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 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 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 for topological sortingvector<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 codeint 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;} |
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Topological Sorting using DFS:
In this implementation, we use DFS based algorithm for Topological Sort.
// 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 Sortbool 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;} |
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Reference: https://leetcode.com/problems/course-schedule-ii/
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