Word formation using concatenation of two dictionary words
Given a dictionary find out if given word can be made by two words in the dictionary.
Note: Words in the dictionary must be unique and the word to be formed should not be a repetition of same words that are present in the Trie.
Examples:
Input : dictionary[] = {"news", "abcd", "tree",
"geeks", "paper"}
word = "newspaper"
Output : Yes
We can form "newspaper" using "news" and "paper"
Input : dictionary[] = {"geeks", "code", "xyz",
"forgeeks", "paper"}
word = "geeksforgeeks"
Output : Yes
Input : dictionary[] = {"geek", "code", "xyz",
"forgeeks", "paper"}
word = "geeksforgeeks"
Output : No
The idea is store all words of dictionary in a Trie. We do prefix search for given word. Once we find a prefix, we search for rest of the word.
Algorithm :
1- Store all the words of the dictionary in a Trie. 2- Start searching for the given word in Trie. If it partially matched then split it into two parts and then search for the second part in the Trie. 3- If both found, then return true. 4- Otherwise return false.
Below is the implementation of above idea.
C++
// C++ program to check if a string can be// formed by concatenating two words#include<bits/stdc++.h>using namespace std; // Converts key current character into index// use only 'a' through 'z'#define char_int(c) ((int)c - (int)'a') // Alphabet size#define SIZE (26) // Trie Nodestruct TrieNode{ TrieNode *children[SIZE]; // isLeaf is true if the node represents // end of a word bool isLeaf;}; // Returns new trie node (initialized to NULLs)TrieNode *getNode(){ TrieNode * newNode = new TrieNode; newNode->isLeaf = false; for (int i =0 ; i< SIZE ; i++) newNode->children[i] = NULL; return newNode;} // If not present, inserts key into Trie// If the key is prefix of trie node, just// mark leaf nodevoid insert(TrieNode *root, string Key){ int n = Key.length(); TrieNode * pCrawl = root; for (int i=0; i<n; i++) { int index = char_int(Key[i]); if (pCrawl->children[index] == NULL) pCrawl->children[index] = getNode(); pCrawl = pCrawl->children[index]; } // make last node as leaf node pCrawl->isLeaf = true;} // Searches a prefix of key. If prefix is present,// returns its ending position in string. Else// returns -1.int findPrefix(struct TrieNode *root, string key){ int pos = -1, level; struct TrieNode *pCrawl = root; for (level = 0; level < key.length(); level++) { int index = char_int(key[level]); if (pCrawl->isLeaf == true) pos = level; if (!pCrawl->children[index]) return pos; pCrawl = pCrawl->children[index]; } if (pCrawl != NULL && pCrawl->isLeaf) return level;} // Function to check if word formation is possible// or notbool isPossible(struct TrieNode* root, string word){ // Search for the word in the trie and // store its position upto which it is matched int len = findPrefix(root, word); // print not possible if len = -1 i.e. not // matched in trie if (len == -1) return false; // If word is partially matched in the dictionary // as another word // search for the word made after splitting // the given word up to the length it is // already,matched string split_word(word, len, word.length()-(len)); int split_len = findPrefix(root, split_word); // check if word formation is possible or not return (len + split_len == word.length());} // Driver program to test above functionint main(){ // Let the given dictionary be following vector<string> dictionary = {"geeks", "forgeeks", "quiz", "geek"}; string word = "geeksquiz"; //word to be formed // root Node of trie TrieNode *root = getNode(); // insert all words of dictionary into trie for (int i=0; i<dictionary.size(); i++) insert(root, dictionary[i]); isPossible(root, word) ? cout << "Yes": cout << "No"; return 0;} |
Java
import java.util.ArrayList;import java.util.List; // Java program to check if a string can be // formed by concatenating two words public class GFG { // Alphabet size final static int SIZE = 26; // Trie Node static class TrieNode { TrieNode[] children = new TrieNode[SIZE]; // isLeaf is true if the node represents // end of a word boolean isLeaf; // constructor public TrieNode() { isLeaf = false; for (int i =0 ; i< SIZE ; i++) children[i] = null; } } static TrieNode root; // If not present, inserts key into Trie // If the key is prefix of trie node, just // mark leaf node static void insert(TrieNode root, String Key) { int n = Key.length(); TrieNode pCrawl = root; for (int i=0; i<n; i++) { int index = Key.charAt(i) - 'a'; if (pCrawl.children[index] == null) pCrawl.children[index] = new TrieNode(); pCrawl = pCrawl.children[index]; } // make last node as leaf node pCrawl.isLeaf = true; } // Searches a prefix of key. If prefix is present, // returns its ending position in string. Else // returns -1. static List<Integer> findPrefix(TrieNode root, String key) { List<Integer> prefixPositions = new ArrayList<Integer>(); int level; TrieNode pCrawl = root; for (level = 0; level < key.length(); level++) { int index = key.charAt(level) - 'a'; if (pCrawl.isLeaf == true) prefixPositions.add(level); if (pCrawl.children[index] == null) return prefixPositions; pCrawl = pCrawl.children[index]; } if (pCrawl != null && pCrawl.isLeaf) prefixPositions.add(level); return prefixPositions; } // Function to check if word formation is possible // or not static boolean isPossible(TrieNode root, String word) { // Search for the word in the trie and get its prefix positions // upto which there is matched List<Integer> prefixPositions1 = findPrefix(root, word); // Word formation is not possible if the word did not have // at least one prefix match if (prefixPositions1.isEmpty()) return false; // Search for rest of substring for each prefix match for (Integer len1 : prefixPositions1) { String restOfSubstring = word.substring(len1, word.length()); List<Integer> prefixPositions2 = findPrefix(root, restOfSubstring); for (Integer len2 : prefixPositions2) { // check if word formation is possible if (len1 + len2 == word.length()) return true; } } return false; } // Driver program to test above function public static void main(String args[]) { // Let the given dictionary be following String[] dictionary = {"news", "newspa", "paper", "geek"}; String word = "newspaper"; //word to be formed // root Node of trie root = new TrieNode(); // insert all words of dictionary into trie for (int i=0; i<dictionary.length; i++) insert(root, dictionary[i]); if(isPossible(root, word)) System.out.println( "Yes"); else System.out.println("No"); } }// This code is contributed by Sumit Ghosh// Updated by Narendra Jha |
C#
// C# program to check if a string can be // formed by concatenating two words using System;using System.Collections.Generic; class GFG { // Alphabet size readonly public static int SIZE = 26; // Trie Node public class TrieNode { public TrieNode []children = new TrieNode[SIZE]; // isLeaf is true if the node // represents end of a word public bool isLeaf; // constructor public TrieNode() { isLeaf = false; for (int i = 0 ; i < SIZE ; i++) children[i] = null; } } static TrieNode root; // If not present, inserts key into Trie // If the key is prefix of trie node, just // mark leaf node static void insert(TrieNode root, String Key) { int n = Key.Length; TrieNode pCrawl = root; for (int i = 0; i < n; i++) { int index = Key[i] - 'a'; if (pCrawl.children[index] == null) pCrawl.children[index] = new TrieNode(); pCrawl = pCrawl.children[index]; } // make last node as leaf node pCrawl.isLeaf = true; } // Searches a prefix of key. If prefix // is present, returns its ending position // in string. Else returns -1. static List<int> findPrefix(TrieNode root, String key) { List<int> prefixPositions = new List<int>(); int level; TrieNode pCrawl = root; for (level = 0; level < key.Length; level++) { int index = key[level] - 'a'; if (pCrawl.isLeaf == true) prefixPositions.Add(level); if (pCrawl.children[index] == null) return prefixPositions; pCrawl = pCrawl.children[index]; } if (pCrawl != null && pCrawl.isLeaf) prefixPositions.Add(level); return prefixPositions; } // Function to check if word // formation is possible or not static bool isPossible(TrieNode root, String word) { // Search for the word in the trie // and get its prefix positions // upto which there is matched List<int> prefixPositions1 = findPrefix(root, word); // Word formation is not possible // if the word did not have // at least one prefix match if (prefixPositions1.Count==0) return false; // Search for rest of substring // for each prefix match foreach (int len1 in prefixPositions1) { String restOfSubstring = word.Substring(len1, word.Length-len1); List<int> prefixPositions2 = findPrefix(root, restOfSubstring); foreach (int len2 in prefixPositions2) { // check if word formation is possible if (len1 + len2 == word.Length) return true; } } return false; } // Driver code public static void Main(String []args) { // Let the given dictionary be following String[] dictionary = {"news", "newspa", "paper", "geek"}; // word to be formed String word = "newspaper"; // root Node of trie root = new TrieNode(); // insert all words of dictionary into trie for (int i = 0; i < dictionary.Length; i++) insert(root, dictionary[i]); if(isPossible(root, word)) Console.WriteLine( "Yes"); else Console.WriteLine("No"); } } // This code is contributed by 29AjayKumar |
Output:
Yes
Exercise :
A generalized version of the problem is to check if a given word can be formed using concatenation of 1 or more dictionary words. Write code for the generalized version.
This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
