Sorting array of strings (or words) using Trie
Given an array of strings, print them in alphabetical (dictionary) order. If there are duplicates in input array, we need to print them only once.
Examples:
Input : "abc", "xy", "bcd" Output : abc bcd xy Input : "Beginner", "for", "Beginner", "a", "portal", "to", "learn", "can", "be", "computer", "science", "zoom", "yup", "fire", "in", "data" Output : a be can computer data fire for Beginner in learn portal science to yup zoom
Trie is an efficient data structure used for storing data like strings. To print the string in alphabetical order we have to first insert in the trie and then perform preorder traversal to print in alphabetical order.
Implementation:
CPP
// C++ program to sort an array of strings // using Trie #include <bits/stdc++.h> using namespace std; const int MAX_CHAR = 26; struct Trie { // index is set when node is a leaf // node, otherwise -1; int index; Trie* child[MAX_CHAR]; /*to make new trie*/ Trie() { for ( int i = 0; i < MAX_CHAR; i++) child[i] = NULL; index = -1; } }; /* function to insert in trie */ void insert(Trie* root, string str, int index) { Trie* node = root; for ( int i = 0; i < str.size(); i++) { /* taking ascii value to find index of child node */ char ind = str[i] - 'a' ; /* making new path if not already */ if (!node->child[ind]) node->child[ind] = new Trie(); // go to next node node = node->child[ind]; } // Mark leaf (end of word) and store // index of word in arr[] node->index = index; } /* function for preorder traversal */ bool preorder(Trie* node, string arr[]) { if (node == NULL) return false ; for ( int i = 0; i < MAX_CHAR; i++) { if (node->child[i] != NULL) { /* if leaf node then print key*/ if (node->child[i]->index != -1) cout << arr[node->child[i]->index] << endl; preorder(node->child[i], arr); } } } void printSorted(string arr[], int n) { Trie* root = new Trie(); // insert all keys of dictionary into trie for ( int i = 0; i < n; i++) insert(root, arr[i], i); // print keys in lexicographic order preorder(root, arr); } // Driver code int main() { string arr[] = { "abc" , "xy" , "bcd" }; int n = sizeof (arr) / sizeof (arr[0]); printSorted(arr, n); return 0; } |
Java
// Java program to sort an array of strings using Trie // Author : Rohit Jain // GFG user_id : @rj03012002 import java.util.*; public class GFG { // Alphabet size static final int MAX_CHAR = 26 ; // trie node static class Trie { // index is set when node is a leaf // node, otherwise -1; int index; Trie child[] = new Trie[MAX_CHAR]; /*to make new trie*/ Trie() { for ( int i = 0 ; i < MAX_CHAR; i++) child[i] = null ; index = - 1 ; } } /* function to insert in trie */ static void insert(Trie root, String str, int index) { Trie node = root; for ( int i = 0 ; i < str.length(); i++) { /* taking ascii value to find index of child node */ int ind = str.charAt(i) - 'a' ; /* making new path if not already */ if (node.child[ind] == null ) node.child[ind] = new Trie(); // go to next node node = node.child[ind]; } // Mark leaf (end of word) and store // index of word in arr[] node.index = index; } /* function for preorder traversal */ static boolean preorder(Trie node, String arr[]) { if (node == null ) { return false ; } for ( int i = 0 ; i < MAX_CHAR; i++) { if (node.child[i] != null ) { /* if leaf node then print key*/ if (node.child[i].index != - 1 ) { System.out.print( arr[node.child[i].index] + " " ); } preorder(node.child[i], arr); } } return false ; } static void printSorted(String arr[], int n) { Trie root = new Trie(); // insert all keys of dictionary into trie for ( int i = 0 ; i < n; ++i) { insert(root, arr[i], i); } // print keys in lexicographic order preorder(root, arr); } public static void main(String[] args) { String arr[] = { "abc" , "xy" , "bcd" }; int n = arr.length; printSorted(arr, n); } } |
Python3
# Python3 program to sort an array of strings # using Trie MAX_CHAR = 26 class Trie: # index is set when node is a leaf # node, otherwise -1; # to make new trie def __init__( self ): self .child = [ None for i in range (MAX_CHAR)] self .index = - 1 # def to insert in trie def insert(root, str ,index): node = root for i in range ( len ( str )): # taking ascii value to find index of # child node ind = ord ( str [i]) - ord ( 'a' ) # making new path if not already if (node.child[ind] = = None ): node.child[ind] = Trie() # go to next node node = node.child[ind] # Mark leaf (end of word) and store # index of word in arr[] node.index = index # function for preorder traversal def preorder(node, arr): if (node = = None ): return False for i in range (MAX_CHAR): if (node.child[i] ! = None ): # if leaf node then print key if (node.child[i].index ! = - 1 ): print (arr[node.child[i].index]) preorder(node.child[i], arr) def printSorted(arr,n): root = Trie() # insert all keys of dictionary into trie for i in range (n): insert(root, arr[i], i) # print keys in lexicographic order preorder(root, arr) # Driver code arr = [ "abc" , "xy" , "bcd" ] n = len (arr) printSorted(arr, n) # This code is contributed by shinjanpatra |
C#
// C# program to sort an array of strings using Trie using System; class GFG { static int MAX_CHAR = 26; // trie node public class Trie { // index is set when node is a leaf node, otherwise -1 public int index; public Trie[] child = new Trie[MAX_CHAR]; // constructor to make a new trie node public Trie() { for ( int i = 0; i < MAX_CHAR; i++) child[i] = null ; index = -1; } } /* function to insert in trie */ static void Insert(Trie root, string str, int index) { Trie node = root; for ( int i = 0; i < str.Length; i++) { /* taking ASCII value to find the index of the child node */ int ind = str[i] - 'a' ; /* making a new path if not already */ if (node.child[ind] == null ) node.child[ind] = new Trie(); // go to the next node node = node.child[ind]; } // mark the leaf node (end of the word) and store the index of the word in arr[] node.index = index; } /* function for preorder traversal */ static void Preorder(Trie node, string [] arr) { if (node == null ) { return ; } for ( int i = 0; i < MAX_CHAR; i++) { if (node.child[i] != null ) { /* if leaf node, then print the key */ if (node.child[i].index != -1) { Console.WriteLine(arr[node.child[i].index] + " " ); } Preorder(node.child[i], arr); } } } static void PrintSorted( string [] arr, int n) { Trie root = new Trie(); // insert all keys of the dictionary into the trie for ( int i = 0; i < n; i++) { Insert(root, arr[i], i); } // print the keys in lexicographic order Preorder(root, arr); } static void Main( string [] args) { string [] arr = { "abc" , "xy" , "bcd" }; int n = arr.Length; PrintSorted(arr, n); } } // This code is contributed by Aman Kumar. |
Javascript
<script> // JavaScript program to sort an array of strings // using Trie const MAX_CHAR = 26; class Trie { // index is set when node is a leaf // node, otherwise -1; /*to make new trie*/ constructor() { this .child = new Array(MAX_CHAR).fill( null ); this .index = -1; } } /* function to insert in trie */ function insert(root,str,index) { let node = root; for (let i = 0; i < str.length; i++) { /* taking ascii value to find index of child node */ let ind = str.charCodeAt(i) - 'a' .charCodeAt(0); /* making new path if not already */ if (node.child[ind] == null ) node.child[ind] = new Trie(); // go to next node node = node.child[ind]; } // Mark leaf (end of word) and store // index of word in arr[] node.index = index; } /* function for preorder traversal */ function preorder(node, arr) { if (node == null ) return false ; for (let i = 0; i < MAX_CHAR; i++) { if (node.child[i] != null ) { /* if leaf node then print key*/ if (node.child[i].index != -1) document.write(arr[node.child[i].index], "</br>" ); preorder(node.child[i], arr); } } } function printSorted(arr,n) { let root = new Trie(); // insert all keys of dictionary into trie for (let i = 0; i < n; i++) insert(root, arr[i], i); // print keys in lexicographic order preorder(root, arr); } // Driver code let arr = [ "abc" , "xy" , "bcd" ]; let n = arr.length; printSorted(arr, n); // This code is contributed by shinjanpatra </script> |
Output
abc bcd xy
Time Complexity: O(n*m) where n is the length of the array and m is the length of the longest word.
Auxiliary Space: O(n*m)