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Find whether it is possible to finish all tasks or not from given dependencies

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  • Difficulty Level : Medium
  • Last Updated : 20 Apr, 2022

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.
 

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 not
bool 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.


Output

Possible to finish all tasks

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.
 

CPP




// 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 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 to check whether possible to finish all tasks or not
bool 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.


Output

Possible to finish all tasks

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

Using Union-find 

Another approach which is Union-find method can also be used to solve this problem. Each pair we get can be considered in a parent-child relation.

 As soon as we find a pair of tasks which are supposed to be in a parent-child relation, we check if they already have a common parent and hence it means there is a dependency cycle in the tasks which will never end and hence solving such set of tasks is not possible.

Java




/*
* A union-find based solution for stating whether
* the given set of tasks with respective prerequisites
* can be finished completely or not.
*/
import java.io.*;
import java.util.*;
 
//Driver code
class GFG {
    public static void main(String args[]) throws IOException
    {
             
            int prerequisites[][] = new int[][]{ {1,0}, {2,1}, {3,2} };
         
            Solution ob = new Solution();
       
            if(ob.isPossible(4, prerequisites))
            {
                System.out.println("Yes");
            }
            else{
                System.out.println("No");
            }
    }
}
//Driver Code Ends
 
 
//Solution code
class Solution {
  //returns true/false stating whether tasks can be finished or not 
  public boolean isPossible(int N, int[][] prerequisites)
    {
        //make object of Union class for N tasks
        Union u = new Union(N);
         
    //traverse through pre-requisites array
        for(int i = 0; i < prerequisites.length; i++){
          //check whether given pre-requisite pair
          //already have a common pre-requisite(parent)
          if(u.findParent(prerequisites[i][0]) ==
            u.findParent(prerequisites[i][1])) {
                 
                //tasks cannot be completed because there was
                //a cyclic condition in the tasks
                return false;
            }
             
          //make parent-child relation between pre-requisite task
          //and the task dependent on it
            u.makeParent(prerequisites[i][0], prerequisites[i][1]);
        }
         
        //if there was no cycle found, tasks can be completed
        return true;
    }
     
}
class Union {
    //to store the parents of respective tasks
    int[] arr;
   
    //parameterized constructor for Union class
    public Union(int n){
       
        arr = new int[n];
        //Initially, everyone is their own child
        for(int i = 0; i < n; i++){
            arr[i] = i;    
        }
    }
     
    public void makeParent(int a, int b){
        //find parent of b and make it a's parent
        arr[a] = findParent(b);
    }
    public int findParent(int c){
        //when an independent task is found 
        if(c == arr)
        return c;
        //recursively find the parent of given task
        return findParent(arr);
    }
}


Output

Yes

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