vec<i32>>) -> bool {
    let num_course = num_course as usize;
    // The graph representation
    let mut graph: Vec<Vec<i32>> = vec![vec![]; num_course];
    // Record the in degree for each node
    let mut in_degree_vec: Vec<i32> = vec![0; num_course];
    let mut q: VecDeque<usize> = VecDeque::new();
    let mut count = 0;

    // Initialize the graph & in degree vector
    for p in &prerequisites {
      let (from, to) = (p[0], p[1]);
      graph[from as usize].push(to);
      in_degree_vec[to as usize] += 1;
    }

    // Enqueue the first batch of nodes with in degree 0
    for i in 0..num_course {
      if in_degree_vec[i] == 0 {
        q.push_back(i);
      }
    }

    // Begin the traverse & update through the graph
    while !q.is_empty() {
      // Get the current node index
      let index = q.front().unwrap().clone();
      // This course can be finished
      count += 1;
      q.pop_front();
      for i in &graph[index] {
        // Update the in degree for the current node
        in_degree_vec[*i as usize] -= 1;
        // See if can be enqueued
        if in_degree_vec[*i as usize] == 0 {
          q.push_back(*i as usize);
        }
      }
    }

    count == num_course
  }
}

public class Solution {
  public bool CanFinish(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 cnt = 0;
    while (q.Count > 0) {
      int i = q.Dequeue();
      ++cnt;
      foreach (int j in g[i]) {
        if (--indeg[j] == 0) {
          q.Enqueue(j);
        }
      }
    }
    return cnt == numCourses;
  }
}