Starup Interview Question for SDE-2s

Country: India
Interview Type: Written Test

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of 0 vote


1. There will always be 2 arrays that will describe the connection and they will always be in the same length.

2. The arrays arranged in the correct order, the next pair of connection will be the continue of the previous.

For more professional way of detection I think that creating the structure of network in linked list data structure is a better way.

package com.cracking.starup;

public class DetectNetwrokTopology {
	public static enum TopologyType {
		TopologyType(String description) {
			this.description = description;
		public String description;
	public static final int[] srcNodes = {2,3,4,5};
	public static final int[] destNodes = {1,1,1,1};

	public static void main(String[] args) {
		TopologyType type = getTopolgyType(srcNodes, destNodes);
		System.out.println("Topolgy = " + type.description);
	public static TopologyType getTopolgyType(int[] srcNodes, int[] destNodes) {
		if(isTopologyStar(srcNodes, destNodes)) return TopologyType.STAR;
		if(isTopologyCircular(srcNodes, destNodes)) return TopologyType.CIRCULAR;
		if(isTopologyBus(srcNodes, destNodes)) return TopologyType.BUS;
		return TopologyType.UNKNOWN;

	public static boolean isTopologyBus(int[] srcNodes, int[] destNodes) {	
		for(int i=0;i<srcNodes.length && i<destNodes.length; i++) {
			if( (i+1)<srcNodes.length && (i+1)<destNodes.length ) {
				int expected = srcNodes[i+1];
				int actual = destNodes[i];
				if(actual!=expected) return false;
		return true;
	public static boolean isTopologyCircular(int[] srcNodes, int[] destNodes) {
		int first = srcNodes[0];
		int last = destNodes[destNodes.length-1];
		return first == last && isTopologyBus(srcNodes,destNodes);
	public static boolean isTopologyStar(int[] srcNodes, int[] destNodes) {
		for(int i=1; i<destNodes.length; i++) {
			if(destNodes[0] != destNodes[i]) return false;
		return true;

- ProTechMulti October 16, 2017 | Flag Reply
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of 0 vote


using namespace std;

class Connection {
		Connection(int from, int to)
			from_ = from;
			to_ = to;
		int from_, to_;

string Topology(vector<Connection> const &a)
	unordered_map<int, int> nodes;
	for (auto c : a) {
		if (++nodes[c.from_] > 2 ||
			++nodes[c.to_] > 2)
			return "star";
	if (a.size() == nodes.size() - 1) {
		return "bus";
	if (a.size() == nodes.size()) {
		return "ring";
	return "something else";

int main()
	cout << Topology({Connection(1, 2), Connection(2, 3)}) << "\n";
	cout << Topology({Connection(1, 2), Connection(2, 3), Connection(3, 1)}) << "\n";
	cout << Topology({Connection(2, 1), Connection(3, 1), Connection(4, 1), Connection(5, 1)}) << "\n";

	return 0;

- Alex November 01, 2017 | Flag Reply
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of 0 vote

If it is sure that it should be one of the topologies(star, bus, circular)

if(no of nodes == no of connections)
else if(all connections point to same destination or the second array has only one unique element)

- Gugan February 08, 2018 | Flag Reply

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