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.


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 C++ implementation of above idea.

// C++ program to check if a string can be
// formed by concatenating two words
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 Node
struct 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 node
void 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 not
bool 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 function
int 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;



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.

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