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Внедрете телефонен указател

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Има списък с контакти, които съществуват в телефонен указател. Задачата е да се въведе заявка за търсене в телефонния указател. Заявката за търсене на низ „ ул “ показва всички контакти с префикс „ ул ’. Едно специално свойство на функцията за търсене е, че когато потребител търси контакт от списъка с контакти, предложенията (Контакти с префикс като низ, въведен така за) се показват, след като потребителят въведе всеки знак.
Забележка: Контактите в списъка се състоят само от малки букви. Пример:

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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' 

Препоръчително: Моля, решете го на ПРАКТИКА първо, преди да преминете към решението.

Телефонният указател може да бъде ефективно внедрен с помощта на Опитай Структура на данните. Вмъкваме всички контакти в Trie. Обикновено заявката за търсене на Trie е да се определи дали низът присъства или не в trie, но в този случай от нас се иска да намерим всички низове с всеки префикс на „str“. Това е еквивалентно на извършване на a DFS обхождане на графика . От Trie възел посетете съседни Trie възли и правете това рекурсивно, докато няма повече съседни. Тази рекурсивна функция ще приеме 2 аргумента, един като Trie Node, който сочи към текущия Trie Node, който е посетен, и друг като низ, който съхранява намерения до момента низ с префикс като „str“. Всеки възел Trie съхранява булева променлива „isLast“, която е вярна, ако възелът представлява края на контакт (дума).

// 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)

Потребителят ще въвежда низа символ по знак и ние трябва да покажем предложения с префикса, образуван след всеки въведен знак. Така че един подход за намиране на префикса, започващ с образувания низ, е да проверите дали префиксът съществува в Trie, ако да, след това да извикате функцията displayContacts(). При този подход след всеки въведен знак проверяваме дали низът съществува в Trie. Вместо да проверяваме отново и отново, можем да поддържаме указател prevNode ', който сочи към TrieNode, който съответства на последния въведен знак от потребителя, сега трябва да проверим дъщерния възел за 'prevNode', когато потребителят въведе друг знак, за да провери дали той съществува в Trie. Ако новият префикс не е в Trie, тогава всички низове, които са образувани чрез въвеждане на знаци след 'prefix', също не могат да бъдат намерени в Trie. Така че прекъсваме цикъла, който се използва за генериране на префикси един по един и отпечатваме „Няма намерен резултат“ за всички останали знаци. 



C++
// C++ Program to Implement a Phone // Directory Using Trie Data Structure #include    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 ';  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<CharacterTrieNode> child;  // 'isLast' is true if the node represents  // end of a contact  boolean isLast;  // Default Constructor  public TrieNode()  {  child = new HashMap<CharacterTrieNode>();  // Initialize all the Trie nodes with NULL  for (char i = 'a'; i <= 'z'; i++)  child.put(inull);  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);  } } 
Python3
# Python Program to Implement a Phone # Directory Using Trie Data Structure class TrieNode: def __init__(self): # Each Trie Node contains a Map 'child' # where each alphabet points to a Trie # Node. self.child = {} self.is_last = False # Making root NULL for ease so that it doesn't # have to be passed to all functions. root = TrieNode() # Insert a Contact into the Trie def insert(string): # 'itr' is used to iterate the Trie Nodes itr = root for char in string: # Check if the s[i] is already present in # Trie if char not in itr.child: # If not found then create a new TrieNode itr.child[char] = TrieNode() # Move the iterator('itr') to point to next # Trie Node itr = itr.child[char] # If its the last character of the string 's' # then mark 'isLast' as true itr.is_last = True # This function simply displays all dictionary words # going through current node. String 'prefix' # represents string corresponding to the path from # root to curNode. def display_contacts_util(cur_node prefix): # Check if the string 'prefix' ends at this Node # If yes then display the string found so far if cur_node.is_last: print(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 i in range(ord('a') ord('z') + 1): char = chr(i) next_node = cur_node.child.get(char) if next_node: display_contacts_util(next_node prefix + char) # Display suggestions after every character enter by # the user for a given query string 'str' def displayContacts(string): prev_node = root prefix = '' # Display the contact List for string formed # after entering every character for i char in enumerate(string): # 'prefix' stores the string formed so far prefix += char # Find the Node corresponding to the last # character of 'prefix' which is pointed by # prevNode of the Trie cur_node = prev_node.child.get(char) # If nothing found then break the loop as # no more prefixes are going to be present. if not cur_node: print(f'No Results Found for {prefix}n') break # If present in trie then display all # the contacts with given prefix. print(f'Suggestions based on {prefix} are 'end=' ') display_contacts_util(cur_node prefix) print() # Change prevNode for next prefix prev_node = cur_node # Once search fails for a prefix we print # 'Not Results Found' for all remaining # characters of current query string 'str'. for char in string[i+1:]: prefix += char print(f'No Results Found for {prefix}n') # Insert all the Contacts into the Trie def insertIntoTrie(contacts): # Insert each contact into the trie for contact in contacts: insert(contact) # Driver program to test above functions # Contact list of the User  contacts=['gforgeeks''geeksquiz'] # Size of the Contact List n=len(contacts) # Insert all the Contacts into Trie insertIntoTrie(contacts) 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) # This code is contributed by Aman Kumar 
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('nSuggestions 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 
JavaScript
<script>  // Javascript Program to Implement a Phone  // Directory Using Trie Data Structure  class TrieNode {  constructor() {  // 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.  this.child = {};  // 'isLast' is true if the node represents  // end of a contact  this.isLast = false;  }  }  // Making root NULL for ease so that it doesn't  // have to be passed to all functions.  let root = null;    // Insert a Contact into the Trie  function insert(s) {  const len = s.length;  // 'itr' is used to iterate the Trie Nodes  let itr = root;  for (let i = 0; i < len; i++) {  // Check if the s[i] is already present in  // Trie  const char = s[i];  let nextNode = itr.child[char];  if (nextNode === undefined) {  // If not found then create a new TrieNode  nextNode = new TrieNode();  // Insert into the Map  itr.child[char] = 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.  function displayContactsUtil(curNode prefix) {  // Check if the string 'prefix' ends at this Node  // If yes then display the string found so far  if (curNode.isLast) {  document.write(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 (let i = 97; i <= 122; i++) { const char = String.fromCharCode(i); const nextNode = curNode.child[char]; if (nextNode !== undefined) { displayContactsUtil(nextNode prefix + char); } } } // Display suggestions after every character enter by // the user for a given query string 'str' function displayContacts(str) { let prevNode = root; let prefix = ''; const len = str.length; // Display the contact List for string formed // after entering every character let i; for (i = 0; i < len; i++) { // 'prefix' stores the string formed so far prefix += str[i]; // Get the last character entered const lastChar = prefix[i]; // Find the Node corresponding to the last // character of 'prefix' which is pointed by // prevNode of the Trie const curNode = prevNode.child[lastChar]; // If nothing found then break the loop as // no more prefixes are going to be present. if (curNode === undefined) { document.write(`No Results Found for ${prefix}`+'
'
); i++; break; } // If present in trie then display all // the contacts with given prefix. document.write(`Suggestions based on ${prefix} are `); displayContactsUtil(curNode prefix); document.write('
'
); // Change prevNode for next prefix prevNode = curNode; } document.write('
'
); // 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 += str[i]; document.write('No Results Found for ' + prefix + '
'
); } } // Insert all the Contacts into the Trie function insertIntoTrie(contacts) { // Initialize root Node root = new TrieNode(); const n = contacts.length; // Insert each contact into the trie for (let i = 0; i < n; i++) { insert(contacts[i]); } } // Driver program to test above functions // Contact list of the User const contacts = ['gforgeeks' 'geeksquiz']; //Insert all the Contacts into Trie insertIntoTrie(contacts); const 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); // This code is contributed by Utkarsh Kumar. </script>

Изход
Suggestions based on gare geeksquiz gforgeeks Suggestions based on geare geeksquiz No Results Found for gek No Results Found for gekk

Времева сложност: O(n*m), където n е броят на контактите, а m е максималната дължина на низ от контакти.
Помощно пространство: O(n*m)