Skip to main content

Detect cycle in Directed Graph with code in c++

 Detect cycle in a directed graph

Given a Directed Graph with V vertices (Numbered from 0 to V-1) and E edges, check whether it contains any cycle or not.


Example 1:

Input:



Output: 1
Explanation: 3 -> 3 is a cycle


Example 2:

Input:


Output: 0
Explanation: no cycle in the graph


Your task:
You dont need to read input or print anything. Your task is to complete the function isCyclic() which takes the integer V denoting the number of vertices and adjacency list as input parameters and returns a boolean value denoting if the given directed graph contains a cycle or not.


Expected Time Complexity: O(V + E)
Expected Auxiliary Space: O(V)


Constraints:
1 ≤ V, E ≤ 105


solution :


  To detect cycle in directed graph , we uses two Boolean array . first which take record of stacks of recursive function  and second which take record of how much of vertex are visited till now.

it basically uses the concept of DFS alogithm and backtracking also.


The code is here :



class Solution {

  public:

    // Function to detect cycle in a directed graph.

    bool dfs(vector<bool>&visited , vector<bool>&visiteddfs , vector<int>adj[]  , int i)

    {

        visited[i] = true;

        visiteddfs[i] = true;

        vector<int>x = adj[i];

        for(int k = 0 ; k < x.size() ; k++)

        {

            if(!visited[x[k]])

            {

                if(dfs(visited, visiteddfs , adj , x[k]))

                    return true;

            }

            else

            {

                if(visiteddfs[x[k]])

                 return true;

            }

               

        }

        visiteddfs[i] = false;

        return false;

    }

    bool isCyclic(int V, vector<int> adj[]) {

        

        vector<bool>visited(V ,false);

        vector<bool>visiteddfs(V , false);

        for(int i = 0 ; i < V ; i++)

        {     

            if(!visited[i])

              { 

                  if(dfs(visited, visiteddfs , adj , i))

                    return true;

                     

              }

            

        }

        return false;

    }

};



References : https://practice.geeksforgeeks.org/problems/detect-cycle-in-a-directed-graph/1

Comments

Popular posts from this blog

leetcode 48 solution

  48 .  Rotate Image You are given an  n x n  2D  matrix  representing an image, rotate the image by  90  degrees (clockwise). You have to rotate the image  in-place , which means you have to modify the input 2D matrix directly.  DO NOT  allocate another 2D matrix and do the rotation.   Example 1: Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [[7,4,1],[8,5,2],[9,6,3]] Example 2: Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]] Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]   Constraints: n == matrix.length == matrix[i].length 1 <= n <= 20 -1000 <= matrix[i][j] <= 1000 solution: class Solution { public:     void swap(int& a , int &b)     {         int c ;         c = a;         a = b;         b = c;     }     void transpose (vector<vector<int>...

Regular Expression Matching Leetcode Solution

Regular Expression Matching Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where: '.' Matches any single character.​​​​ '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). Example 1: Input: s = "aa", p = "a"  Output: false  Explanation: "a" does not match the entire string "aa". Example 2: Input: s = "aa", p = "a*"  Output: true  Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa". Example 3: Input: s = "ab", p = ".*"  Output: true  Explanation: ".*" means "zero or more (*) of any character (.)". Constraints: 1 <= s.length <= 20 1 <= p.length <= 20 s contains only lowercase English letters. p contains only lowercase Englis...

2485. Find the Pivot Integer | Binary search

  Given a positive integer   n , find the   pivot integer   x   such that: The sum of all elements between  1  and  x  inclusively equals the sum of all elements between  x  and  n  inclusively. Return  the pivot integer  x . If no such integer exists, return  -1 . It is guaranteed that there will be at most one pivot index for the given input.   Example 1: Input: n = 8 Output: 6 Explanation: 6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21. Example 2: Input: n = 1 Output: 1 Explanation: 1 is the pivot integer since: 1 = 1. Example 3: Input: n = 4 Output: -1 Explanation: It can be proved that no such integer exist.   Constraints: 1 <= n <= 1000 Solution : class Solution { publ ic:     int pivotInteger( int n ) {         int sum = (( n )*( n + 1 ))/ 2 ;         int i = 1 ;         int j =...