Revise Complete Graphs For Interview | C++

Revision en13, by harshiscoding, 2021-09-29 22:18:17

Vertices,edges,undirected and directed graph,cyclic & acyclic graph,degree indegree & outdegree and weights. In an undirected graph, the sum of degrees of all vertices is double the vertices (We consider degree=indegree=outdegree in an undirected graph).

Graph representation in C++

Let us say there are n nodes and m edges. And adj[i][j] represents there is an edge from node i to j.

There are two ways to represent a graph

1. Adjacency matrix:

code

Time Complexity:O(m)**** Space Complexity:**O(n*n)**

2. Adjacency List

code

Time Complexity:**O(m)** Space Complexity:**O(n*n)**

If there are also weights in edges, we create the adjacency list as ' vector <vector < pair <int,int> > > ' where first value is node(v) and second value represents weight b/w the nodes(u and v).

Since there can be multiple disconnected components in a graph, while graph traversal we have to call BFS/DFS from every unvisited node. To avoid repetition we store information if a node is visited or not by creating a boolean array where is_visited[i]==true represents node 'i' is visited.

for(int i=1i<=n;i++)
{
if(!is_visited[i]) BFS(i);
}

BFS

code

Time Complexity:**O(n+e)** ( O(n+e): n time taken for visiting n nodes and e time taken for traversing through adjacent nodes)

Space Complexity:**O(n)** ( O(n) for visiting array and O(n) for queue )

DFS

code

Time Complexity:**O(n+e)** ( O(n+e): n time taken for visiting n nodes and e time taken for traversing through adjacent nodes)

Space Complexity:**O(n)** ( O(n) for visiting array and O(n) for stack space )

Cycle Detection in Graphs

Case 1: Cycle detection in undirected graph using dfs

If a node adjacent to a node ( that is not its parent node ) is already visited, the component contains a cycle.

code

Case 2:Cycle detection in the undirected graph using BFS


code

Case 3:Cycle detection in the Directed graph using DFS-

code

Bipartite graph:

A graph will not be a bipartite iff it has a cycle with an odd no. of nodes.

Check whether a graph is bipartite or not using BFS

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Check whether a graph is bipartite or not using DFS

code

Topological Sorting:

Only possible in DAG. It is a linear order of vertices such that for every directed edge u-->v, vertex u comes before v in that order. A DAG has at least one node with indegree=0.

Topological sorting using BFS

First, calculate indegree of all nodes and store it in vector.Then push all nodes with degree==0 in a queue.Take each node one by one out of queue and for all its child ,decrease their indegree by 1. After this if any child has indegree==0, push them in the queue.

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Topological sorting using DFS

Topological order is the order of nodes in decreasing order of finishing time i.e. the node that gets finished at last comes first in order.

code

History

 
 
 
 
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en43 English harshiscoding 2022-09-03 16:10:01 36
en42 English harshiscoding 2022-09-03 16:08:44 1 Tiny change: 'SU**\n\n1. Finding th' -> 'SU**\n\n1.Finding th'
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en28 English harshiscoding 2021-12-11 18:51:41 18 Tiny change: 'ouble the vertices (We con' -> 'ouble the the no. of edges (We con'
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en23 English harshiscoding 2021-10-02 01:28:45 11261 Reverted to en21
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en9 English harshiscoding 2021-09-29 20:39:11 2367 Tiny change: '\n==================\n' -> '\n========\n'
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en2 English harshiscoding 2021-09-27 16:42:44 15 Tiny change: 'ologies:\n\nbhbm\n==================' -> 'ologies:\n==================\n\nhvh'
en1 English harshiscoding 2021-09-27 16:41:52 90 Initial revision (saved to drafts)