Juspay Interview Question for Software Developers


Country: India




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s

- Anonymous January 18, 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

function largestCycle(edges)

{

var result, visitedFrom, startCell, cell, cells;

result = [];

visitedFrom = Array(edges.length).fill(-1);

for (startCell = 0; startCell<edges.length; startCell++)

{

cells = [];

for

(cell=startCell;cell>-1&&visitedFrom[cell]===-1; cell = edges[cell])

{

visitedFrom[cell] = startCell;

cells.push(cell);

}

if (cell > -1 && visitedFrom[cell] ===startCell)

{

cells =cells.slice(cells.indexOf(cell));

if (cells.length > result.length) result= cells;

}

}

return result;

}

var input = document.querySelector('textarea');

var output = document.querySelector('span');

(input.oninput = function () {

varedges=input.value.trim().split(/\s+/).map(Number);

var cycle = largestCycle(edges);

output.textContent = cycle.length + ': ' + JSON.stringify(cycle); })();

This is complexity with O(N) runtime where each edge of which there's at most N nodes is followed at most 3 times in the graph and the cache is updated exactly once for each and every node in the graph. It's uses O(N) additional storage.




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- Anonymous January 12, 2022 | Flag Reply
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