Dada una lista de contactos que existen en un directorio telefónico. La tarea es implementar la consulta de búsqueda para el directorio telefónico. La consulta de búsqueda en una string ‘ str ‘ muestra todos los contactos que tienen el prefijo ‘ str ‘. Una propiedad especial de la función de búsqueda es que, cuando un usuario busca un contacto de la lista de contactos, se muestran sugerencias (contactos con prefijo como la string ingresada) después de que el usuario ingresa cada carácter.
Nota: los contactos en la lista consisten solo en minúsculas.
Ejemplo:
Input : contacts [] = {“gforgeeks” , “geeksquiz” }
Query String = “gekk”
Output : Suggestions based on "g" are
geeksquiz
gforgeeks
Suggestions based on "ge" are
geeksquiz
No Results Found for "gek"
No Results Found for "gekk"
El directorio telefónico se puede implementar de manera eficiente utilizando Trie Data Structure. Insertamos todos los contactos en Trie.
Generalmente, la consulta de búsqueda en un Trie es para determinar si la string está presente o no en el trie, pero en este caso se nos pide encontrar todas las strings con cada prefijo de ‘str’. Esto es equivalente a hacer un recorrido DFS en un gráfico . Desde un Node Trie, visite los Nodes Trie adyacentes y hágalo recursivamente hasta que no haya más adyacentes. Esta función recursiva tomará 2 argumentos, uno como Trie Node que apunta al Trie Node actual que se está visitando y otro como la string que almacena la string encontrada hasta el momento con el prefijo ‘str’.
Cada Trie Node almacena una variable booleana ‘isLast’ que es verdadera si el Node representa el final de un contacto (palabra).
// This function displays all words with given
// prefix. "node" represents last node when
// path from root follows characters of "prefix".
displayContacts (TreiNode node, string prefix)
If (node.isLast is true)
display prefix
// finding adjacent nodes
for each character ‘i’ in lower case Alphabets
if (node.child[i] != NULL)
displayContacts(node.child[i], prefix+i)
El usuario ingresará la string carácter por carácter y necesitamos mostrar sugerencias con el prefijo formado después de cada carácter ingresado.
Entonces, un enfoque para encontrar el prefijo que comienza con la string formada es verificar si el prefijo existe en el Trie, si es así, llame a la función displayContacts(). En este enfoque, después de cada carácter ingresado, verificamos si la string existe en el Trie.
En lugar de verificar una y otra vez, podemos mantener un puntero prevNode‘ que apunta al TrieNode que corresponde al último carácter ingresado por el usuario, ahora debemos verificar el Node secundario para el ‘prevNode’ cuando el usuario ingresa otro carácter para verificar si existe en el Trie. Si el nuevo prefijo no está en Trie, entonces toda la string que se forma ingresando caracteres después de ‘prefijo’ tampoco se puede encontrar en Trie. Entonces rompemos el bucle que se usa para generar prefijos uno por uno e imprimimos «No se encontraron resultados» para todos los caracteres restantes.
C++
// C++ Program to Implement a Phone
// Directory Using Trie Data Structure
#include <bits/stdc++.h>
using namespace std;
struct TrieNode
{
// Each Trie Node contains a Map 'child'
// where each alphabet points to a Trie
// Node.
// We can also use a fixed size array of
// size 256.
unordered_map<char,TrieNode*> child;
// 'isLast' is true if the node represents
// end of a contact
bool isLast;
// Default Constructor
TrieNode()
{
// Initialize all the Trie nodes with NULL
for (char i = 'a'; i <= 'z'; i++)
child[i] = NULL;
isLast = false;
}
};
// Making root NULL for ease so that it doesn't
// have to be passed to all functions.
TrieNode *root = NULL;
// Insert a Contact into the Trie
void insert(string s)
{
int len = s.length();
// 'itr' is used to iterate the Trie Nodes
TrieNode *itr = root;
for (int i = 0; i < len; i++)
{
// Check if the s[i] is already present in
// Trie
TrieNode *nextNode = itr->child[s[i]];
if (nextNode == NULL)
{
// If not found then create a new TrieNode
nextNode = new TrieNode();
// Insert into the Map
itr->child[s[i]] = nextNode;
}
// Move the iterator('itr') ,to point to next
// Trie Node
itr = nextNode;
// If its the last character of the string 's'
// then mark 'isLast' as true
if (i == len - 1)
itr->isLast = true;
}
}
// This function simply displays all dictionary words
// going through current node. String 'prefix'
// represents string corresponding to the path from
// root to curNode.
void displayContactsUtil(TrieNode *curNode, string prefix)
{
// Check if the string 'prefix' ends at this Node
// If yes then display the string found so far
if (curNode->isLast)
cout << prefix << endl;
// Find all the adjacent Nodes to the current
// Node and then call the function recursively
// This is similar to performing DFS on a graph
for (char i = 'a'; i <= 'z'; i++)
{
TrieNode *nextNode = curNode->child[i];
if (nextNode != NULL)
displayContactsUtil(nextNode, prefix + (char)i);
}
}
// Display suggestions after every character enter by
// the user for a given query string 'str'
void displayContacts(string str)
{
TrieNode *prevNode = root;
string prefix = "";
int len = str.length();
// Display the contact List for string formed
// after entering every character
int i;
for (i=0; i<len; i++)
{
// 'prefix' stores the string formed so far
prefix += (char)str[i];
// Get the last character entered
char lastChar = prefix[i];
// Find the Node corresponding to the last
// character of 'prefix' which is pointed by
// prevNode of the Trie
TrieNode *curNode = prevNode->child[lastChar];
// If nothing found, then break the loop as
// no more prefixes are going to be present.
if (curNode == NULL)
{
cout << "nNo Results Found for "" << prefix
<< "" n";
i++;
break;
}
// If present in trie then display all
// the contacts with given prefix.
cout << "nSuggestions based on "" << prefix
<< "" are n";
displayContactsUtil(curNode, prefix);
// Change prevNode for next prefix
prevNode = curNode;
}
// Once search fails for a prefix, we print
// "Not Results Found" for all remaining
// characters of current query string "str".
for (; i<len; i++)
{
prefix += (char)str[i];
cout << "nNo Results Found for "" << prefix
<< "" n";
}
}
// Insert all the Contacts into the Trie
void insertIntoTrie(string contacts[],int n)
{
// Initialize root Node
root = new TrieNode();
// Insert each contact into the trie
for (int i = 0; i < n; i++)
insert(contacts[i]);
}
// Driver program to test above functions
int main()
{
// Contact list of the User
string contacts[] = {"gforgeeks" , "geeksquiz"};
// Size of the Contact List
int n = sizeof(contacts)/sizeof(string);
// Insert all the Contacts into Trie
insertIntoTrie(contacts, n);
string query = "gekk";
// Note that the user will enter 'g' then 'e', so
// first display all the strings with prefix as 'g'
// and then all the strings with prefix as 'ge'
displayContacts(query);
return 0;
}
Java
// Java Program to Implement a Phone
// Directory Using Trie Data Structure
import java.util.*;
class TrieNode
{
// Each Trie Node contains a Map 'child'
// where each alphabet points to a Trie
// Node.
HashMap<Character,TrieNode> child;
// 'isLast' is true if the node represents
// end of a contact
boolean isLast;
// Default Constructor
public TrieNode()
{
child = new HashMap<Character,TrieNode>();
// Initialize all the Trie nodes with NULL
for (char i = 'a'; i <= 'z'; i++)
child.put(i,null);
isLast = false;
}
}
class Trie
{
TrieNode root;
// Insert all the Contacts into the Trie
public void insertIntoTrie(String contacts[])
{
root = new TrieNode();
int n = contacts.length;
for (int i = 0; i < n; i++)
{
insert(contacts[i]);
}
}
// Insert a Contact into the Trie
public void insert(String s)
{
int len = s.length();
// 'itr' is used to iterate the Trie Nodes
TrieNode itr = root;
for (int i = 0; i < len; i++)
{
// Check if the s[i] is already present in
// Trie
TrieNode nextNode = itr.child.get(s.charAt(i));
if (nextNode == null)
{
// If not found then create a new TrieNode
nextNode = new TrieNode();
// Insert into the HashMap
itr.child.put(s.charAt(i),nextNode);
}
// Move the iterator('itr') ,to point to next
// Trie Node
itr = nextNode;
// If its the last character of the string 's'
// then mark 'isLast' as true
if (i == len - 1)
itr.isLast = true;
}
}
// This function simply displays all dictionary words
// going through current node. String 'prefix'
// represents string corresponding to the path from
// root to curNode.
public void displayContactsUtil(TrieNode curNode,
String prefix)
{
// Check if the string 'prefix' ends at this Node
// If yes then display the string found so far
if (curNode.isLast)
System.out.println(prefix);
// Find all the adjacent Nodes to the current
// Node and then call the function recursively
// This is similar to performing DFS on a graph
for (char i = 'a'; i <= 'z'; i++)
{
TrieNode nextNode = curNode.child.get(i);
if (nextNode != null)
{
displayContactsUtil(nextNode, prefix + i);
}
}
}
// Display suggestions after every character enter by
// the user for a given string 'str'
void displayContacts(String str)
{
TrieNode prevNode = root;
// 'flag' denotes whether the string entered
// so far is present in the Contact List
String prefix = "";
int len = str.length();
// Display the contact List for string formed
// after entering every character
int i;
for (i = 0; i < len; i++)
{
// 'str' stores the string entered so far
prefix += str.charAt(i);
// Get the last character entered
char lastChar = prefix.charAt(i);
// Find the Node corresponding to the last
// character of 'str' which is pointed by
// prevNode of the Trie
TrieNode curNode = prevNode.child.get(lastChar);
// If nothing found, then break the loop as
// no more prefixes are going to be present.
if (curNode == null)
{
System.out.println("nNo Results Found for ""
+ prefix + """);
i++;
break;
}
// If present in trie then display all
// the contacts with given prefix.
System.out.println("nSuggestions based on ""
+ prefix + "" are");
displayContactsUtil(curNode, prefix);
// Change prevNode for next prefix
prevNode = curNode;
}
for ( ; i < len; i++)
{
prefix += str.charAt(i);
System.out.println("nNo Results Found for ""
+ prefix + """);
}
}
}
// Driver code
class Main
{
public static void main(String args[])
{
Trie trie = new Trie();
String contacts [] = {"gforgeeks", "geeksquiz"};
trie.insertIntoTrie(contacts);
String query = "gekk";
// Note that the user will enter 'g' then 'e' so
// first display all the strings with prefix as 'g'
// and then all the strings with prefix as 'ge'
trie.displayContacts(query);
}
}
C#
// C# Program to Implement a Phone
// Directory Using Trie Data Structure
using System;
using System.Collections.Generic;
class TrieNode
{
// Each Trie Node contains a Map 'child'
// where each alphabet points to a Trie
// Node.
public Dictionary<char, TrieNode> child;
// 'isLast' is true if the node represents
// end of a contact
public bool isLast;
// Default Constructor
public TrieNode()
{
child = new Dictionary<char, TrieNode>();
// Initialize all the Trie nodes with NULL
for (char i = 'a'; i <= 'z'; i++)
child.Add(i, null);
isLast = false;
}
}
class Trie
{
public TrieNode root;
// Insert all the Contacts into the Trie
public void insertIntoTrie(String []contacts)
{
root = new TrieNode();
int n = contacts.Length;
for (int i = 0; i < n; i++)
{
insert(contacts[i]);
}
}
// Insert a Contact into the Trie
public void insert(String s)
{
int len = s.Length;
// 'itr' is used to iterate the Trie Nodes
TrieNode itr = root;
for (int i = 0; i < len; i++)
{
// Check if the s[i] is already present in
// Trie
TrieNode nextNode = itr.child[s[i]];
if (nextNode == null)
{
// If not found then create a new TrieNode
nextNode = new TrieNode();
// Insert into the Dictionary
if(itr.child.ContainsKey(s[i]))
itr.child[s[i]] = nextNode;
else
itr.child.Add(s[i], nextNode);
}
// Move the iterator('itr') ,to point to next
// Trie Node
itr = nextNode;
// If its the last character of the string 's'
// then mark 'isLast' as true
if (i == len - 1)
itr.isLast = true;
}
}
// This function simply displays all dictionary words
// going through current node. String 'prefix'
// represents string corresponding to the path from
// root to curNode.
public void displayContactsUtil(TrieNode curNode,
String prefix)
{
// Check if the string 'prefix' ends at this Node
// If yes then display the string found so far
if (curNode.isLast)
Console.WriteLine(prefix);
// Find all the adjacent Nodes to the current
// Node and then call the function recursively
// This is similar to performing DFS on a graph
for (char i = 'a'; i <= 'z'; i++)
{
TrieNode nextNode = curNode.child[i];
if (nextNode != null)
{
displayContactsUtil(nextNode, prefix + i);
}
}
}
// Display suggestions after every character enter by
// the user for a given string 'str'
public void displayContacts(String str)
{
TrieNode prevNode = root;
// 'flag' denotes whether the string entered
// so far is present in the Contact List
String prefix = "";
int len = str.Length;
// Display the contact List for string formed
// after entering every character
int i;
for (i = 0; i < len; i++)
{
// 'str' stores the string entered so far
prefix += str[i];
// Get the last character entered
char lastChar = prefix[i];
// Find the Node corresponding to the last
// character of 'str' which is pointed by
// prevNode of the Trie
TrieNode curNode = prevNode.child[lastChar];
// If nothing found, then break the loop as
// no more prefixes are going to be present.
if (curNode == null)
{
Console.WriteLine("\nNo Results Found for'"
+ prefix + "'");
i++;
break;
}
// If present in trie then display all
// the contacts with given prefix.
Console.WriteLine("Suggestions based on '"
+ prefix + "' are");
displayContactsUtil(curNode, prefix);
// Change prevNode for next prefix
prevNode = curNode;
}
for ( ; i < len; i++)
{
prefix += str[i];
Console.WriteLine("\nNo Results Found for '"
+ prefix + "'");
}
}
}
// Driver code
public class GFG
{
public static void Main(String []args)
{
Trie trie = new Trie();
String []contacts = {"gforgeeks", "geeksquiz"};
trie.insertIntoTrie(contacts);
String query = "gekk";
// Note that the user will enter 'g' then 'e' so
// first display all the strings with prefix as 'g'
// and then all the strings with prefix as 'ge'
trie.displayContacts(query);
}
}
// This code is contributed by PrinciRaj1992
Producción:
Suggestions based on "g" are geeksquiz gforgeeks Suggestions based on "ge" are geeksquiz No Results Found for "gek" No Results Found for "gekk"
Este artículo es una contribución de Chirag Agarwal . Si le gusta GeeksforGeeks y le gustaría contribuir, también puede escribir un artículo usando contribuya.geeksforgeeks.org o envíe su artículo por correo a contribuya@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA