leetcode-210-course-schedule-ii.swift

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https://leetcode.com/problems/course-schedule-ii/


There are a total of numCourses courses you have to take, labeled from 0 to numCourses-1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?

 

Example 1:

There are a total of n courses you have to take, labeled from 0 to n-1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

Example 1:

Input: 2, [[1,0]] 
Output: [0,1]
Explanation: There are a total of 2 courses to take. To take course 1 you should have finished   
             course 0. So the correct course order is [0,1] .
Example 2:

Input: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both     
             courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. 
             So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3] .
Note:

The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
You may assume that there are no duplicate edges in the input prerequisites.

*/

/**
Big O Annotation
Time complexity O(n) where n is the amount of elements from 0...numCourses.
Space complexity O(n) where n is the amount of elements from 0...numCourses.
*/
func canFinish(_ numCourses: Int, _ prerequisites: [[Int]]) -> Bool {
        
    /**
    Create a graph of empty arrays for each course.
    The graph represents each courses goal.

    For example: I need this course to be able to do XY.
    I need to finish course 2 to be able to do courses [3,4]
    */
    var graph: [[Int]] = Array(repeating: [], count: numCourses)
    
    // Iterate through the prerequisites to fill the graph
    for element in prerequisites {

        // Set for each requirement the course you can do once done
        graph[element[1]].append(element[0])
    }
    
    /**
    Init a visiting and visited array.
    Both representing the total amount of courses and both
    implemented false to begin with.
    */
    var visiting: [Bool] =  Array(repeating: false, count: numCourses)
    var visited: [Bool] =  Array(repeating: false, count: numCourses)
    var result: [Int] = []
    
    // Iterate through each course
    for course in 0..<numCourses {
        
        // And check if we can find a circle
        if foundCircle(in: graph, at: course, visiting: &visiting, visited: &visited, result: &result) {
            return []
        }
    }
    
    /**
    The result array is in descending order 
    starting from the last node of the DFS.
    So we need to reverse it.
    */
    return result.reversed()
}

private func foundCircle(in graph: [[Int]], 
                        at course: Int, 
                        visiting: inout [Bool], 
                        visited: inout [Bool],
                        result: inout [Int]) -> Bool {
    
    /**
    If we are currently visiting the neighbors of this course,
    it means that we have found a circle.
    Example: Course 0 requires course 1, and course 1 course 0.
    */
    if visiting[course] == true {
        return true 
    }
    
    /**
    If we have already visited this node, we know it's fine
    and we can return true. There can't be a circle
    otherwise we would have never successfully visited it.
    */
    if visited[course] == true {
        return false
    }
    
    /**
    Set the current node to visiting,
    to be able to catch it above on the recursion
    */
    visiting[course] = true
    
    // Iterate through the current node's neighbors
    for neighbor in graph[course] {
        
        // If we find a circle, return true
        if foundCircle(in: graph, at: neighbor, visiting: &visiting, visited: &visited, result: &result) {
            return true 
        }
    }

    // Append the current node
    result.append(course)
    
    // Set the visiting back to false
    visiting[course] = false

    // And update the visitation
    visited[course] = true
    return false
}