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 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)