Check if the given string of words can be formed from words present in the dictionary

Given a string array of M words and a dictionary of N words. The task is to check if the given string of words can be formed from words present in the dictionary. 

Examples:

dict[] = { find, a, geeks, all, for, on, geeks, answers, inter } 
Input: str[] = { “find”, “all”, “answers”, “on”, “geeks”, “for”, “geeks” }; 
Output: YES 
all words of str[] are present in the dictionary so the output is YES

Input: str = {“find”, “a”, “geek”} 
Output: NO 
In str[], “find” and “a” were present in the dictionary but “geek” is not present in the dictionary so the output is NO

A naive Approach will be to match all words of the input sentence separately with each of the words in the dictionary and maintain a count of the number of occurrence of all words in the dictionary. So if the number of words in dictionary be n and no of words in the sentence be m this algorithm will take O(M*N) time. 

A better approach will be to use the modified version of the advanced data structure Trie the time complexity can be reduced to O(M * t) where t is the length of longest word in the dictionary which is lesser than n. So here a modification has been done to the trie node such that the isEnd variable is now an integer storing the count of occurrence of the word ending on this node. Also, the search function has been modified to find a word in the trie and once found decrease the count of isEnd of that node so that for multiple occurrences of a word in a sentence each is matched with a separate occurrence of that word in the dictionary.

Below is the illustration of the above approach:  

C++

// C++ program to check if a sentence
// can be formed from a given set of words.
#include <bits/stdc++.h>
using namespace std;
const int ALPHABET_SIZE = 26;

// here isEnd is an integer that will store
// count of words ending at that node
struct trieNode {
    trieNode* t[ALPHABET_SIZE];
    int isEnd;
};

// utility function to create a new node
trieNode* getNode()
{
    trieNode* temp = new (trieNode);

    // Initialize new node with null
    for (int i = 0; i < ALPHABET_SIZE; i++)
        temp->t[i] = NULL;
    temp->isEnd = 0;
    return temp;
}

// Function to insert new words in trie
void insert(trieNode* root, string key)
{
    trieNode* trail;
    trail = root;

    // Iterate for the length of a word
    for (int i = 0; i < key.length(); i++) {

        // If the next key does not contains the character
        if (trail->t[key[i] - 'a'] == NULL) {
            trieNode* temp;
            temp = getNode();
            trail->t[key[i] - 'a'] = temp;
        }
        trail = trail->t[key[i] - 'a'];
    }

    // isEnd is increment so not only the word but its count is also stored
    (trail->isEnd)++;
}
// Search function to find a word of a sentence
bool search_mod(trieNode* root, string word)
{
    trieNode* trail;
    trail = root;

    // Iterate for the complete length of the word
    for (int i = 0; i < word.length(); i++) {

        // If the character is not present then word
        // is also not present
        if (trail->t[word[i] - 'a'] == NULL)
            return false;

        // If present move to next character in Trie
        trail = trail->t[word[i] - 'a'];
    }

    // If word foundthen decrement count of the word
    if ((trail->isEnd) > 0 && trail != NULL) {
        // if the word is found decrement isEnd showing one
        // occurrence of this word is already taken so
        (trail->isEnd)--;
        return true;
    }
    else
        return false;
}
// Function to check if string can be
// formed from the sentence
void checkPossibility(string sentence[], int m, trieNode* root)
{
    int flag = 1;

    // Iterate for all words in the string
    for (int i = 0; i < m; i++) {

        if (search_mod(root, sentence[i]) == false) {

            // if a word is not found in a string then the
            // sentence cannot be made from this dictionary of words
            cout << "NO";

            return;
        }
    }

    // If possible
    cout << "YES";
}

// Function to insert all the words of dictionary in the Trie
void insertToTrie(string dictionary[], int n,
                  trieNode* root)
{

    for (int i = 0; i < n; i++)
        insert(root, dictionary[i]);
}

// Driver Code
int main()
{
    trieNode* root;
    root = getNode();

    // Dictionary of words
    string dictionary[] = { "find", "a", "geeks",
                            "all", "for", "on",
                            "geeks", "answers", "inter" };
    int N = sizeof(dictionary) / sizeof(dictionary[0]);

    // Calling Function to insert words of dictionary to tree
    insertToTrie(dictionary, N, root);

    // String to be checked
    string sentence[] = { "find", "all", "answers", "on",
                          "geeks", "for", "geeks" };

    int M = sizeof(sentence) / sizeof(sentence[0]);

    // Function call to check possibility
    checkPossibility(sentence, M, root);

    return 0;
}

Python3

# Python3 program to check if a sentence
# can be formed from a given set of words.
#include <bits/stdc++.h>

ALPHABET_SIZE = 26;

# here isEnd is an integer that will store
# count of words ending at that node
class trieNode:

    def __init__(self):

        self.t = [None for i in range(ALPHABET_SIZE)]
        self.isEnd = 0

# utility function to create a new node
def getNode():

    temp = trieNode()
    return temp;

# Function to insert new words in trie
def insert(root, key):

    trail = None
    trail = root;

    # Iterate for the length of a word
    for i in range(len(key)):

        # If the next key does not contains the character
        if (trail.t[ord(key[i]) - ord('a')] == None):
            temp = None
            temp = getNode();
            trail.t[ord(key[i]) - ord('a')] = temp;

        trail = trail.t[ord(key[i]) - ord('a')];

    # isEnd is increment so not only
    # the word but its count is also stored
    (trail.isEnd) += 1

# Search function to find a word of a sentence
def search_mod(root, word):

    trail = root;

    # Iterate for the complete length of the word
    for i in range(len(word)):

        # If the character is not present then word
        # is also not present
        if (trail.t[ord(word[i]) - ord('a')] == None):
            return False;

        # If present move to next character in Trie
        trail = trail.t[ord(word[i]) - ord('a')];

    # If word found then decrement count of the word
    if ((trail.isEnd) > 0 and trail != None):

        # if the word is found decrement isEnd showing one
        # occurrence of this word is already taken so
        (trail.isEnd) -= 1
        return True;
    else:
        return False;

# Function to check if string can be
# formed from the sentence
def checkPossibility(sentence, m, root):
    flag = 1;

    # Iterate for all words in the string
    for i in range(m):

        if (search_mod(root, sentence[i]) == False):

            # if a word is not found in a string then the
            # sentence cannot be made from this dictionary of words
            print('NO', end='')

            return;

    # If possible
    print('YES')

# Function to insert all the words of dict in the Trie
def insertToTrie(dictionary, n, root):
    for i in range(n):
        insert(root, dictionary[i]);

# Driver Code
if __name__=='__main__':

    root = getNode();

    # Dictionary of words
    dictionary = [ "find", "a", "geeks",
                            "all", "for", "on",
                            "geeks", "answers", "inter" ]

    N = len(dictionary)

    # Calling Function to insert words of dictionary to tree
    insertToTrie(dictionary, N, root);

    # String to be checked
    sentence = [ "find", "all", "answers", "on",
                          "geeks", "for", "geeks" ]

    M = len(sentence)

    # Function call to check possibility
    checkPossibility(sentence, M, root);

# This code is contributed by pratham76

Java

// Java program to check if a sentence
// can be formed from a given set of words.
import java.util.*;

public class Trie {
    // Trie Node class
    static class TrieNode {
        TrieNode[] t;
        int isEnd;
        // Constructor
        TrieNode() {
            t = new TrieNode[26];
            isEnd = 0;
        }
    }

    // Root of Trie
    static TrieNode root;

    // Constructor
    Trie() {
        root = new TrieNode();
    }

    // Function to insert a word
    public static void insert(String key)
    {
        TrieNode trail = root;
        int len = key.length();
        char[] ch = key.toCharArray();
        for (int i = 0; i < len; i++) {
            int index = ch[i] - 'a';
            if (trail.t[index] == null) {
                trail.t[index] = new TrieNode();
            }
            trail = trail.t[index];
        }
        trail.isEnd++;
    }

    // Function to search a word
    public static boolean search_mod(String key)
    {
        TrieNode trail = root;
        int len = key.length();
        char[] ch = key.toCharArray();
        for (int i = 0; i < len; i++) {
            int index = ch[i] - 'a';
            if (trail.t[index] == null)
                return false;
            trail = trail.t[index];
        }
        if (trail.isEnd > 0 && trail != null) {
            trail.isEnd--;
            return true;
        }
        else
            return false;
    }

    // Function to check if string can be
    // formed from the sentence
    public static void checkPossibility(String[] sentence,
                                       int m, TrieNode root)
    {
        int flag = 1;
        for (int i = 0; i < m; i++) {
            if (search_mod(sentence[i]) == false) {
                System.out.println("NO");
                return;
            }
        }
        System.out.println("YES");
    }

    // Driver Code
    public static void main(String[] args)
    {
        // Dictionary of words
        String dictionary[] = { "find", "a", "geeks",
                "all", "for", "on",
                "geeks", "answers", "inter" };
        int N = dictionary.length;
        Trie trie = new Trie();
        for (int i = 0; i < N; i++)
            insert(dictionary[i]);
        String sentence[] = { "find", "all", "answers", "on",
                "geeks", "for", "geeks" };
        int M = sentence.length;
        checkPossibility(sentence, M, root);
    }
}

C#

// C# program to check if a sentence
// can be formed from a given set of words.
using System;
using System.Collections.Generic;

// Trie Node class
public class TrieNode
{
    public TrieNode[] t;
    public int isEnd;

    // Constructor
    public TrieNode()
    {
        t = new TrieNode[26];
        isEnd = 0;
    }
}

public class Trie
{
    // Root of Trie
    public TrieNode root;

    // Constructor
    public Trie()
    {
        root = new TrieNode();
    }

    // Function to insert a word
    public void Insert(string key)
    {
        TrieNode trail = root;
        int len = key.Length;
        char[] ch = key.ToCharArray();
        for (int i = 0; i < len; i++)
        {
            int index = ch[i] - 'a';
            if (trail.t[index] == null)
            {
                trail.t[index] = new TrieNode();
            }
            trail = trail.t[index];
        }
        trail.isEnd++;
    }

    // Function to search a word
    public bool search_mod(string key)
    {
        TrieNode trail = root;
        int len = key.Length;
        char[] ch = key.ToCharArray();
        for (int i = 0; i < len; i++)
        {
            int index = ch[i] - 'a';
            if (trail.t[index] == null)
            {
                return false;
            }
            trail = trail.t[index];
        }
        if (trail.isEnd > 0 && trail != null)
        {
            trail.isEnd--;
            return true;
        }
        else
        {
            return false;
        }
    }

    // Function to check if string can be
    // formed from the sentence
    public void CheckPossibility(string[] sentence, int m, TrieNode root)
    {
        int flag = 1;
        for (int i = 0; i < m; i++)
        {
            if (!search_mod(sentence[i]))
            {
                Console.WriteLine("NO");
                return;
            }
        }
        Console.WriteLine("YES");
    }
}

public class Program
{
    // Driver Code
    public static void Main(string[] args)
    {
        // Dictionary of words
        string[] dictionary = { "find", "a", "geeks", "all", "for", "on", "geeks", "answers", "inter" };
        int N = dictionary.Length;
        Trie trie = new Trie();
        for (int i = 0; i < N; i++)
        {
            trie.Insert(dictionary[i]);
        }
        string[] sentence = { "find", "all", "answers", "on", "geeks", "for", "geeks" };
        int M = sentence.Length;
        trie.CheckPossibility(sentence, M, trie.root);
    }
}

// This code is contributed by Utkarsh Kumar.

Javascript

// Javascript code for the above approach

// Define the alphabet size
const ALPHABET_SIZE = 26;

// Define a trie node structure to store words
class TrieNode {
  constructor() {
  this.t = new Array(ALPHABET_SIZE).fill(null);
  this.isEnd = 0;
  }
}

// Function to insert new words in trie
function insert(root, key) {
    let trail = root;

    // Iterate for the length of a word
    for (let i = 0; i < key.length; i++) {


    // If the next key does not contain the character
    if (trail.t[key.charCodeAt(i) - 'a'.charCodeAt(0)] === null) {
      let temp = new TrieNode();
      trail.t[key.charCodeAt(i) - 'a'.charCodeAt(0)] = temp;
    }
      trail = trail.t[key.charCodeAt(i) - 'a'.charCodeAt(0)];
    }

    // Increment isEnd so not only the word but its count is also stored
    trail.isEnd++;
}

// Search function to find a word of a sentence
function searchMod(root, word) {
    let trail = root;

    // Iterate for the complete length of the word
    for (let i = 0; i < word.length; i++) {

    // If the character is not present then word
    // is also not present
    if (trail.t[word.charCodeAt(i) - 'a'.charCodeAt(0)] === null)
      return false;

    // If present move to next character in Trie
    trail = trail.t[word.charCodeAt(i) - 'a'.charCodeAt(0)];
    }

    // If word found then decrement count of the word
    if (trail.isEnd > 0 && trail != null) {
    // if the word is found decrement isEnd showing one
    // occurrence of this word is already taken so
    trail.isEnd--;
    return true;
    } else
    return false;
}

    // Function to check if string can be
    // formed from the sentence
    function checkPossibility(sentence, m, root) {
    let flag = 1;

    // Iterate for all words in the string
    for (let i = 0; i < m; i++) {


    if (searchMod(root, sentence[i]) === false) {

      // if a word is not found in a string then the
      // sentence cannot be made from this dictionary of words
      console.log("NO");

  return;
}


}

// If possible
console.log("YES");
}

// Function to insert all the words of dictionary in the Trie
function insertToTrie(dictionary, n, root) {
    for (let i = 0; i < n; i++)
    insert(root, dictionary[i]);
    }

    // Driver Code
    function main() {
    let root = new TrieNode();

    // Dictionary of words
    let dictionary = [
    "find",
    "a",
    "geeks",
    "all",
    "for",
    "on",
    "geeks",
    "answers",
    "inter",
    ];
    let N = dictionary.length;

    // Calling Function to insert words of dictionary to tree
    insertToTrie(dictionary, N, root);

    // String to be checked
    let sentence = [
    "find",
    "all",
    "answers",
    "on",
    "geeks",
    "for",
    "geeks",
    ];

    let M = sentence.length;

    // Function call to check possibility
    checkPossibility(sentence, M, root);
}

main();

// this code is contributed by bhardwajji

Output:

YES

Time Complexity: O(M*N)
M is the length of the sentence and N is the number of words in the dictionary.

Space Complexity: O(N)
N is the number of words in the dictionary.

An efficient approach will be to use map. Keep the count of words in the map, iterate in the string and check if the word is present in the map. If present, then decrease the count of the word in the map. If it is not present, then it is not possible to make the given string from the given dictionary of words. 

Below is the implementation of above approach :  

C++

// C++ program to check if a sentence
// can be formed from a given set of words.
#include <bits/stdc++.h>
using namespace std;

// Function to check if the word
// is in the  dictionary or not
bool match_words(string dictionary[], string sentence[],
                 int n, int m)
{
    // map to store all words in
    // dictionary with their count
    unordered_map<string, int> mp;

    // adding all words in map
    for (int i = 0; i < n; i++) {
        mp[dictionary[i]]++;
    }

    // search in map for all
    // words in the sentence
    for (int i = 0; i < m; i++) {
        if (mp[sentence[i]])
            mp[sentence[i]] -= 1;
        else
            return false;
    }

    // all words of sentence are present
    return true;
}

// Driver Code
int main()
{
    string dictionary[] = { "find", "a", "geeks",
                            "all", "for", "on",
                            "geeks", "answers", "inter" };

    int n = sizeof(dictionary) / sizeof(dictionary[0]);

    string sentence[] = { "find", "all", "answers", "on",
                          "geeks", "for", "geeks" };

    int m = sizeof(sentence) / sizeof(sentence[0]);

    // Calling function to check if words are
    // present in the dictionary or not
    if (match_words(dictionary, sentence, n, m))
        cout << "YES";
    else
        cout << "NO";

    return 0;
}

Java

// Java program to check if a sentence
// can be formed from a given set of words.
import java.util.*;

class GFG
{

// Function to check if the word
// is in the dictionary or not
static boolean match_words(String dictionary[], String sentence[],
                                                    int n, int m)
{
    // map to store all words in
    // dictionary with their count
    Map<String,Integer> mp = new HashMap<>();

    // adding all words in map
    for (int i = 0; i < n; i++)
    {
        if(mp.containsKey(dictionary[i]))
        {
            mp.put(dictionary[i], mp.get(dictionary[i])+1);
        }
        else
        {
            mp.put(dictionary[i], 1);
        }
    }

    // search in map for all
    // words in the sentence
    for (int i = 0; i < m; i++)
    {
        if (mp.containsKey(sentence[i]))
            mp.put(sentence[i],mp.get(sentence[i])-1);
        else
            return false;
    }

    // all words of sentence are present
    return true;
}

// Driver Code
public static void main(String[] args)
{
    String dictionary[] = { "find", "a", "geeks",
                            "all", "for", "on",
                            "geeks", "answers", "inter" };

    int n = dictionary.length;

    String sentence[] = { "find", "all", "answers", "on",
                        "geeks", "for", "geeks" };

    int m = sentence.length;

    // Calling function to check if words are
    // present in the dictionary or not
    if (match_words(dictionary, sentence, n, m))
        System.out.println("YES");
    else
        System.out.println("NO");
    }
}

// This code is contributed by Princi Singh

Python3

# Python3 program to check if a sentence
# can be formed from a given set of words.

# Function to check if the word
# is in the dictionary or not
def match_words(dictionary, sentence, n, m):

    # map to store all words in
    # dictionary with their count
    mp = dict()

    # adding all words in map
    for i in range(n):
        mp[dictionary[i]] = mp.get(dictionary[i], 0) + 1

    # search in map for all
    # words in the sentence
    for i in range(m):
        if (mp[sentence[i]]):
            mp[sentence[i]] -= 1
        else:
            return False

    # all words of sentence are present
    return True

# Driver Code
dictionary = ["find", "a", "geeks", "all", "for",
              "on", "geeks", "answers", "inter"]

n = len(dictionary)

sentence = ["find", "all", "answers", "on",
            "geeks", "for", "geeks"]

m = len(sentence)

# Calling function to check if words are
# present in the dictionary or not
if (match_words(dictionary, sentence, n, m)):
    print("YES")
else:
    print("NO")

# This code is contributed by mohit kumar

C#

// C# program to check if a sentence
// can be formed from a given set of words.
using System;
using System.Collections.Generic;

class GFG
{

// Function to check if the word
// is in the dictionary or not
static Boolean match_words(String []dictionary,
                           String []sentence,
                           int n, int m)
{
    // map to store all words in
    // dictionary with their count
    Dictionary<String,
               int> mp = new Dictionary<String,
                                        int>();

    // adding all words in map
    for (int i = 0; i < n; i++)
    {
        if(mp.ContainsKey(dictionary[i]))
        {
            mp[dictionary[i]] = mp[dictionary[i]] + 1;
        }
        else
        {
            mp.Add(dictionary[i], 1);
        }
    }

    // search in map for all
    // words in the sentence
    for (int i = 0; i < m; i++)
    {
        if (mp.ContainsKey(sentence[i]) && mp[sentence[i]] > 0)
            mp[sentence[i]] = mp[sentence[i]] - 1;
        else
            return false;
    }

    // all words of sentence are present
    return true;
}

// Driver Code
public static void Main(String[] args)
{
    String []dictionary = { "find", "a", "geeks",
                            "all", "for", "on",
                            "geeks", "answers", "inter" };

    int n = dictionary.Length;

    String []sentence = { "find", "all", "answers", "on",
                          "geeks", "for", "geeks", "geeks" };

    int m = sentence.Length;

    // Calling function to check if words are
    // present in the dictionary or not
    if (match_words(dictionary, sentence, n, m))
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// Javascript program to check if a sentence
// can be formed from a given set of words.

// Function to check if the word
// is in the  dictionary or not
function match_words(dictionary, sentence, n, m)
{

    // map to store all words in
    // dictionary with their count
    let mp = new Map();

    // Adding all words in map
    for(let i = 0; i < n; i++)
    {
        if(mp.has(dictionary[i]))
        {
            mp.set(dictionary[i],
            mp.get(dictionary[i]) + 1);
        }
        else
        {
            mp.set(dictionary[i], 1);
        }
    }

    // Search in map for all
    // words in the sentence
    for(let i = 0; i < m; i++)
    {
        if (mp.has(sentence[i]))
            mp.set(sentence[i],
            mp.get(sentence[i]) - 1);
        else
            return false;
    }

    // All words of sentence are present
    return true;
}

// Driver code
let dictionary = [ "find", "a", "geeks",
                   "all", "for", "on",
                   "geeks", "answers", "inter" ];

let n = dictionary.length;

let sentence = [ "find", "all", "answers", "on",
                 "geeks", "for", "geeks" ];

let m = sentence.length;

// Calling function to check if words are
// present in the dictionary or not
if (match_words(dictionary, sentence, n, m))
    document.write("YES");
else
    document.write("NO");

// This code is contributed by patel2127

</script>

Output:

YES

Time Complexity: O(M)
Space Complexity: O(N) where N is no of words in a dictionary