Lab126 Interview Question for Software Engineer / Developers


Country: United States
Interview Type: In-Person




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4
of 6 vote

Hmm, seems to be a good application case for UnionFind.
en.m.wikipedia.org/wiki/Disjoint-set_data_structure

- gameboy1024 October 13, 2014 | Flag Reply
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0
of 0 votes

Could you provide the solution? What complexity will it have?
I see a lot of pluses, though it does not seem to be the linear solution. Check my O(n) solution with graphs and DFS below

- alisovenko October 14, 2014 | Flag
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1
of 1 vote

Find all connected components of the graph where nodes are all the distinct characters, here (A,B,C,D,E,F,G,H) and edges are the given pairs. Here A-B , C-D ... are edges.

- khunima October 13, 2014 | Flag Reply
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0
of 0 votes

That is a nice approach but maintaining Graph just for evaluating this. Will it not be expensive ? I am asking.

- Anonymous October 13, 2014 | Flag
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0
of 0 vote

I think it is just bucket printing. Create bucket and do the rest.

- Soulslayer October 13, 2014 | Flag Reply
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0
of 0 vote

Use Hashset

- Soulslayer October 13, 2014 | Flag Reply
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0
of 0 votes

HashSet can have only unique entries. Here we have mapping A-B and A-D..

- Anonymous October 13, 2014 | Flag
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0
of 0 vote

Read about disjoint set data structure

- Rahul Sharma October 13, 2014 | Flag Reply
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0
of 0 vote

There are two algorithms here : quickunion and quickfind.
If we use quickfind, it will be kind of slow, if we have to add the pairs.
it will be a better choice if we use quickunion!

- vishal.cs.bits October 13, 2014 | Flag Reply
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0
of 0 vote

Union-find seems to be bad variant as it has the api "are two points connected" and "connect two points". So this would lead to O(n^2) algorigthm, where N is the number of pairs.
For ASCI string we can use a simple graph that can be checked for connectivity using DFS. And although the implementation is a bit massive, it's very simple:
1. create graph as adjacency list
2. dfs through it, all dfs cycle discovers the connected elements

This is effectively O(n) implementation

public static void clusterize(char[][] pairs) {
        // build graph
        List<Character>[] graph = buildGraph(pairs);

        char[] marked = new char[255];

        for (int i = 0; i < graph.length; i++) {
            if (marked[i] == 0 && graph[i] != null) {
                // we have not come to this letter yet
                dfsAndPrint(i, graph, marked);
                System.out.println();
            }
        }
    }

    private static void dfsAndPrint(int i, List<Character>[] graph, char[] marked) {
        if (marked[i] > 0) {
            return;
        }
        System.out.printf("%s, ", (char)i);
        marked[i] = 1;

        if (graph[i] != null) {
            for (Character c : graph[i]) {
                dfsAndPrint(c, graph, marked);
            }
        }
    }

    private static List<Character>[] buildGraph(char[][] pairs) {
        List<Character>[] graph = new ArrayList[255];

        for (final char[] pair : pairs) {
            addToGraph(graph, pair[0], pair[1]);
            addToGraph(graph, pair[1], pair[0]);
        }
        return graph;
    }

    public static void addToGraph(List<Character>[] graph, char v1, char v2) {
        List<Character> characters = graph[v1];
        if (characters == null) {
            characters = new ArrayList<>();
            graph[v1] = characters;
        }
        characters.add(v2);
    }

    public static void main(String[] args) {
        char[][] input = {
                {'a', 'b'},
                {'c', 'd'},
                {'e', 'f'},
                {'g', 'h'},
                {'a', 'd'},
                {'f', 'g'}
        };
        clusterize(input);
    }

- alisovenko October 14, 2014 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Union find approach would be a O(n) and should be relatively simple:

public static ArrayList<char[]> getSets(char[][] pairs){
  HashMap<Character, Character> unionFindMap = new HashMap<Character, Character>();
  //add the mappings of left to right
  for(char[] pair : pairs){
    union(unionFindMap, pair[0], pair[1]);
  }

  //organize group the mapped content
  HashMap<Character, ArrayList<Character>> resultCollator = new HashMap<Character, ArrayList<Character>>();
  for(Entry<Character, Character> entry : unionFindMap.entrySet()){
    Character val = entry.getValue();
    Character key = entry.getKey();
    if(val == null){
      val = key;
    }
    ArrayList<Character> list = resultCollator.get(val);
    if(list == null){
      list = new ArrayList<Character>();
      resultCollator.put(val, list);
    }
    list.add(key);
  }

  //make the output
  ArrayList<char[]> results = new ArrayList<char[]>(resultCollator.size());
  for(ArrayList<Character> list : resultCollator.getValue()){
    char[] set = new char[list.size()];
    for(int i = 0; i < list.size(); i++){
      set[i] = list.get(i);
    }
    results.add(set);
  }
  return results;
}  

private void union(HashMap<Character, Character> unionFindMap, char c1, char c2){
  if(!unionFindMap.contains(c2)){
    unionFindMap.put(c2, null);
  }
  char dest =  find(unionFindMap, c2);
  unionFindMap.put(c1, c2);
}

private char find(HashMap<Character, Character> unionFindMap, char c){
  Character dest = unionFindMap.get(c);
  if(dest == null){
    return c;
  }
  char newDest = find(unionFindMap, dest);
  unionFindMap.put(c, newDest);
  return newDest;
}

- Zortlord October 15, 2014 | Flag Reply
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0
of 0 votes

Your code output for example input:
[b]
[a, c, d]
[e]
[f]
[g, h]

- alisovenko October 15, 2014 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

Implemented a few things incorrectly including collating the results:

public static ArrayList<char[]> getSets(char[][] pairs){
        HashMap<Character, Character> unionFindMap = new HashMap<Character, Character>();
        //add the mappings of left to right
        for(char[] pair : pairs){
          union(unionFindMap, pair[0], pair[1]);
        }

        //organize group the mapped content
        HashMap<Character, ArrayList<Character>> resultCollator = new HashMap<Character, ArrayList<Character>>();
        for(Character key : unionFindMap.keySet()){
          Character val = find(unionFindMap,key);
          if(val == null){
            val = key;
          }
          ArrayList<Character> list = resultCollator.get(val);
          if(list == null){
            list = new ArrayList<Character>();
            resultCollator.put(val, list);
          }
          list.add(key);
        }

        //make the output
        ArrayList<char[]> results = new ArrayList<char[]>(resultCollator.size());
        for(ArrayList<Character> list : resultCollator.values()){
          char[] set = new char[list.size()];
          for(int i = 0; i < list.size(); i++){
            set[i] = list.get(i);
          }
          results.add(set);
        }
        return results;
      }  

      private static void union(HashMap<Character, Character> unionFindMap, char c1, char c2){
        if(!unionFindMap.containsKey(c2)){
          unionFindMap.put(c2, null);
        }
        if(!unionFindMap.containsKey(c1)){
          unionFindMap.put(c1, null);
        }
        char dest1 = find(unionFindMap, c1);
        char dest2 = find(unionFindMap, c2);
        unionFindMap.put(dest2, dest1);        
      }

      private static char find(HashMap<Character, Character> unionFindMap, char c){
        Character dest = unionFindMap.get(c);
        if(dest == null){
          return c;
        }
        char newDest = find(unionFindMap, dest);
        unionFindMap.put(c, newDest);
        return newDest;
      }

- Zortlord October 17, 2014 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

How about using double linked list.
like this :
(find right end of first and left end of second then connect them.)
a<->b, c<->d, e<->f, g<->h
a-d
(right end of 'a' is 'b' and left end of 'd' is 'c' so connect 'b' and 'c')
a<->b<->c<->d
:
:
Here's the code :
(In fact the code has a problem :
if you add pairs in same group alreday, the code will make cirular link
so need to add the code checking the case)

import java.util.HashMap;
import java.util.Map;

public class Main {

	public static class DoubleLink { 
		Character left;
		Character right;
	}

	static Map <Character, DoubleLink> map = new HashMap<Character, DoubleLink>();

	static void addLink(Character a, Character b) {
		Character LeftEndB = getLeftEnd(b);
		Character RightEndA = getRightEnd(a);
		map.get(LeftEndB).left = RightEndA;
		map.get(RightEndA).right = LeftEndB;
	}

	static Character getLeftEnd(Character a) {
		if (!map.containsKey(a)) {
			map.put(a, new DoubleLink());
			return a;
		}
		Character end = a;
		while (map.get(end).left != null) {
			end = map.get(end).left;
		}
		return end;
	}

	static Character getRightEnd(Character a) {
		if (!map.containsKey(a)) {
			map.put(a, new DoubleLink());
			return a;
		}
		Character end = a;
		while (map.get(end).right != null) {
			end = map.get(end).right;
		}
		return end;
	}

	public static void main(String[] args) {
		addLink('a','b');
		addLink('c','d');
		addLink('e','f');
		addLink('g','h');
		addLink('a','d');
		addLink('f','g');

		for (Character key : map.keySet()) {
			if (map.get(key).left == null) {
				System.out.print(key);
				while (true) {
					key = map.get(key).right;
					if (key == null) {
						break;
					}
					System.out.print(key);
				}
				System.out.println();
			}
		}
	}
}

- curiosus October 20, 2014 | Flag Reply


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