Design a datastructure for telephone directory - algorithm

Question

Design a datastructure for telephone directory for storing the name and phone number so that we can search for key given name and vice versa.
We can use 2 hash maps as follows
Map<String,int>
Map<int,String>
But it requires twice the memory.Can anyone suggest any other solution?

Answer 1

One person can have more than one number, and one number can belong to more than one person (members of a family). And as Nick said, a telephone number in general case can have non-numeric characters. All considered, instead of Map<String,int> you might be using Map<String,List<String>>, or to have only pointers to strings (in C++ terms), to avoid redundancy: Map<String*,List<String*>>.

Answer 2

A bimap (or "bidirectional map") is a
map that preserves the uniqueness of
its values as well as that of its
keys. This constraint enables bimaps
to support an "inverse view", which is
another bimap containing the same
entries as this bimap but with
reversed keys and values.
http://guava-libraries.googlecode.com/svn/trunk/javadoc/com/google/common/collect/BiMap.html
BiMap<String, Integer> biMap = HashBiMap.create();
biMap.put("Mike", 123);
biMap.put("John", 456);
biMap.put("Steven", 789);
BiMap<Integer, String> invertedBiMap = biMap.inverse();
Edit: Multimaps
Multimap<String, String> multimap = HashMultimap.create();
multimap.put("John", "111111111");
multimap.put("John", "222222222");
multimap.put("Mike", "333333333");
System.out.println(multimap.get("John")); //[222222222, 111111111]
for(Map.Entry<String, String> entry : multimap.entries()){
if(entry.getValue().equals("222222222")){
System.out.println(entry.getKey()); //John
}
}
//or
Multimap<String, String> inverted = HashMultimap.create();
Multimaps.invertFrom(multimap, inverted);
System.out.println(inverted.get("222222222")); //[John]

Answer 3

another possibility is storing the Strings in a trie, and have a '$' sign indicate the end of each string. use doubly linked pointers for each step, and hold a double linked pointer from each '$' (end of name) to its number (in an array or a list).
now, when you want to get a phone from name:
find the '$' indicating the end of the word for this string.
it is connected to a number in a list - that is your number.
when you want to get a name from a phone:
find the number, it is connected to a '$' sign.
follow the up-links all the way to the root, this will get you the name (reversed).
reverse it and you are done.
also, as I said in the comment (regarding the double map approach): assuming your strings are pretty large, and the map holds a pointer/reference to the string (and not the actual string), you can assume the storage space needed will not be double, but something much better.

Answer 4

Using a binary search tree for implementing the telephone directory is the best way to do it. Think about the practical implementation of the mobile phone telephone contact list. It is sorted alphabetically. If one uses map template then we shall not get a sorted list. You can not sort the elements of the map, and it's not effective.
The only way to do it is the binary tree way. Since, while adding a new entry it's inserted in a ordered way. So, no more sorting is needed. It's already ordered. Remember, left_tree < root, and root < right_tree in case of binary tree.