Dada una serie de palabras, imprima todos los anagramas juntos. Por ejemplo, si la array dada es {“gato”, “perro”, “tac”, “dios”, “acto”}, entonces la salida puede ser “gato tac acto perro dios”.
Hemos discutido dos métodos diferentes en la publicación anterior . En esta publicación, se discute una solución más eficiente.
Trie estructura de datos se puede utilizar para una solución más eficiente. Inserte el orden ordenado de cada palabra en el trie. Ya que todos los anagramas terminarán en el mismo Node hoja. Podemos comenzar una lista enlazada en los Nodes hoja donde cada Node representa el índice de la array original de palabras. Finalmente, atraviesa el Trie. Mientras recorre el Trie, recorra cada lista enlazada una línea a la vez. Los siguientes son los pasos detallados.
1) Cree un Trie vacío
2) Una por una, tome todas las palabras de la secuencia de entrada. Haga lo siguiente para cada palabra
… a) Copie la palabra en un búfer.
… b)Ordenar el búfer
… c) Insertar el búfer ordenado y el índice de esta palabra en Trie. Cada Node hoja de Trie es el encabezado de una lista de índice. La lista de índice almacena el índice de palabras en la secuencia original. Si el búfer ordenado ya está presente, insertamos el índice de esta palabra en la lista de índice.
3) Triángulo transversal. Mientras atraviesa, si llega a un Node de hoja, recorra la lista de índice. E imprima todas las palabras utilizando el índice obtenido de la lista de índices.
A continuación se muestra la implementación del enfoque anterior:
C++
// An efficient program to print all anagrams together
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#define NO_OF_CHARS 26
// Structure to represent list node for indexes of words in
// the given sequence. The list nodes are used to connect
// anagrams at leaf nodes of Trie
struct IndexNode
{
int index;
struct IndexNode* next;
};
// Structure to represent a Trie Node
struct TrieNode
{
bool isEnd; // indicates end of word
struct TrieNode* child[NO_OF_CHARS]; // 26 slots each for 'a' to 'z'
struct IndexNode* head; // head of the index list
};
// A utility function to create a new Trie node
struct TrieNode* newTrieNode()
{
struct TrieNode* temp = new TrieNode;
temp->isEnd = 0;
temp->head = NULL;
for (int i = 0; i < NO_OF_CHARS; ++i)
temp->child[i] = NULL;
return temp;
}
/* Following function is needed for library function qsort(). Refer
http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */
int compare(const void* a, const void* b)
{ return *(char*)a - *(char*)b; }
/* A utility function to create a new linked list node */
struct IndexNode* newIndexNode(int index)
{
struct IndexNode* temp = new IndexNode;
temp->index = index;
temp->next = NULL;
return temp;
}
// A utility function to insert a word to Trie
void insert(struct TrieNode** root, char* word, int index)
{
// Base case
if (*root == NULL)
*root = newTrieNode();
if (*word != '\0')
insert( &( (*root)->child[tolower(*word) - 'a'] ), word+1, index );
else // If end of the word reached
{
// Insert index of this word to end of index linked list
if ((*root)->isEnd)
{
IndexNode* pCrawl = (*root)->head;
while( pCrawl->next )
pCrawl = pCrawl->next;
pCrawl->next = newIndexNode(index);
}
else // If Index list is empty
{
(*root)->isEnd = 1;
(*root)->head = newIndexNode(index);
}
}
}
// This function traverses the built trie. When a leaf node is reached,
// all words connected at that leaf node are anagrams. So it traverses
// the list at leaf node and uses stored index to print original words
void printAnagramsUtil(struct TrieNode* root, char *wordArr[])
{
if (root == NULL)
return;
// If a lead node is reached, print all anagrams using the indexes
// stored in index linked list
if (root->isEnd)
{
// traverse the list
IndexNode* pCrawl = root->head;
while (pCrawl != NULL)
{
printf( "%s ", wordArr[ pCrawl->index ] );
pCrawl = pCrawl->next;
}
}
for (int i = 0; i < NO_OF_CHARS; ++i)
printAnagramsUtil(root->child[i], wordArr);
}
// The main function that prints all anagrams together. wordArr[] is input
// sequence of words.
void printAnagramsTogether(char* wordArr[], int size)
{
// Create an empty Trie
struct TrieNode* root = NULL;
// Iterate through all input words
for (int i = 0; i < size; ++i)
{
// Create a buffer for this word and copy the word to buffer
int len = strlen(wordArr[i]);
char *buffer = new char[len+1];
strcpy(buffer, wordArr[i]);
// Sort the buffer
qsort( (void*)buffer, strlen(buffer), sizeof(char), compare );
// Insert the sorted buffer and its original index to Trie
insert(&root, buffer, i);
}
// Traverse the built Trie and print all anagrams together
printAnagramsUtil(root, wordArr);
}
// Driver program to test above functions
int main()
{
char* wordArr[] = {"cat", "dog", "tac", "god", "act", "gdo"};
int size = sizeof(wordArr) / sizeof(wordArr[0]);
printAnagramsTogether(wordArr, size);
return 0;
}
Java
// An efficient program to print all
// anagrams together
import java.util.Arrays;
import java.util.LinkedList;
public class GFG
{
static final int NO_OF_CHARS = 26;
// Class to represent a Trie Node
static class TrieNode
{
boolean isEnd; // indicates end of word
// 26 slots each for 'a' to 'z'
TrieNode[] child = new TrieNode[NO_OF_CHARS];
// head of the index list
LinkedList<Integer> head;
// constructor
public TrieNode()
{
isEnd = false;
head = new LinkedList<>();
for (int i = 0; i < NO_OF_CHARS; ++i)
child[i] = null;
}
}
// A utility function to insert a word to Trie
static TrieNode insert(TrieNode root,String word,
int index, int i)
{
// Base case
if (root == null)
{
root = new TrieNode();
}
if (i < word.length() )
{
int index1 = word.charAt(i) - 'a';
root.child[index1] = insert(root.child[index1],
word, index, i+1 );
}
else // If end of the word reached
{
// Insert index of this word to end of
// index linked list
if (root.isEnd == true)
{
root.head.add(index);
}
else // If Index list is empty
{
root.isEnd = true;
root.head.add(index);
}
}
return root;
}
// This function traverses the built trie. When a leaf
// node is reached, all words connected at that leaf
// node are anagrams. So it traverses the list at leaf
// node and uses stored index to print original words
static void printAnagramsUtil(TrieNode root,
String wordArr[])
{
if (root == null)
return;
// If a lead node is reached, print all anagrams
// using the indexes stored in index linked list
if (root.isEnd)
{
// traverse the list
for(Integer pCrawl: root.head)
System.out.println(wordArr[pCrawl]);
}
for (int i = 0; i < NO_OF_CHARS; ++i)
printAnagramsUtil(root.child[i], wordArr);
}
// The main function that prints all anagrams together.
// wordArr[] is input sequence of words.
static void printAnagramsTogether(String wordArr[],
int size)
{
// Create an empty Trie
TrieNode root = null;
// Iterate through all input words
for (int i = 0; i < size; ++i)
{
// Create a buffer for this word and copy the
// word to buffer
char[] buffer = wordArr[i].toCharArray();
// Sort the buffer
Arrays.sort(buffer);
// Insert the sorted buffer and its original
// index to Trie
root = insert(root, new String(buffer), i, 0);
}
// Traverse the built Trie and print all anagrams
// together
printAnagramsUtil(root, wordArr);
}
// Driver program to test above functions
public static void main(String args[])
{
String wordArr[] = {"cat", "dog", "tac", "god",
"act", "gdo"};
int size = wordArr.length;
printAnagramsTogether(wordArr, size);
}
}
// This code is contributed by Sumit Ghosh
C#
// An efficient C# program to print all
// anagrams together
using System;
using System.Collections.Generic;
class GFG
{
static readonly int NO_OF_CHARS = 26;
// Class to represent a Trie Node
public class TrieNode
{
// indicates end of word
public bool isEnd;
// 26 slots each for 'a' to 'z'
public TrieNode[] child = new TrieNode[NO_OF_CHARS];
// head of the index list
public List<int> head;
// constructor
public TrieNode()
{
isEnd = false;
head = new List<int>();
for (int i = 0; i < NO_OF_CHARS; ++i)
child[i] = null;
}
}
// A utility function to insert a word to Trie
static TrieNode insert(TrieNode root,String word,
int index, int i)
{
// Base case
if (root == null)
{
root = new TrieNode();
}
if (i < word.Length )
{
int index1 = word[i] - 'a';
root.child[index1] = insert(root.child[index1],
word, index, i + 1 );
}
// If end of the word reached
else
{
// Insert index of this word to end of
// index linked list
if (root.isEnd == true)
{
root.head.Add(index);
}
// If Index list is empty
else
{
root.isEnd = true;
root.head.Add(index);
}
}
return root;
}
// This function traverses the built trie.
// When a leaf node is reached, all words
// connected at that leaf node are anagrams.
// So it traverses the list at leaf node
// and uses stored index to print original words
static void printAnagramsUtil(TrieNode root,
String []wordArr)
{
if (root == null)
return;
// If a lead node is reached,
// print all anagrams using the
// indexes stored in index linked list
if (root.isEnd)
{
// traverse the list
foreach(int pCrawl in root.head)
Console.WriteLine(wordArr[pCrawl]);
}
for (int i = 0; i < NO_OF_CHARS; ++i)
printAnagramsUtil(root.child[i], wordArr);
}
// The main function that prints
// all anagrams together. wordArr[]
// is input sequence of words.
static void printAnagramsTogether(String []wordArr,
int size)
{
// Create an empty Trie
TrieNode root = null;
// Iterate through all input words
for (int i = 0; i < size; ++i)
{
// Create a buffer for this word
// and copy the word to buffer
char[] buffer = wordArr[i].ToCharArray();
// Sort the buffer
Array.Sort(buffer);
// Insert the sorted buffer and
// its original index to Trie
root = insert(root, new String(buffer), i, 0);
}
// Traverse the built Trie and
// print all anagrams together
printAnagramsUtil(root, wordArr);
}
// Driver code
public static void Main(String []args)
{
String []wordArr = {"cat", "dog", "tac", "god",
"act", "gdo"};
int size = wordArr.Length;
printAnagramsTogether(wordArr, size);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// An efficient program to print all
// anagrams together
let NO_OF_CHARS = 26;
// Class to represent a Trie Node
class TrieNode
{
constructor()
{
this.isEnd = false; // indicates end of word
// 26 slots each for 'a' to 'z'
this.child = new Array(NO_OF_CHARS);
for (let i = 0; i < NO_OF_CHARS; ++i)
this.child[i] = null;
// head of the index list
this.head=[];
}
}
// A utility function to insert a word to Trie
function insert(root,word,index,i)
{
// Base case
if (root == null)
{
root = new TrieNode();
}
if (i < word.length )
{
let index1 = word[i].charCodeAt(0) - 'a'.charCodeAt(0);
root.child[index1] = insert(root.child[index1],
word, index, i+1 );
}
else // If end of the word reached
{
// Insert index of this word to end of
// index linked list
if (root.isEnd == true)
{
root.head.push(index);
}
else // If Index list is empty
{
root.isEnd = true;
root.head.push(index);
}
}
return root;
}
// This function traverses the built trie. When a leaf
// node is reached, all words connected at that leaf
// node are anagrams. So it traverses the list at leaf
// node and uses stored index to print original words
function printAnagramsUtil(root,wordArr)
{
if (root == null)
return;
// If a lead node is reached, print all anagrams
// using the indexes stored in index linked list
if (root.isEnd)
{
// traverse the list
for(let pCrawl=0;pCrawl<root.head.length;pCrawl++)
document.write(wordArr[root.head[pCrawl]]+"<br>");
}
for (let i = 0; i < NO_OF_CHARS; ++i)
printAnagramsUtil(root.child[i], wordArr);
}
// The main function that prints all anagrams together.
// wordArr[] is input sequence of words.
function printAnagramsTogether(wordArr,size)
{
// Create an empty Trie
let root = null;
// Iterate through all input words
for (let i = 0; i < size; ++i)
{
// Create a buffer for this word and copy the
// word to buffer
let buffer = wordArr[i].split("");
// Sort the buffer
(buffer).sort();
// Insert the sorted buffer and its original
// index to Trie
root = insert(root, (buffer).join(""), i, 0);
}
// Traverse the built Trie and print all anagrams
// together
printAnagramsUtil(root, wordArr);
}
// Driver program to test above functions
let wordArr=["cat", "dog", "tac", "god",
"act", "gdo"];
let size = wordArr.length;
printAnagramsTogether(wordArr, size);
// This code is contributed by rag2127
</script>
Producción:
cat tac act dog god gdo
Complejidad de tiempo: O(MN+N*MlogM) tiempo donde-
N = Número de cuerdas
M = Longitud de la cuerda más grande
Insertar todas las palabras en el trie toma tiempo O(MN) y ordenar el búfer toma tiempo O(N*MlogM)
Espacio auxiliar: Para almacenar todas las strings necesitamos asignar O(26*M*N) ~ O(MN) espacio para el Trie.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA