Google Interview Question for SDE1s


Country: United States
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
1
of 1 vote

Python solution:

def helper(adj, visited, start, end):
	if start == end:
		return True
	if len(adj[start])==0:
		return False
	visited[start] = True
	return all([helper(adj, visited, child, end) for child in adj[start] if not visited[child]])

def allPathsConnect(adj, start, end):
	v = {node:False for node in adj}
	return helper(adj, v, start, end)

- Angus January 01, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

You need to find a path from start that does not end in the end node. You can brute force all paths for from start. You can optimize if you store for a given node if it has a path to end so you do not redo the same work over and over. Recursive solution should be straight forward.

- Chris January 02, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

I'd do something like this,

class Node {
    int num;
    HashSet<Node> next = new HashSet<>();
}

boolean allPathsConnect(Node curr, HashSet<Node> visited, Node end) {
    if (curr == end) return true;
    if (visited.contains(curr)) return true;
    else if (curr.next.isEmpty()) return false;
    else {
        visited.add(curr);
        for (Node i : curr.next) {
            if (!allPathsConnect(i, visited, end)) return false;
        }
        return true;
    }
}

- PeyarTheriyaa January 03, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I believe you're asked about direct graph, so in this case you can build dominator tree and check whether all nodes dominate end node.

- mem January 01, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Python solution!

def helper(adj, visited, start, end):
	if start == end:
		return True
	if len(adj[start])==0:
		return False
	visited[start] = True
	return all([helper(adj, visited, child, end) for child in adj[start] if not visited[child]])

def allPathsConnect(adj, start, end):
	v = {node:False for node in adj}
	return helper(adj, v, start, end)

- Angus January 01, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

This would be a DFS where you check two things 1) you are not going to a node you already visited and 2) if the final node is not the ending node return false.

Use a hashset to keep track of nodes already visited, then when you are on a node go to it's children that you haven't visited yet. If that node doesn't have any children that you haven't visited then return false.

- mojojo January 01, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

It is a DFS. If reached means if it is end node, then return true. As soon as the program reaches null means end of path, return false.

- Victor January 02, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Actually, I made a mistake here as I assumed it was a tree. Since it is a graph, the error condition is that if all the neighbors of "a" current node are visited and the current node is not the end node, then return false. Meaning that there is a path that does not lead to end node.

- Victor January 05, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

Was this a phone screen or onsite?

- Ram January 02, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

This was onsite!

- nikki January 03, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

The only certain answer to this question is BFS.

From the start node, you have to scan all path and as we traverse down till we either reach the end node or flag it false for when you have checked all nodes.

- hprem991 January 02, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Brute force:
For each path starting from the start point, we do DFS to see if it can reach the end point. If all of the paths can reach end point, then we are good. Otherwise, return false;

Optimization:
Because we did a lot of redundant work during the DFS phase, we can memorize the successful paths for the previous DFS. And the next DFS is based on all of the previous DFS results, if we reach to a point which is a node in the successful path, then we are done.

- Anonymous January 03, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

can you revisit vertices in a path?
lets say start was 1, end was 4. Is this a valid path :
1 - 2 - 3 - 5 - 6 - 2 - 4

- Anonymous January 04, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

Can you revisit vertices in a path? Lets say start is 1 and end is 4.
Is this a valid path?
1 - 2 - 3 - 5 - 6 - 2 - 4

- Kaz January 04, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

compute the max flow with start node and end node, if it is equal to number of edges coming out of source then report 'true' else 'false'.

- ravan January 06, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

find maxflow between start node and end node, if its equal to number of edges coming out of start node then report true else false.

- ravan January 06, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

This is how I would do it... some sort of DFS with a little dynamic programming.
So, if for a current vertex that is visited all the path for all adjacent vertices goes T, then all the paths from the current vertex go to T also. If this is done recursively... it should give you the right answer. Basically, if conn[S] == True then NECESSARILY CONNECTED.

connected(v, T)
	visited[v] = True
	
	flag = False
	for u in adj(v) do
		if  (u == T) OR // this is the ending vertex, so True
			(visited[u] == True AND conn[u] == True) then // this is visited and also the paths from it goes to T
			flag = True
		
		if visited[u] == False then
			flag = connected(u, T)
			// no need to check further...
			if flag == False then
				break;

	conn[v] = flag
	return flag
	
MAIN:
	flag = connected(S, T)
	if flag == False then
		NOT NECESSARILY CONNECTED
	else
		NECESSARILY CONNECTED

- Anonymous January 11, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

This is how I would do it... some sort of DFS with a little dynamic programming.
So, if for a current vertex that is visited all the path for all adjacent vertices goes T, then all the paths from the current vertex go to T also. If this is done recursively... it should give you the right answer. Basically, if conn[S] == True then NECESSARILY CONNECTED.

connected(v, T)
	visited[v] = True
	
	flag = False
	for u in adj(v) do
		if  (u == T) OR // this is the ending vertex, so True
			(visited[u] == True AND conn[u] == True) then // this is visited and also the paths from it goes to T
			flag = True
		
		if visited[u] == False then
			flag = connected(u, T)
			// no need to check further...
			if flag == False then
				break;

	conn[v] = flag
	return flag
	
MAIN:
	flag = connected(S, T)
	if flag == False then
		NOT NECESSARILY CONNECTED
	else
		NECESSARILY CONNECTED

- p.andrey January 11, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- lad February 07, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

Assuming that if cycles are detected than the graph would not be considered Necesssarily connected, firstly using DFS find out if any cycle is present, if yes then directly return false.

We need to find all the nodes which are reachable from Source node before we reach desstination node. These are the nodes present in all the paths between sorce and destination node. For this graph to be necessarily connected none of these should be a sink node. All paths paasing through them should sink at destination node. So for all these nodes check if any node is a Sink node, i.e. has no out going edges then the graph is not 'necessarily connected'. If no such node found then graph is 'necessarily connected'

Complexity will be: O(n^2) , n: no. of vertices in the graph

- Shruti January 12, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Your statement: "Assuming that if cycles are detected then the graph would not be considered Necessarily connected..." - so, you say if a graph that contains a cycle along a path from S->T then S->T are not necessarily connected? If that is true, then yes, the DFS solution is incomplete, but it can be adapted extremely easy and avoid O(n^2).

- Anonymous January 12, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

was there any info given on how the graph is stored or it was left upto you to decide however you want to store the graph?

- jadonv January 13, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

to me it's BFS..

begin with "start node" and review all adjcent node and again and again.. if any of those does not go to "end node", then return false. otherwise, true. there are many youtube out there explaining this. ex) find shortest path.. or all connected?

Google's nature is search company. at least people will need to know all search algorithm (including data structure) very well along with Big O.

frankly speaking, if this was really the interview question in on-site, i would say it was easy one. (don't feel bad but i'm just saying and telling you experience from my friends). because some that i know said.. their interview question was "do code some sort of game".

- lee19856 January 14, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Consider the graph with the destination node deleted (remove all edges going in and out of the destination, too). This graph represents the paths that can be formed before hitting the destination node (paths that are eligible for failure). 1) In this graph, is there a path from the source to a zero-outdegree vertex? If so, this is a failing path. 2) Is there a path from the source to a cycle? If so, this is a failing path.

If neither case happens, there are no failing paths and the source and destination are necessarily connected.

The case of a cycle from the source can be checked with DFS easily enough. The case of a path to a zero-outdegree edge can be part of the same DFS, or can be done separately with BFS (but why if you're already doing DFS?)

- Anonymous January 23, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

test

- test February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

testesteestestste

- test February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- guest February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- guest February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- guest February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- a passing guy February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- a passing guy February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

{{I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.}}

- a passing guy February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- a passing lad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- ncielad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- passinglad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- lad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- lad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- lad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test3 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test3 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test3 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
{{
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
}}
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- test3 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- lad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- lad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- lad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                         -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaabbbdddddd February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaabbbdddddd February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For example, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- Anonymous February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- aaa February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)

- qqq February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

- qqq February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.

- qqq February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- lad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- 123 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- 123 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- 123 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- 123 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- 123 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- lad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- lad February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (

node1->node2->node5

where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- Anonymous February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (

node1->node2->node5

where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- 123 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. (

node1->node2->node5

where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- 123 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,

node1(start) -> node 2 -> node3 -< node4(end)
                                     -> node5

it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. ( node1, node2, node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- Anonymous February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I don't know any of the solutions above works.
My understanding of the question is, not just all direct paths from Start node. For exmaple, if the graph is like this,
node1(start) -> node 2 -> node3 -< node4(end)
-> node5
it is not necessarily connected even though there is a path from node1(start) to node4(end) when "all paths" are not connected. ( node1->node2->node5 where it is terminated)
If we just use DFS or BFS, it will find a path to end(node4) and return true.
I don't think google's interview question is so easy to run dfs against direct edges from start node only.

- Anonymous February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

123

- 132 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

1231231312312

- 123 February 07, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

12312321312
1231323

- 123 February 07, 2019 | Flag Reply


Add a Comment
Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

Books

is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.

Learn More

Videos

CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.

Learn More

Resume Review

Most engineers make critical mistakes on their resumes -- we can fix your resume with our custom resume review service. And, we use fellow engineers as our resume reviewers, so you can be sure that we "get" what you're saying.

Learn More

Mock Interviews

Our Mock Interviews will be conducted "in character" just like a real interview, and can focus on whatever topics you want. All our interviewers have worked for Microsoft, Google or Amazon, you know you'll get a true-to-life experience.

Learn More