Phonebook using a trie
$begingroup$
I implemented a PhoneBook utilizing a Trie
data structure.
Could I ask you to evaluate it?
Did I apply Trie
correctly for such case or is it better to store data somehow else?
I'm looking not only for working solution but also for an optimal one in terms of run-time and storage.
From my calculations, it takes NlogT to add()
and get()
where N is a number of chars and T is a number of children.
Am I right?
Here is the logic
public final class PhoneBook {
private final Node name;
private final Node surname;
private final Comparator<Record> comparator;
public PhoneBook() {
this.name = new Node();
this.surname = new Node();
comparator = (r1, r2) -> {
int result = r1.getName().compareTo(r2.getName());
if (result == 0) {
result = r1.getSurname().compareTo(r2.getSurname());
if (result == 0) {
result = r1.getNumber().compareTo(r2.getNumber());
}
}
return result;
};
}
public void add(final Record record) {
add(record.getName().toLowerCase(), record, name);
add(record.getSurname().toLowerCase(), record, surname);
}
public SortedSet<Record> get(final String prefix) {
final String lc = prefix.toLowerCase();
final List<Record> recordsRetrievedByName = get(lc, name);
final List<Record> recordsRetrievedBySurname = get(lc, surname);
final SortedSet<Record> result = new TreeSet<>(comparator);
result.addAll(recordsRetrievedByName);
result.addAll(recordsRetrievedBySurname);
return result;
}
private List<Record> get(final String prefix, final Node ancestor) {
Node node = ancestor;
for (final char c: prefix.toCharArray()) {
final Node child = node.children.get(c);
if (child == null) {
return Collections.emptyList();
}
node = child;
}
return node.records;
}
private void add(final String str, final Record record, final Node ancestor) {
Node node = ancestor;
for (final char c: str.toCharArray()) {
final Node child;
if (node.children.containsKey(c)) {
child = node.children.get(c);
} else {
child = new Node();
node.children.put(c, child);
}
child.records.add(record);
node = child;
}
}
private static final class Node {
private final Map<Character, Node> children = new TreeMap<>();
private final List<Record> records = new ArrayList<>();
}
}
And the Record
immutable object
public final class Record {
private final String name;
private final String surname;
private final String number;
//constructor, getters, toString
}
java algorithm tree complexity trie
New contributor
$endgroup$
add a comment |
$begingroup$
I implemented a PhoneBook utilizing a Trie
data structure.
Could I ask you to evaluate it?
Did I apply Trie
correctly for such case or is it better to store data somehow else?
I'm looking not only for working solution but also for an optimal one in terms of run-time and storage.
From my calculations, it takes NlogT to add()
and get()
where N is a number of chars and T is a number of children.
Am I right?
Here is the logic
public final class PhoneBook {
private final Node name;
private final Node surname;
private final Comparator<Record> comparator;
public PhoneBook() {
this.name = new Node();
this.surname = new Node();
comparator = (r1, r2) -> {
int result = r1.getName().compareTo(r2.getName());
if (result == 0) {
result = r1.getSurname().compareTo(r2.getSurname());
if (result == 0) {
result = r1.getNumber().compareTo(r2.getNumber());
}
}
return result;
};
}
public void add(final Record record) {
add(record.getName().toLowerCase(), record, name);
add(record.getSurname().toLowerCase(), record, surname);
}
public SortedSet<Record> get(final String prefix) {
final String lc = prefix.toLowerCase();
final List<Record> recordsRetrievedByName = get(lc, name);
final List<Record> recordsRetrievedBySurname = get(lc, surname);
final SortedSet<Record> result = new TreeSet<>(comparator);
result.addAll(recordsRetrievedByName);
result.addAll(recordsRetrievedBySurname);
return result;
}
private List<Record> get(final String prefix, final Node ancestor) {
Node node = ancestor;
for (final char c: prefix.toCharArray()) {
final Node child = node.children.get(c);
if (child == null) {
return Collections.emptyList();
}
node = child;
}
return node.records;
}
private void add(final String str, final Record record, final Node ancestor) {
Node node = ancestor;
for (final char c: str.toCharArray()) {
final Node child;
if (node.children.containsKey(c)) {
child = node.children.get(c);
} else {
child = new Node();
node.children.put(c, child);
}
child.records.add(record);
node = child;
}
}
private static final class Node {
private final Map<Character, Node> children = new TreeMap<>();
private final List<Record> records = new ArrayList<>();
}
}
And the Record
immutable object
public final class Record {
private final String name;
private final String surname;
private final String number;
//constructor, getters, toString
}
java algorithm tree complexity trie
New contributor
$endgroup$
$begingroup$
It's look fine. I am also a beginner so i don't have very good knowledge about it. So i suggest to you please check this link for more geeksforgeeks.org/trie-insert-and-search
$endgroup$
– Ved Prakash
10 mins ago
add a comment |
$begingroup$
I implemented a PhoneBook utilizing a Trie
data structure.
Could I ask you to evaluate it?
Did I apply Trie
correctly for such case or is it better to store data somehow else?
I'm looking not only for working solution but also for an optimal one in terms of run-time and storage.
From my calculations, it takes NlogT to add()
and get()
where N is a number of chars and T is a number of children.
Am I right?
Here is the logic
public final class PhoneBook {
private final Node name;
private final Node surname;
private final Comparator<Record> comparator;
public PhoneBook() {
this.name = new Node();
this.surname = new Node();
comparator = (r1, r2) -> {
int result = r1.getName().compareTo(r2.getName());
if (result == 0) {
result = r1.getSurname().compareTo(r2.getSurname());
if (result == 0) {
result = r1.getNumber().compareTo(r2.getNumber());
}
}
return result;
};
}
public void add(final Record record) {
add(record.getName().toLowerCase(), record, name);
add(record.getSurname().toLowerCase(), record, surname);
}
public SortedSet<Record> get(final String prefix) {
final String lc = prefix.toLowerCase();
final List<Record> recordsRetrievedByName = get(lc, name);
final List<Record> recordsRetrievedBySurname = get(lc, surname);
final SortedSet<Record> result = new TreeSet<>(comparator);
result.addAll(recordsRetrievedByName);
result.addAll(recordsRetrievedBySurname);
return result;
}
private List<Record> get(final String prefix, final Node ancestor) {
Node node = ancestor;
for (final char c: prefix.toCharArray()) {
final Node child = node.children.get(c);
if (child == null) {
return Collections.emptyList();
}
node = child;
}
return node.records;
}
private void add(final String str, final Record record, final Node ancestor) {
Node node = ancestor;
for (final char c: str.toCharArray()) {
final Node child;
if (node.children.containsKey(c)) {
child = node.children.get(c);
} else {
child = new Node();
node.children.put(c, child);
}
child.records.add(record);
node = child;
}
}
private static final class Node {
private final Map<Character, Node> children = new TreeMap<>();
private final List<Record> records = new ArrayList<>();
}
}
And the Record
immutable object
public final class Record {
private final String name;
private final String surname;
private final String number;
//constructor, getters, toString
}
java algorithm tree complexity trie
New contributor
$endgroup$
I implemented a PhoneBook utilizing a Trie
data structure.
Could I ask you to evaluate it?
Did I apply Trie
correctly for such case or is it better to store data somehow else?
I'm looking not only for working solution but also for an optimal one in terms of run-time and storage.
From my calculations, it takes NlogT to add()
and get()
where N is a number of chars and T is a number of children.
Am I right?
Here is the logic
public final class PhoneBook {
private final Node name;
private final Node surname;
private final Comparator<Record> comparator;
public PhoneBook() {
this.name = new Node();
this.surname = new Node();
comparator = (r1, r2) -> {
int result = r1.getName().compareTo(r2.getName());
if (result == 0) {
result = r1.getSurname().compareTo(r2.getSurname());
if (result == 0) {
result = r1.getNumber().compareTo(r2.getNumber());
}
}
return result;
};
}
public void add(final Record record) {
add(record.getName().toLowerCase(), record, name);
add(record.getSurname().toLowerCase(), record, surname);
}
public SortedSet<Record> get(final String prefix) {
final String lc = prefix.toLowerCase();
final List<Record> recordsRetrievedByName = get(lc, name);
final List<Record> recordsRetrievedBySurname = get(lc, surname);
final SortedSet<Record> result = new TreeSet<>(comparator);
result.addAll(recordsRetrievedByName);
result.addAll(recordsRetrievedBySurname);
return result;
}
private List<Record> get(final String prefix, final Node ancestor) {
Node node = ancestor;
for (final char c: prefix.toCharArray()) {
final Node child = node.children.get(c);
if (child == null) {
return Collections.emptyList();
}
node = child;
}
return node.records;
}
private void add(final String str, final Record record, final Node ancestor) {
Node node = ancestor;
for (final char c: str.toCharArray()) {
final Node child;
if (node.children.containsKey(c)) {
child = node.children.get(c);
} else {
child = new Node();
node.children.put(c, child);
}
child.records.add(record);
node = child;
}
}
private static final class Node {
private final Map<Character, Node> children = new TreeMap<>();
private final List<Record> records = new ArrayList<>();
}
}
And the Record
immutable object
public final class Record {
private final String name;
private final String surname;
private final String number;
//constructor, getters, toString
}
java algorithm tree complexity trie
java algorithm tree complexity trie
New contributor
New contributor
edited 3 hours ago
200_success
130k17153419
130k17153419
New contributor
asked 3 hours ago
PavelPavel
1163
1163
New contributor
New contributor
$begingroup$
It's look fine. I am also a beginner so i don't have very good knowledge about it. So i suggest to you please check this link for more geeksforgeeks.org/trie-insert-and-search
$endgroup$
– Ved Prakash
10 mins ago
add a comment |
$begingroup$
It's look fine. I am also a beginner so i don't have very good knowledge about it. So i suggest to you please check this link for more geeksforgeeks.org/trie-insert-and-search
$endgroup$
– Ved Prakash
10 mins ago
$begingroup$
It's look fine. I am also a beginner so i don't have very good knowledge about it. So i suggest to you please check this link for more geeksforgeeks.org/trie-insert-and-search
$endgroup$
– Ved Prakash
10 mins ago
$begingroup$
It's look fine. I am also a beginner so i don't have very good knowledge about it. So i suggest to you please check this link for more geeksforgeeks.org/trie-insert-and-search
$endgroup$
– Ved Prakash
10 mins ago
add a comment |
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$begingroup$
It's look fine. I am also a beginner so i don't have very good knowledge about it. So i suggest to you please check this link for more geeksforgeeks.org/trie-insert-and-search
$endgroup$
– Ved Prakash
10 mins ago