Problem statement: You are giv

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Problem statement: You are given a maze with N cells. Each cell may have multiple entry points but not more than one exit (ie. entry/exit points are unidirectional doors like valves). The cells are named with an integer value from 0 to N-1. You need to find the following :

The length of the largest cycle in the maze. Return -1 if there are no cycles.

Note: Aim for O(N) solution.

INPUT FORMAT - First line has the number of cells N

Second line has list of N values of the edge[] array. edge[i] contains the cell number that can be reached from of cell ‘i’ in one step. edge[i] is -1 if the ‘i’th cell doesn’t have an exit.

OUTPUT FORMAT - length of the largest cycle.
- sarthakbansal19 September 23, 2016 in India | Report Duplicate | Flag | PURGE
Juspay Software Developer Data Structures

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of 0 vote

- Anonymous January 18, 2017 | Flag Reply

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of 2 vote

144

- anona January 21, 2017 | Flag Reply

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ccvfd

- ffcdscd January 21, 2017 | Flag Reply

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class path 
{
public static void main(String [] args)
{
    Scanner scnr=new Scanner(System.in);
    int n=scnr.nextInt();
    int a[]=new int[n];
    for(int i=0;i<n;i++)
    {
        a[i]=scnr.nextInt();
    }
    int path=-1;

    int j;
    for(int i=0;i<n;i++)
    {   j=i;
        int count=0;
        while(true)
        {   
            count++;
            if(a[j]==-1)
            {
                break;
            }
            else if(i==a[j])
            {
                if(count>path)
                path=count;
                break;
            }
            else
            {   int temp=j;
                j=a[j];
                a[temp]=-1;                 
            }
        }
 }
    System.out.println("my path: "+path);
}
}

- Sangeet Moy August 07, 2017 | Flag Reply

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of 0 vote

class path
{
public static void main(String [] args)
{
Scanner scnr=new Scanner(System.in);
int n=scnr.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++)
{
a[i]=scnr.nextInt();
}
int path=-1;

int j;
for(int i=0;i<n;i++)
{ j=i;
int count=0;
while(true)
{
count++;
if(a[j]==-1)
{
break;
}
else if(i==a[j])
{
if(count>path)
path=count;
break;
}
else
{ int temp=j;
j=a[j];
a[temp]=-1;
}
}
}
System.out.println("my path: "+path);
}
}
- Sangeet Moy August 07, 2017 | FlagReply

- Anonymous September 08, 2020 | Flag Reply

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of 0 vote

#include<bits/stdc++.h>
using namespace std;
int dist[23];
bool vis[23];

void addEdge(vector<int> adj[], int u, int v)
{
	adj[u].push_back(v);
	// adj[v].push_back(u);
}


void dfs(vector<int> adj[], int node, int d, int src)
{
	vis[node] = true;
	for (auto child : adj[node]) {
		if (!vis[child])dfs(adj, child, d + 1, src);
		else if (child == src)dist[child] = max(dist[child], d + 1);
	}
	vis[node] = false;
}

int main()
{
	int V;
	cin >> V;
	// int V = 7;
	memset(vis, false, sizeof(vis));
	memset(dist, false, sizeof(dist));
	vector<int> adj[V];
	for (int i = 0; i < V; ++i)
	{
		int to; cin >> to;
		if (to != -1)addEdge(adj, i, to);
	}
	for (int i = 0; i < V; ++i)
	{
		dfs(adj, i, 0, i);
	}
	int ans = INT_MIN;
	for (int i = 0; i < V; ++i)
	{
		ans = max(ans, dist[i]);
	}
	cout << ans;
	// printGraph(adj, V);
	return 0;
}

- black panther January 03, 2021 | Flag Reply

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of 0 vote

#include<bits/stdc++.h>
using namespace std;
int dist[23];
bool vis[23];

void addEdge(vector<int> adj[], int u, int v)
{
adj[u].push_back(v);
// adj[v].push_back(u);
}

void dfs(vector<int> adj[], int node, int d, int src)
{
vis[node] = true;
for (auto child : adj[node]) {
if (!vis[child])dfs(adj, child, d + 1, src);
else if (child == src)dist[child] = max(dist[child], d + 1);
}
vis[node] = false;
}

int main()
{
int V;
cin >> V;
// int V = 7;
memset(vis, false, sizeof(vis));
memset(dist, false, sizeof(dist));
vector<int> adj[V];
for (int i = 0; i < V; ++i)
{
int to; cin >> to;
if (to != -1)addEdge(adj, i, to);
}
for (int i = 0; i < V; ++i)
{
dfs(adj, i, 0, i);
}
int ans = INT_MIN;
for (int i = 0; i < V; ++i)
{
ans = max(ans, dist[i]);
}
cout << ans;
// printGraph(adj, V);
return 0;
}

- Anonymous April 16, 2021 | Flag Reply

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of 0 vote

- Anonymous January 12, 2022 | Flag Reply

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of 0 vote

In this problem, a nagging React newbie B is constantly troubling React expert A. React Expert A needs to know the minimum set of people following him he needs to remove from his network to stop B from reaching out to him.

The input is given as follows:

The first line is a number which is the number of people in the social network. The next following lines are the actual members of the social network whose names are written in separate lines. Following that is the number of connections, followed by a new line separated list of connections given as <follower> <following> Then on the next line is the id of the person who wants to our React Expert (A) and on the next line is the id of the person who is the newbie (B)

- Anushka karane November 03, 2022 | Flag Reply

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of 0 vote

N=int(input())
dic={}
def addnode(u,v):
dic[u].append(v)
for i in range(N):
n1=int(input())
dic[n1]=[]

edge=int(input())
for i in range(edge):
u,v=map(int,input().split())
addnode(u,v)
start=int(input())
end=int(input())
tr=[]
tr.append(start)
def traverse(dic,start):
for i in dic[start]:
if(i==end):
return
if(len(dic[start])==0):
return
else:
tr.append(i)
traverse(dic,i)

traverse(dic,start)
tr=sorted(tr)
for i in tr:
print(i,end=" ")

- Anonymous March 03, 2023 | Flag Reply

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-1

of 1 vote

The Malicious Haskeller

3 4

import sys

5 def soluti 6

7

In this problem, a nagging React newbie B is constantly troubling React expert A. React Expert A needs to know the minimum set of people following him he needs to remove from his network to stop B from reaching out to him.

8

return

9 membersCour

10 memberIds 11 for in ra

12

The input is given as follows:

The first line is a number which is the number of people in the social network. The next following liner are the actual members of the social network whose names are written in separate lines. Following that is the number of connections, followed by a new line separated list of connections given as <follower> <following Then on the next line is the id of the person who wants to our React Expert (A) and on the next line is the id of the person who is the newbie (B)

13

memberl

14 edgesCount 15 edgeFrom 16 edgeTo []

17 for 18

19

INPUT FORMAT

- Anonymous October 05, 2022 | Flag Reply

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