Print completed tasks at end according to Dependencies

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 representation 
class 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 graph 
void 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 DFS 
void Graph::printTasks() 
{ 
    for (int i = 0; i < 26; i++) { 
        if (visited[i]) 
            cout << char(i + 65) << " "; 
    } 
    cout << endl; 
} 
  
// Driver code 
int 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 approach 
from 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() 

Output:

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)

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Python3

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