# Swift solution using BFS with full comments

• This is a solution based on the best of several approaches discussed. It uses BFS (note the queue) and should make sense to most readers.

``````class Solution {
func findOrder(_ numCourses: Int, _ prerequisites: [[Int]]) -> [Int] {
// sanity check
if numCourses == 0 {
return []
}

// create count prereqs by course
var coursePrereqCount = [Int](repeating: 0, count: numCourses)
for i in 0..<prerequisites.count {
coursePrereqCount[prerequisites[i][0]] += 1
}

// our course order result
var courseOrder: [Int] = []

// find courses with zero prereqs
// - add to the course taken order
// - queue taken course to count against prereqs
var courseTakenQueue: [Int] = []
for i in 0..<numCourses {
// if no prereqs,
if coursePrereqCount[i] == 0 {
courseOrder.append(i)
courseTakenQueue.append(i)
}
}

// when the queue of satisfied prereqs is empty, we're done
while !courseTakenQueue.isEmpty{
let courseTaken = courseTakenQueue.removeFirst()
// decrements the prereq count for any courses dependent on the course
for i in 0..<prerequisites.count {
if prerequisites[i][1] == courseTaken {
// found a dependency
let course = prerequisites[i][0]
coursePrereqCount[course] -= 1
if coursePrereqCount[course] == 0 {
// no more prereqs for this course
// - add to the course taken order
// - queue taken course to count against prereqs
courseOrder.append(course)
courseTakenQueue.append(course)
}
}
}
}

// if we have an order which includes all numCourses, return it, otherwise empty
return courseOrder.count == numCourses ? courseOrder : []
}
}
``````

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