Find the ordering of tasks from given dependencies
There are a total of n tasks you have to pick, labeled from 0 to n-1. Some tasks may have prerequisites tasks, for example to pick task 0 you have to first finish tasks 1, which is expressed as a pair: [0, 1] Given the total number of tasks and a list of prerequisite pairs, return the ordering of tasks you should pick to finish all tasks. There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all tasks, return an empty array.
Examples:
Input: 2, [[1, 0]]
Output: [0, 1]
Explanation: There are a total of 2 tasks to pick. To pick task 1 you should have finished task 0. So the correct task order is [0, 1] .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 tasks to pick. To pick task 3 you should have finished both tasks 1 and 2. Both tasks 1 and 2 should be pick after you finished task 0. So one correct task order is [0, 1, 2, 3]. Another correct ordering is [0, 2, 1, 3].
Asked In: Google, Twitter, Amazon and many more companies.
Solution: We can consider this problem as a graph (related to topological sorting) problem. All tasks are nodes of the graph and if task u is a prerequisite of task v, we will add a directed edge from node u to node v. Now, this problem is equivalent to finding a topological ordering of nodes/tasks (using topological sorting) in the graph represented by prerequisites. If there is a cycle in the graph, then it is not possible to finish all tasks (because in that case there is no any topological order of tasks). Both BFS and DFS can be used for topological sorting to solve it. Since pair is inconvenient for the implementation of graph algorithms, we first transform it to a graph. If task u is a prerequisite of task v, we will add a directed edge from node u to node v.
Topological Sorting using BFS Here we use Kahn’s algorithm for topological sorting. BFS uses the indegrees of each node. We will first try to find a node with 0 indegree. If we fail to do so, there must be a cycle in the graph and we return false. Otherwise we have found one. We set its indegree to be -1 to prevent from visiting it again and reduce the indegrees of all its neighbors by 1. This process will be repeated for n (number of nodes) times.
CPP
// CPP program to find order to process tasks // so that all tasks can be finished. This program // mainly uses Kahn's algorithm. #include <bits/stdc++.h> using namespace std; // Returns adjacency list representation of graph from // given set of pairs. vector<unordered_set< int > > make_graph( int numTasks, vector<pair< int , int > >& prerequisites) { vector<unordered_set< int > > graph(numTasks); for ( auto pre : prerequisites) graph[pre.second].insert(pre.first); return graph; } // Computes in-degree of every vertex vector< int > compute_indegree(vector<unordered_set< int > >& graph) { vector< int > degrees(graph.size(), 0); for ( auto neighbors : graph) for ( int neigh : neighbors) degrees[neigh]++; return degrees; } // main function for topological sorting vector< int > findOrder( int numTasks, vector<pair< int , int > >& prerequisites) { // Create an adjacency list vector<unordered_set< int > > graph = make_graph(numTasks, prerequisites); // Find vertices of zero degree vector< int > degrees = compute_indegree(graph); queue< int > zeros; for ( int i = 0; i < numTasks; i++) if (!degrees[i]) zeros.push(i); // Find vertices in topological order // starting with vertices of 0 degree // and reducing degrees of adjacent. vector< int > toposort; for ( int i = 0; i < numTasks; i++) { if (zeros.empty()) return {}; int zero = zeros.front(); zeros.pop(); toposort.push_back(zero); for ( int neigh : graph[zero]) { if (!--degrees[neigh]) zeros.push(neigh); } } return toposort; } // Driver code int main() { int numTasks = 4; vector<pair< int , int > > prerequisites; // for prerequisites: [[1, 0], [2, 1], [3, 2]] prerequisites.push_back(make_pair(1, 0)); prerequisites.push_back(make_pair(2, 1)); prerequisites.push_back(make_pair(3, 2)); vector< int > v = findOrder(numTasks, prerequisites); for ( int i = 0; i < v.size(); i++) { cout << v[i] << " " ; } return 0; } |
Java
// Java program to find order to process tasks // so that all tasks can be finished. This program // mainly uses Kahn's algorithm. import java.util.ArrayList; import java.util.HashSet; import java.util.LinkedList; import java.util.Queue; public class Dep { // Returns adjacency list representation of graph from // given set of pairs. static ArrayList<HashSet<Integer> > make_graph( int numTasks, int [][] prerequisites) { ArrayList<HashSet<Integer> > graph = new ArrayList<HashSet<Integer>>(numTasks); for ( int i = 0 ; i < numTasks; i++) graph.add( new HashSet<Integer>()); for ( int [] pre : prerequisites) graph.get(pre[ 1 ]).add(pre[ 0 ]); return graph; } // Computes in-degree of every vertex static int [] compute_indegree( ArrayList<HashSet<Integer> > graph) { int [] degrees = new int [graph.size()]; for (HashSet<Integer> neighbors : graph) for ( int neigh : neighbors) degrees[neigh]++; return degrees; } // main function for topological sorting static ArrayList<Integer> findOrder( int numTasks, int [][] prerequisites) { // Create an adjacency list ArrayList<HashSet<Integer> > graph = make_graph(numTasks, prerequisites); // Find vertices of zero degree int [] degrees = compute_indegree(graph); Queue<Integer> zeros = new LinkedList<Integer>(); for ( int i = 0 ; i < numTasks; i++) if (degrees[i] == 0 ) zeros.add(i); // Find vertices in topological order // starting with vertices of 0 degree // and reducing degrees of adjacent. ArrayList<Integer> toposort = new ArrayList<Integer>(); for ( int i = 0 ; i < numTasks; i++) { if (zeros.isEmpty()) return new ArrayList<Integer>(); int zero = zeros.peek(); zeros.poll(); toposort.add(zero); for ( int neigh : graph.get(zero)) { if (--degrees[neigh] == 0 ) zeros.add(neigh); } } return toposort; } // Driver code public static void main(String[] args) { int numTasks = 4 ; int [][] prerequisites = { { 1 , 0 }, { 2 , 1 }, { 3 , 2 } }; ArrayList<Integer> v = findOrder(numTasks, prerequisites); for ( int i = 0 ; i < v.size(); i++) { System.out.print(v.get(i) + " " ); } } } // This code is contributed by Lovely Jain |
Python3
# Python program to find order to process tasks # so that all tasks can be finished. This program # mainly uses Kahn's algorithm. from collections import deque # Returns adjacency list representation of graph from # given set of pairs. def make_graph(numTasks, prerequisites): graph = [ set () for _ in range (numTasks)] for u, v in prerequisites: graph[v].add(u) return graph # Computes in-degree of every vertex def compute_indegree(graph): indegrees = [ 0 ] * len (graph) for neighbors in graph: for neigh in neighbors: indegrees[neigh] + = 1 return indegrees # main function for topological sorting def findOrder(numTasks, prerequisites): # Create an adjacency list graph = make_graph(numTasks, prerequisites) # Find vertices of zero degree indegrees = compute_indegree(graph) zeros = deque([i for i, degree in enumerate (indegrees) if degree = = 0 ]) # Find vertices in topological order # starting with vertices of 0 degree # and reducing degrees of adjacent. toposort = [] while zeros: zero = zeros.popleft() toposort.append(zero) for neigh in graph[zero]: indegrees[neigh] - = 1 if indegrees[neigh] = = 0 : zeros.append(neigh) return toposort if len (toposort) = = numTasks else [] # Driver code if __name__ = = '__main__' : numTasks = 4 # for prerequisites: [[1, 0], [2, 1], [3, 2]] prerequisites = [( 1 , 0 ), ( 2 , 1 ), ( 3 , 2 )] toposort = findOrder(numTasks, prerequisites) print (toposort) # This code is contributed by Aman Kumar. |
C#
using System; using System.Collections.Generic; class Dep { // Returns adjacency list representation of graph from // given set of pairs. static List<HashSet< int >> MakeGraph( int numTasks, int [][] prerequisites) { List<HashSet< int >> graph = new List<HashSet< int >>(numTasks); for ( int i = 0; i < numTasks; i++) graph.Add( new HashSet< int >()); foreach ( int [] pre in prerequisites) graph[pre[1]].Add(pre[0]); return graph; } // Computes in-degree of every vertex static int [] ComputeIndegree(List<HashSet< int >> graph) { int [] degrees = new int [graph.Count]; foreach (HashSet< int > neighbors in graph) foreach ( int neigh in neighbors) degrees[neigh]++; return degrees; } // Main function for topological sorting static List< int > FindOrder( int numTasks, int [][] prerequisites) { // Create an adjacency list List<HashSet< int >> graph = MakeGraph(numTasks, prerequisites); // Find vertices of zero degree int [] degrees = ComputeIndegree(graph); Queue< int > zeros = new Queue< int >(); for ( int i = 0; i < numTasks; i++) if (degrees[i] == 0) zeros.Enqueue(i); // Find vertices in topological order // starting with vertices of 0 degree // and reducing degrees of adjacent. List< int > toposort = new List< int >(); for ( int i = 0; i < numTasks; i++) { if (zeros.Count == 0) return new List< int >(); int zero = zeros.Peek(); zeros.Dequeue(); toposort.Add(zero); foreach ( int neigh in graph[zero]) { if (--degrees[neigh] == 0) zeros.Enqueue(neigh); } } return toposort; } // Driver code static void Main( string [] args) { int numTasks = 4; int [][] prerequisites = new int [][] { new int [] { 1, 0 }, new int [] { 2, 1 }, new int [] { 3, 2 } }; List< int > v = FindOrder(numTasks, prerequisites); Console.WriteLine( string .Join( " " , v)); } } |
Javascript
// JavaScript program to find order to process tasks // so that all tasks can be finished. This program // mainly uses Kahn's algorithm. // Returns adjacency list representation of graph from // given set of pairs. function make_graph(numTasks, prerequisites) { let graph = new Array(numTasks).fill( null ).map(() => new Set()); for (let i = 0; i < prerequisites.length; i++) { let pre = prerequisites[i]; graph[pre[1]].add(pre[0]); } return graph; } // Computes in-degree of every vertex function compute_indegree(graph) { let degrees = new Array(graph.length).fill(0); for (let neighbors of graph) { for (let neigh of neighbors) { degrees[neigh]++; } } return degrees; } // main function for topological sorting function findOrder(numTasks, prerequisites) { // Create an adjacency list let graph = make_graph(numTasks, prerequisites); // Find vertices of zero degree let degrees = compute_indegree(graph); let zeros = []; for (let i = 0; i < numTasks; i++) { if (degrees[i] === 0) { zeros.push(i); } } // Find vertices in topological order // starting with vertices of 0 degree // and reducing degrees of adjacent. let toposort = []; for (let i = 0; i < numTasks; i++) { if (zeros.length === 0) { return []; } let zero = zeros.shift(); toposort.push(zero); for (let neigh of graph[zero]) { if (--degrees[neigh] === 0) { zeros.push(neigh); } } } return toposort; } // Driver code let numTasks = 4; let prerequisites = [[1, 0], [2, 1], [3, 2]]; let v = findOrder(numTasks, prerequisites); for (let i = 0; i < v.length; i++) { console.log(v[i] + " " ); } // This code is contributed by Pushpesh Raj. |
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Topological Sorting using DFS: In this implementation, we use DFS based algorithm for Topological Sort.
CPP
// CPP program to find Topological sorting using // DFS #include <bits/stdc++.h> using namespace std; // Returns adjacency list representation of graph from // given set of pairs. vector<unordered_set< int > > make_graph( int numTasks, vector<pair< int , int > >& prerequisites) { vector<unordered_set< int > > graph(numTasks); for ( auto pre : prerequisites) graph[pre.second].insert(pre.first); return graph; } // Does DFS and adds nodes to Topological Sort bool dfs(vector<unordered_set< int > >& graph, int node, vector< bool >& onpath, vector< bool >& visited, vector< int >& toposort) { if (visited[node]) return false ; onpath[node] = visited[node] = true ; for ( int neigh : graph[node]) if (onpath[neigh] || dfs(graph, neigh, onpath, visited, toposort)) return true ; toposort.push_back(node); return onpath[node] = false ; } // Returns an order of tasks so that all tasks can be // finished. vector< int > findOrder( int numTasks, vector<pair< int , int > >& prerequisites) { vector<unordered_set< int > > graph = make_graph(numTasks, prerequisites); vector< int > toposort; vector< bool > onpath(numTasks, false ), visited(numTasks, false ); for ( int i = 0; i < numTasks; i++) if (!visited[i] && dfs(graph, i, onpath, visited, toposort)) return {}; reverse(toposort.begin(), toposort.end()); return toposort; } int main() { int numTasks = 4; vector<pair< int , int > > prerequisites; // for prerequisites: [[1, 0], [2, 1], [3, 2]] prerequisites.push_back(make_pair(1, 0)); prerequisites.push_back(make_pair(2, 1)); prerequisites.push_back(make_pair(3, 2)); vector< int > v = findOrder(numTasks, prerequisites); for ( int i = 0; i < v.size(); i++) { cout << v[i] << " " ; } return 0; } |
Java
// Java program to find Topological sorting using // DFS import java.util.ArrayList; import java.util.Collections; import java.util.HashSet; public class Dfs1 { // Returns adjacency list representation of graph from // given set of pairs. static ArrayList<HashSet<Integer> > make_graph( int numTasks, int [][] prerequisites) { ArrayList<HashSet<Integer> > graph = new ArrayList(numTasks); for ( int i = 0 ; i < numTasks; i++) graph.add( new HashSet<Integer>()); for ( int [] pre : prerequisites) graph.get(pre[ 1 ]).add(pre[ 0 ]); return graph; } // Does DFS and adds nodes to Topological Sort static boolean dfs(ArrayList<HashSet<Integer> > graph, int node, boolean [] onpath, boolean [] visited, ArrayList<Integer> toposort) { if (visited[node]) return false ; onpath[node] = visited[node] = true ; for ( int neigh : graph.get(node)) if (onpath[neigh] || dfs(graph, neigh, onpath, visited, toposort)) return true ; toposort.add(node); return onpath[node] = false ; } // Returns an order of tasks so that all tasks can be // finished. static ArrayList<Integer> findOrder( int numTasks, int [][] prerequisites) { ArrayList<HashSet<Integer> > graph = make_graph(numTasks, prerequisites); ArrayList<Integer> toposort = new ArrayList<Integer>(); boolean [] onpath = new boolean [numTasks]; boolean [] visited = new boolean [numTasks]; for ( int i = 0 ; i < numTasks; i++) if (!visited[i] && dfs(graph, i, onpath, visited, toposort)) return new ArrayList<Integer>(); Collections.reverse(toposort); return toposort; } // Driver code public static void main(String[] args) { int numTasks = 4 ; int [][] prerequisites = { { 1 , 0 }, { 2 , 1 }, { 3 , 2 } }; ArrayList<Integer> v = findOrder(numTasks, prerequisites); for ( int i = 0 ; i < v.size(); i++) { System.out.print(v.get(i) + " " ); } } } // This code is contributed by Lovely Jain |
Python3
# Python program to find Topological sorting using # DFS # Returns adjacency list representation of graph from # given set of pairs. def make_graph(numTasks, prerequisites): graph = {i: [] for i in range (numTasks)} for pre in prerequisites: graph[pre[ 1 ]].append(pre[ 0 ]) return graph # Does DFS and adds nodes to Topological Sort def dfs(graph, node, onpath, visited, toposort): if visited[node]: return False onpath[node] = visited[node] = True for neigh in graph[node]: if onpath[neigh] or dfs(graph, neigh, onpath, visited, toposort): return True toposort.append(node) return onpath[node] = = False # Returns an order of tasks so that all tasks can be # finished. def findOrder(numTasks, prerequisites): graph = make_graph(numTasks, prerequisites) toposort = [] onpath = [ False for i in range (numTasks)] visited = [ False for i in range (numTasks)] for i in range (numTasks): if not visited[i] and dfs(graph, i, onpath, visited, toposort): return toposort = toposort[:: - 1 ] return toposort # Driver code if __name__ = = "__main__" : numTasks = 4 prerequisites = [[ 1 , 0 ], [ 2 , 1 ], [ 3 , 2 ]] v = findOrder(numTasks, prerequisites) for i in range ( len (v)): print (v[i], end = " " ) |
C#
// C# program to find Topological sorting using // DFS using System; using System.Collections.Generic; using System.Linq; class Solution { // Returns adjacency list representation of graph from // given set of pairs. static List<HashSet< int >> MakeGraph( int numTasks, List<Tuple< int , int >> prerequisites) { var graph = new List<HashSet< int >>(numTasks); for ( int i = 0; i < numTasks; i++) { graph.Add( new HashSet< int >()); } foreach ( var pre in prerequisites) { graph[pre.Item2].Add(pre.Item1); } return graph; } // Does DFS and adds nodes to Topological Sort static bool DFS(List<HashSet< int >> graph, int node, HashSet< int > onPath, HashSet< int > visited, List< int > topoSort) { if (visited.Contains(node)) { return false ; } onPath.Add(node); visited.Add(node); foreach ( var neigh in graph[node]) { if (onPath.Contains(neigh) || DFS(graph, neigh, onPath, visited, topoSort)) { return true ; } } topoSort.Add(node); onPath.Remove(node); return false ; } // Returns an order of tasks so that all tasks can be // finished. static List< int > FindOrder( int numTasks, List<Tuple< int , int >> prerequisites) { var graph = MakeGraph(numTasks, prerequisites); var topoSort = new List< int >(); var onPath = new HashSet< int >(); var visited = new HashSet< int >(); for ( int i = 0; i < numTasks; i++) { if (!visited.Contains(i) && DFS(graph, i, onPath, visited, topoSort)) { return new List< int >(); } } topoSort.Reverse(); return topoSort; } static void Main( string [] args) { int numTasks = 4; // for prerequisites: [[1, 0], [2, 1], [3, 2]] var prerequisites = new List<Tuple< int , int >> { Tuple.Create(1, 0), Tuple.Create(2, 1), Tuple.Create(3, 2) }; var v = FindOrder(numTasks, prerequisites); foreach ( var i in v) { Console.Write(i + " " ); } } } // This code is contributed by Prince Kumar |
Javascript
// Returns adjacency list representation of graph from // given set of pairs. function makeGraph(numTasks, prerequisites) { const graph = {}; for (let i = 0; i < numTasks; i++) { graph[i] = []; } for (let i = 0; i < prerequisites.length; i++) { const pre = prerequisites[i]; graph[pre[1]].push(pre[0]); } return graph; } // Does DFS and adds nodes to Topological Sort function dfs(graph, node, onpath, visited, toposort) { if (visited[node]) { return false ; } onpath[node] = visited[node] = true ; for (let i = 0; i < graph[node].length; i++) { const neigh = graph[node][i]; if (onpath[neigh] || dfs(graph, neigh, onpath, visited, toposort)) { return true ; } } toposort.push(node); return onpath[node] === false ; } // Returns an order of tasks so that all tasks can be // finished. function findOrder(numTasks, prerequisites) { const graph = makeGraph(numTasks, prerequisites); const toposort = []; const onpath = new Array(numTasks).fill( false ); const visited = new Array(numTasks).fill( false ); for (let i = 0; i < numTasks; i++) { if (!visited[i] && dfs(graph, i, onpath, visited, toposort)) { return ; } } return toposort.reverse(); } // Driver code const numTasks = 4; const prerequisites = [[1, 0], [2, 1], [3, 2]]; const v = findOrder(numTasks, prerequisites); temp= "" ; for (let i = 0; i < v.length; i++) { temp = temp +v[i] + " " ; } console.log(temp); |
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Time complexity :- O(N+V+E) The make_graph function creates an adjacency list from the given set of pairs, which takes O(N), where N is the number of prerequisites. The findOrder function does a DFS traversal of the graph, which takes O(V+E) time, where V is the number of tasks and E is the number of prerequisites. Therefore, the overall time complexity is O(N + V + E).
Reference: https://leetcode.com/problems/course-schedule-ii/
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