There are a total of n tasks you have to pick, labelled from 0 to n-1. Some tasks may have prerequisites, for example to pick task 0 you have to first pick task 1, which is expressed as a pair: [0, 1]
Given the total number of tasks and a list of prerequisite pairs, is it possible for you to finish all tasks?
Examples:
Input: 2, [[1, 0]]
Output: true
Explanation: There are a total of 2 tasks to pick. To pick task 1 you should have finished task 0. So it is possible.Input: 2, [[1, 0], [0, 1]]
Output: false
Explanation: There are a total of 2 tasks to pick. To pick task 1 you should have finished task 0, and to pick task 0 you should also have finished task 1. So it is impossible.Input: 3, [[1, 0], [2, 1], [3, 2]]
Output: true
Explanation: There are a total of 3 tasks to pick. To pick tasks 1 you should have finished task 0, and to pick task 2 you should have finished task 1 and to pick task 3 you should have finished task 2. So it is possible.
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 detecting a cycle 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 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.
Prerequisite: Detect Cycle in a Directed Graph
Using DFS For DFS, it will first visit a node, then one neighbor of it, then one neighbor of this neighbor… and so on. If it meets a node which was visited in the current process of DFS visit, a cycle is detected and we will return false. Otherwise it will start from another unvisited node and repeat this process till all the nodes have been visited. Note that you should make two records: one is to record all the visited nodes and the other is to record the visited nodes in the current DFS visit.
The code is as follows. We use a vector visited to record all the visited nodes and another vector onpath to record the visited nodes of the current DFS visit. Once the current visit is finished, we reset the onpath value of the starting node to false.
Implementation:
CPP
// CPP program to check whether we can finish all// tasks or not from given dependencies.#include <bits/stdc++.h>using namespace std;// Returns adjacency list representation from a list// 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;}// A DFS based function to check if there is a cycle// in the directed graph.bool dfs_cycle(vector<unordered_set<int> >& graph, int node, vector<bool>& onpath, vector<bool>& visited){ if (visited[node]) return false; onpath[node] = visited[node] = true; for (int neigh : graph[node]) if (onpath[neigh] || dfs_cycle(graph, neigh, onpath, visited)) return true; return onpath[node] = false;}// Main function to check whether possible to finish all tasks or notbool canFinish(int numTasks, vector<pair<int, int> >& prerequisites){ vector<unordered_set<int> > graph = make_graph(numTasks, prerequisites); vector<bool> onpath(numTasks, false), visited(numTasks, false); for (int i = 0; i < numTasks; i++) if (!visited[i] && dfs_cycle(graph, i, onpath, visited)) return false; return true;}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)); if (canFinish(numTasks, prerequisites)) { cout << "Possible to finish all tasks"; } else { cout << "Impossible to finish all tasks"; } return 0;} |
Java
// Java program to check whether we can finish all// tasks or not from given dependencies.import java.util.*;public class GFG{ // class to store dependencies as a pair static class pair{ int first, second; pair(int first, int second){ this.first = first; this.second = second; } } // Returns adjacency list representation from a list // of pairs. static ArrayList<ArrayList<Integer>> make_graph(int numTasks, Vector<pair> prerequisites) { ArrayList<ArrayList<Integer>> graph = new ArrayList<ArrayList<Integer>>(numTasks); for(int i=0; i<numTasks; i++){ graph.add(new ArrayList<Integer>()); } for (pair pre : prerequisites) graph.get(pre.second).add(pre.first); return graph; } // A DFS based function to check if there is a cycle // in the directed graph. static boolean dfs_cycle(ArrayList<ArrayList<Integer>> graph, int node, boolean onpath[], boolean visited[]) { if (visited[node]) return false; onpath[node] = visited[node] = true; for (int neigh : graph.get(node)) if (onpath[neigh] || dfs_cycle(graph, neigh, onpath, visited)) return true; return onpath[node] = false; } // Main function to check whether possible to finish all tasks or not static boolean canFinish(int numTasks, Vector<pair> prerequisites) { ArrayList<ArrayList<Integer>> graph = make_graph(numTasks, prerequisites); boolean onpath[] = new boolean[numTasks]; boolean visited[] = new boolean[numTasks]; for (int i = 0; i < numTasks; i++) if (!visited[i] && dfs_cycle(graph, i, onpath, visited)) return false; return true; } public static void main(String args[]) { int numTasks = 4; Vector<pair> prerequisites = new Vector<pair>();; // for prerequisites: [[1, 0], [2, 1], [3, 2]] prerequisites.add(new pair(1, 0)); prerequisites.add(new pair(2, 1)); prerequisites.add(new pair(3, 2)); if (canFinish(numTasks, prerequisites)) { System.out.println("Possible to finish all tasks"); } else { System.out.println("Impossible to finish all tasks"); } }}// This code is contributed by adityapande88. |
Python3
# Python3 program to check whether we can finish all# tasks or not from given dependencies.# class to store dependencies as a pairclass pair: def __init__(self, first, second): self.first = first self.second = second# Returns adjacency list representation from a list# of pairs.def make_graph(numTasks, prerequisites): graph = [] for i in range(numTasks): graph.append([]) for pre in prerequisites: graph[pre.second].append(pre.first) return graph# A DFS based function to check if there is a cycle# in the directed graph.def dfs_cycle(graph, node, onpath, visited): if visited[node]: return false onpath[node] = visited[node] = True for neigh in graph[node]: if (onpath[neigh] or dfs_cycle(graph, neigh, onpath, visited)): return true return False# Main function to check whether possible to finish all# tasks or notdef canFinish(numTasks, prerequisites): graph = make_graph(numTasks, prerequisites) onpath = [False]*numTasks visited = [False]*numTasks for i in range(numTasks): if (not visited[i] and dfs_cycle(graph, i, onpath, visited)): return False return True# Driver code to test above functionsnumTasks = 4prerequisites = []prerequisites.append(pair(1, 0))prerequisites.append(pair(2, 1))prerequisites.append(pair(3, 2))if canFinish(numTasks, prerequisites): print("Possible to finish all tasks")else: print("Impossible to finish all tasks")# This code is contributed by Abhijeet Kumar(abhijeet19403) |
C#
// C# program to check whether we can finish all// tasks or not from given dependencies.using System;using System.Collections.Generic;public class GFG { // class to store dependencies as a pair public class pair { public int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } // Returns adjacency list representation from a list // of pairs. static List<List<int> > make_graph(int numTasks, List<pair> prerequisites) { List<List<int> > graph = new List<List<int> >(numTasks); for (int i = 0; i < numTasks; i++) { graph.Add(new List<int>()); } foreach(pair pre in prerequisites) graph[pre.second] .Add(pre.first); return graph; } // A DFS based function to check if there is a cycle // in the directed graph. static bool dfs_cycle(List<List<int> > graph, int node, bool[] onpath, bool[] visited) { if (visited[node]) return false; onpath[node] = visited[node] = true; foreach( int neigh in graph[node]) if (onpath[neigh] || dfs_cycle(graph, neigh, onpath, visited)) return true; //onpath[node] = false; return false; } // Main function to check whether possible to finish all // tasks or not static bool canFinish(int numTasks, List<pair> prerequisites) { List<List<int> > graph = make_graph(numTasks, prerequisites); bool[] onpath = new bool[numTasks]; bool[] visited = new bool[numTasks]; for (int i = 0; i < numTasks; i++) if (!visited[i] && dfs_cycle(graph, i, onpath, visited)) return false; return true; } public static void Main(String[] args) { int numTasks = 4; List<pair> prerequisites = new List<pair>(); ; // for prerequisites: [[1, 0], [2, 1], [3, 2]] prerequisites.Add(new pair(1, 0)); prerequisites.Add(new pair(2, 1)); prerequisites.Add(new pair(3, 2)); if (canFinish(numTasks, prerequisites)) { Console.WriteLine( "Possible to finish all tasks"); } else { Console.WriteLine( "Impossible to finish all tasks"); } }}// This code is contributed by Abhijeet Kumar(abhijeet19403) |
Javascript
// Javascript program to check whether we can finish all// tasks or not from given dependencies.// Returns adjacency list representation from a list// of pairs.function make_graph(numTasks, prerequisites){ let graph = []; for(let i = 0; i < numTasks; i++){ graph.push([]); } for(let i = 0; i < prerequisites.length; i++){ graph[prerequisites[i][1]].push(prerequisites[i][0]); } return graph;}// A DFS based function to check if there is a cycle// in the directed graph.function dfs_cycle(graph, node, onpath, visited){ if (visited[node]) return false; onpath[node] = visited[node] = true; for(let i = 0; i < graph[node].length; i++){ let neigh = graph[node][i]; if (onpath[neigh] == true || dfs_cycle(graph, neigh, onpath, visited) == true) return true; } return false;}// Main function to check whether possible to finish all// tasks or notfunction canFinish(numTasks, prerequisites){ let graph = make_graph(numTasks, prerequisites); let onpath = new Array(numTasks).fill(false); let visited = new Array(numTasks).fill(false); for(let i = 0; i < numTasks; i++){ if(visited[i] == false && dfs_cycle(graph, i, onpath, visited ) == true){ return false; } } return true;}// Driver code to test above functionslet numTasks = 4;let prerequisites = [];prerequisites.push([1, 0]);prerequisites.push([2, 1]);prerequisites.push([3, 2]);if(canFinish(numTasks, prerequisites)) console.log("Possible to finish all tasks");else console.log("Impossible to finish all tasks");// This code is contributed by Nidhi goel |
Output
Possible to finish all tasks
Time Complexity: O(V*(V+E)), where V is the number of vertices and E is the number of edges.
Auxiliary Space: O(V+E)
Using BFS:
BFS can be used to solve it using the idea of topological sort. If topological sorting is possible, it means there is no cycle and it is possible to finish all the tasks.
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. If we have not returned false, we will return true.
Implementation:
C++
// A BFS based solution to check if we can finish// all tasks or not. This solution is mainly based// on Kahn's algorithm.#include <bits/stdc++.h>using namespace std;// Returns adjacency list representation from a list// 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;}// Finds in-degree of every vertexvector<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 to check whether possible to finish all// tasks or notbool canFinish(int numTasks, vector<pair<int, int> >& prerequisites){ vector<unordered_set<int> > graph = make_graph(numTasks, prerequisites); vector<int> degrees = compute_indegree(graph); for (int i = 0; i < numTasks; i++) { int j = 0; for (; j < numTasks; j++) if (!degrees[j]) break; if (j == numTasks) return false; degrees[j] = -1; for (int neigh : graph[j]) degrees[neigh]--; } return true;}int main(){ int numTasks = 4; vector<pair<int, int> > prerequisites; prerequisites.push_back(make_pair(1, 0)); prerequisites.push_back(make_pair(2, 1)); prerequisites.push_back(make_pair(3, 2)); if (canFinish(numTasks, prerequisites)) { cout << "Possible to finish all tasks"; } else { cout << "Impossible to finish all tasks"; } return 0;} |
Java
// A BFS based solution to check if we can finish// all tasks or not. This solution is mainly based// on Kahn's algorithm.import java.util.*;public class GFG { // class to store dependencies as a pair static class pair { int first, second; pair(int first, int second) { this.first = first; this.second = second; } } // Returns adjacency list representation from a list // of pairs. static ArrayList<ArrayList<Integer> > make_graph(int numTasks, Vector<pair> prerequisites) { ArrayList<ArrayList<Integer> > graph = new ArrayList<ArrayList<Integer> >(numTasks); for (int i = 0; i < numTasks; i++) { graph.add(new ArrayList<Integer>()); } for (pair pre : prerequisites) graph.get(pre.second).add(pre.first); return graph; } // Finds in-degree of every vertex static int[] compute_indegree( ArrayList<ArrayList<Integer> > graph) { int degrees[] = new int[graph.size()]; for (ArrayList<Integer> neighbors : graph) for (int neigh : neighbors) degrees[neigh]++; return degrees; } // Main function to check whether possible to finish all // tasks or not static boolean canFinish(int numTasks, Vector<pair> prerequisites) { ArrayList<ArrayList<Integer> > graph = make_graph(numTasks, prerequisites); int degrees[] = compute_indegree(graph); for (int i = 0; i < numTasks; i++) { int j = 0; for (; j < numTasks; j++) if (degrees[j] == 0) break; if (j == numTasks) return false; degrees[j] = -1; for (int neigh : graph.get(j)) degrees[neigh]--; } return true; } public static void main(String args[]) { int numTasks = 4; Vector<pair> prerequisites = new Vector<pair>(); prerequisites.add(new pair(1, 0)); prerequisites.add(new pair(2, 1)); prerequisites.add(new pair(3, 2)); if (canFinish(numTasks, prerequisites)) { System.out.println( "Possible to finish all tasks"); } else { System.out.println( "Impossible to finish all tasks"); } }}// This code is contributed by adityapande88. |
Python3
# A BFS based solution to check if we can finish# all tasks or not. This solution is mainly based# on Kahn's algorithm.# class to store dependencies as a pairclass pair: def __init__(self, first, second): self.first = first self.second = second# Returns adjacency list representation from a list# of pairs.def make_graph(numTasks, prerequisites): graph = [] for i in range(numTasks): graph.append([]) for pre in prerequisites: graph[pre.second].append(pre.first) return graph# Finds in-degree of every vertexdef compute_indegree(graph): degrees = [0]*len(graph) for neighbors in graph: for neigh in neighbors: degrees[neigh] += 1 return degrees# Main function to check whether possible to finish all tasks or notdef canFinish(numTasks, prerequisites): graph = make_graph(numTasks, prerequisites) degrees = compute_indegree(graph) for i in range(numTasks): j = 0 while(j < numTasks): if (degrees[j] == 0): break j += 1 if (j == numTasks): return False degrees[j] = -1 for neigh in graph[j]: degrees[neigh] -= 1 return TruenumTasks = 4prerequisites = []prerequisites.append(pair(1, 0))prerequisites.append(pair(2, 1))prerequisites.append(pair(3, 2))if (canFinish(numTasks, prerequisites)): print("Possible to finish all tasks")else: print("Impossible to finish all tasks")# This code is contributed by Lovely Jain |
C#
// A BFS based solution to check if we can finish// all tasks or not. This solution is mainly based// on Kahn's algorithm.using System;using System.Collections.Generic;public class GFG { // class to store dependencies as a pair public class pair { public int first, second; public pair(int first, int second) { this.first = first; this.second = second; } } // Returns adjacency list representation from a list // of pairs. static List<List<int> > make_graph(int numTasks, List<pair> prerequisites) { List<List<int> > graph = new List<List<int> >(numTasks); for (int i = 0; i < numTasks; i++) { graph.Add(new List<int>()); } foreach(pair pre in prerequisites) graph[pre.second].Add(pre.first); return graph; } // Finds in-degree of every vertex static int[] compute_indegree( List<List<int> > graph) { int[] degrees = new int[graph.Count]; foreach(List<int> neighbors in graph) foreach(int neigh in neighbors) degrees[neigh]++; return degrees; } // Main function to check whether possible to finish all // tasks or not static bool canFinish(int numTasks, List<pair> prerequisites) { List<List<int> > graph = make_graph(numTasks, prerequisites); int[] degrees = compute_indegree(graph); for (int i = 0; i < numTasks; i++) { int j = 0; for (; j < numTasks; j++) if (degrees[j] == 0) break; if (j == numTasks) return false; degrees[j] = -1; foreach(int neigh in graph[j]) degrees[neigh]--; } return true; } public static void Main(String[] args) { int numTasks = 4; List<pair> prerequisites = new List<pair>(); prerequisites.Add(new pair(1, 0)); prerequisites.Add(new pair(2, 1)); prerequisites.Add(new pair(3, 2)); if (canFinish(numTasks, prerequisites)) { Console.WriteLine( "Possible to finish all tasks"); } else { Console.WriteLine( "Impossible to finish all tasks"); } }}// This code is contributed by Abhijeet Kumar(abhijeet19403) |
Javascript
<script> // Linked list implementation in javascript // Creating a nodeclass Node { constructor(){ this.value = 0; this.next = null; } }let head = null;let one = null;let two = null;let three = null;// allocate 3 nodes in the heapone = new Node();two = new Node();three = new Node();// Assign value valuesone.value = 1;two.value = 2;three.value = 3;// Connect nodesone.next = two;two.next = three;three.next = null;// print the linked list valuehead = one;while (head != null) { document.write(head.value); head = head.next;}// The code is contributed by gautam goel. </script> |
Output
Possible to finish all tasks
Time Complexity: O(V+E), where V is the number of vertices and E is the number of edges.
Auxiliary Space: O(V+E)
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