Given N dependencies of the form X Y, where X & Y represents two different tasks. The dependency X Y denotes dependency of the form Y -> X i.e, if task Y happens then task X will happen in other words task Y has to be completed first to initiate task X. Also given M tasks that will initiate first. The task is to print all the tasks that will get completed at the end in the lexicographical order. Note that the tasks will be represented by upper case English letters only.
Input: dep[][] = {{A, B}, {C, B}, {D, A}, {D, C}, {B, E}}, tasks[] = {B, C}
Output: A B C D
Task A happens after task B and task D can only happen
after the completion of tasks A or C.
So, the required order is A B C D.Input: dep[][] = {{Q, P}, {S, Q}, {Q, R}}, tasks[] = {R}
Output: Q R S
Approach: DFS can be used to solve the problem. The dependencies of the form X Y (Y -> X) can be represented as an edge from node Y to node X in the graph. Initiate the DFS from each of the M initial nodes and mark the nodes that are encountered as visited using a boolean array. At last, print the nodes/tasks that are covered using DFS in lexicographical order. The approach works because DFS will cover all the nodes starting from the initial nodes in sequential manner.
Consider the diagram below that represents the first example from the above:
The diagram shows the edges covered during DFS from initial tasks B and C as Red in
color. The nodes thus visited were A, B, C and D.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <cstring>#include <iostream>#include <vector>using namespace std; // Graph class represents a directed graph// using adjacency list representationclass Graph { // Number of vertices int V; // Pointer to an array containing // adjacency lists vector<int>* adj; // Boolean array to mark tasks as visited bool visited[26]; // A recursive function used by DFS void DFSUtil(int v); public: // Constructor Graph() { // There are only 26 English // upper case letters this->V = 26; adj = new vector<int>[26]; } // Function to add an edge to the graph void addEdge(char v, char w); // DFS traversal of the vertices // reachable from v void DFS(char start[], int M); void printTasks();}; // Function to add an edge to the graphvoid Graph::addEdge(char v, char w){ // Add w to v's list adj[v - 65].push_back(w - 65);} void Graph::DFSUtil(int v){ // Mark the current node as visited and // print it visited[v] = true; // Recur for all the vertices adjacent // to this vertex vector<int>::iterator i; for (i = adj[v].begin(); i != adj[v].end(); ++i) if (!visited[*i]) DFSUtil(*i);} // DFS traversal of the vertices reachable// from start nodes// It uses recursive DFSUtil()void Graph::DFS(char start[], int M){ // Mark all the vertices as not visited for (int i = 0; i < V; i++) visited[i] = false; // Call the recursive helper function // to print DFS traversal for (int i = 0; i < M; i++) DFSUtil(start[i] - 65);} // Helper function to print the tasks in// lexicographical order that are completed// at the end of the DFSvoid Graph::printTasks(){ for (int i = 0; i < 26; i++) { if (visited[i]) cout << char(i + 65) << " "; } cout << endl;} // Driver codeint main(){ // Create the graph Graph g; g.addEdge('B', 'A'); g.addEdge('B', 'C'); g.addEdge('A', 'D'); g.addEdge('C', 'D'); g.addEdge('E', 'B'); // Initial tasks to be run char start[] = { 'B', 'C' }; int n = sizeof(start) / sizeof(char); // Start the dfs g.DFS(start, n); // Print the tasks that will get finished g.printTasks(); return 0;} |
Python3
# Python3 implementation of the approachfrom collections import defaultdict # This class represents a directed graph # using adjacency list representation class Graph: # Constructor def __init__(self): # Default dictionary to store the graph self.graph = defaultdict(list) self.visited = [False]*26 # Function to add an edge to the graph def addEdge(self, u, v): self.graph[ord(u)-65].append(ord(v)-65) # A function used by DFS def DFSUtil(self, v): # Mark the current node as visited # and print it self.visited[v]= True # Recur for all the vertices adjacent # to this vertex for i in self.graph[v]: if self.visited[i] == False: self.DFSUtil(i) # Function to perform the DFS traversal # It uses recursive DFSUtil() def DFS(self, start, M): # Total vertices V = len(self.graph) # Call the recursive helper function # to print the DFS traversal starting # from all vertices one by one for i in range(M): self.DFSUtil(ord(start[i])-65) def printOrder(self): for i in range(26): if self.visited[i] == True: print(chr(i + 65), end =" ") print("\n") # Driver code g = Graph() g.addEdge('B', 'A') g.addEdge('B', 'C') g.addEdge('A', 'D') g.addEdge('C', 'D') g.addEdge('E', 'B') g.DFS(['B', 'C'], 2) g.printOrder() |
A B C D
Time Complexity: O(V + E) where V is the number of nodes in the graph and E is the number of edges or dependencies. In this case, since V is always 26, so the time complexity is O(26 + E) or just O(E) in the worst case.
Space Complexity: O(V + E)
