Google Interview Question for SDE-2s


Country: United States




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Traverse the graph with DFS\BFS
For each node check its degree (get the neighbors, if it is its own neighbor, i.e. cycle, add another one)

- Anonymous October 14, 2018 | Flag Reply
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My finding..

1) the number of subtrees of a node is called degree of the node.
2) The degree of a tree is the maximum degree of a node in the tree.
3) A binary tree is degree 2.

then i would to this

1) review all subtree of each node with BFS
2) have one variable to keep updating max num of subtree
3) once complete to review all nodes, then return max num which would be *degree* of the graph or tree

- lee19856 October 15, 2018 | Flag Reply
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@ shivamdurani220 - BTW, are you sure if these are really interview questions from Google interview? recently you uploaded many and looks like some of them are exactly same as one in leetcode.. i'm not trying to be rude but just wondering...

- lee19856 October 15, 2018 | Flag Reply
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struct AdjList{
    unordered_map<int,vector<int> > nodes;

    void insert(int i, int j){
        nodes[i].push_back(j);
        nodes[j].push_back(i);
    }

    vector<int> &getAllNeighbors(int n){
        return nodes[n];
    }
};


int findNodeWithDegree(int n, AdjList &adjList){
    unordered_map<int,int> counts;

    auto bfs = [&](int node){
        deque<int> q;
        counts[node] = 0;
        q.push_back(node);

        while(!q.empty()){
            auto front = q.front();
            q.pop_front();
            for(auto &other : adjList.getAllNeighbors(front)){
                if(counts.find(other) == counts.end()){
                    q.push_back(other);
                }
                counts[other]++;
            }
        }
    };

    for(auto &a : adjList.nodes){
        if(counts.find(a.first) == counts.end()){
            bfs(a.first);
        }
    }

    for(auto &p  : counts){
        if(p.second == n){
            return p.first;
        }
    }

    return -1;
}

int main() {
    AdjList adj;
    adj.insert(0,1);
    adj.insert(0,2);
    adj.insert(0,3);
    adj.insert(1,2);
    adj.insert(1,4);
    adj.insert(1,5);
    adj.insert(4,5);
    adj.insert(2,4);
    adj.insert(2,5);
    adj.insert(2,3);
    adj.nodes[6] = vector<int>();

    cout << findNodeWithDegree(5,adj) << endl;
    return 0;
}

- Guy October 17, 2018 | Flag Reply
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Though the code to calculate vertex degree depends on the implementation of Graph data structure, here is a simple method to calculate a degree of a vertex.

We return TRUE if any of the vertex has a degree equal to N.

public static int degree(Graph G, int v)
{
  int degree = 0;
  for (int w : G.adj(v)) degree++;
  return degree;
}

- Saurabh October 17, 2018 | Flag Reply
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we can make a adjency list and then we can count for each node and say how many nodes have n degree.

- sharadnailwal96 November 25, 2018 | Flag Reply


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