Spell Checker using Trie

Given an array of strings str[] and a string key, the task is to check if the spelling of the key is correct or not. If found to be true, then print “YES”. Otherwise, print the suggested correct spellings.

Examples:

Input:str[] = { “gee”, “geeks”, “ape”, “apple”, “geeksforgeeks” }, key = “geek” 
Output: geeks geeksforgeeks 
Explanation:
The string “geek” not present in the array of strings. 
Therefore, the suggested words are { “geeks”, “geeksforgeeks” }.

Input: str[] = { “gee”, “geeks”, “ape”, “apple”, “arp” }, key = “geeks” 
Output: YES.

Approach:The problem can be solved using Trie. The idea is to traverse the array of string, str[] and insert the string into the Trie such that each node of the Trie contains the character of the string and a boolean value to check if the character is the last character of the string or not. Follow the steps below to solve the problem:

• Initialize a Trie, say root, such that each node of the Trie consists of a character of a string and a boolean value to check if the character is the last character of the string or not.
• Traverse the array of strings arr[], and insert all the strings into the Trie.
• Finally, traverse the string key. For every i^th character, check if the character is present in the Trie or not. If found to be true, then move to the next node of the Trie.
• Otherwise, print all possible strings whose prefix is the string key.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Structure of a Trie node
struct TrieNode {

    // Store address of a character
    TrieNode* Trie[256];

    // Check if the character is
    // last character of a string or not
    bool isEnd;

    // Constructor function
    TrieNode()
    {

        for (int i = 0; i < 256; i++) {

            Trie[i] = NULL;
        }
        isEnd = false;
    }
};

// Function to insert a string into Trie
void InsertTrie(TrieNode* root, string s)
{

    TrieNode* temp = root;

    // Traverse the string, s
    for (int i = 0; i < s.length(); i++) {

        if (temp->Trie[s[i]] == NULL) {

            // Initialize a node
            temp->Trie[s[i]] = new TrieNode();
        }

        // Update temp
        temp = temp->Trie[s[i]];
    }

    // Mark the last character of
    // the string to true
    temp->isEnd = true;
}

// Function to print suggestions of the string
void printSuggestions(TrieNode* root, string res)
{

    // If current character is
    // the last character of a string
    if (root->isEnd == true) {

        cout << res << " ";
    }

    // Iterate over all possible
    // characters of the string
    for (int i = 0; i < 256; i++) {

        // If current character
        // present in the Trie
        if (root->Trie[i] != NULL) {

            // Insert current character
            // into Trie
            res.push_back(i);
            printSuggestions(root->Trie[i], res);
            res.pop_back();
        }
    }
}

// Function to check if the string
// is present in Trie or not
bool checkPresent(TrieNode* root, string key)
{

    // Traverse the string
    for (int i = 0; i < key.length(); i++) {

        // If current character not
        // present in the Trie
        if (root->Trie[key[i]] == NULL) {

            printSuggestions(root, key.substr(0, i));

            return false;
        }

        // Update root
        root = root->Trie[key[i]];
    }
    if (root->isEnd == true) {

        return true;
    }
    printSuggestions(root, key);

    return false;
}

// Driver Code
int main()
{

    // Given array of strings
    vector<string> str = { "gee", "geeks", "ape",
                           "apple", "geeksforgeeks" };

    string key = "geek";

    // Initialize a Trie
    TrieNode* root = new TrieNode();

    // Insert strings to trie
    for (int i = 0; i < str.size(); i++) {
        InsertTrie(root, str[i]);
    }

    if (checkPresent(root, key)) {

        cout << "YES";
    }
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;

// Structure of a Trie node
class TrieNode
{

    // Store address of a character
    TrieNode Trie[];

    // Check if the character is
    // last character of a string or not
    boolean isEnd;

    // Constructor function
    public TrieNode()
    {
        Trie = new TrieNode[256];
        for(int i = 0; i < 256; i++)
        {
            Trie[i] = null;
        }
        isEnd = false;
    }
}

class GFG{

// Function to insert a string into Trie
static void InsertTrie(TrieNode root, String s)
{
    TrieNode temp = root;

    // Traverse the string, s
    for(int i = 0; i < s.length(); i++)
    {
        if (temp.Trie[s.charAt(i)] == null)
        {

            // Initialize a node
            temp.Trie[s.charAt(i)] = new TrieNode();
        }

        // Update temp
        temp = temp.Trie[s.charAt(i)];
    }

    // Mark the last character of
    // the string to true
    temp.isEnd = true;
}

// Function to print suggestions of the string
static void printSuggestions(TrieNode root, String res)
{

    // If current character is
    // the last character of a string
    if (root.isEnd == true)
    {
        System.out.print(res + " ");
    }

    // Iterate over all possible
    // characters of the string
    for(int i = 0; i < 256; i++)
    {

        // If current character
        // present in the Trie
        if (root.Trie[i] != null)
        {

            // Insert current character
            // into Trie
            res += (char)i;
            printSuggestions(root.Trie[i], res);
            res = res.substring(0, res.length() - 2);
        }
    }
}

// Function to check if the string
// is present in Trie or not
static boolean checkPresent(TrieNode root, String key)
{

    // Traverse the string
    for(int i = 0; i < key.length(); i++)
    {

        // If current character not
        // present in the Trie
        if (root.Trie[key.charAt(i)] == null)
        {
            printSuggestions(root, key.substring(0, i));
            return false;
        }

        // Update root
        root = root.Trie[key.charAt(i)];
    }
    if (root.isEnd == true)
    {
        return true;
    }
    printSuggestions(root, key);

    return false;
}

// Driver Code
public static void main(String[] args)
{

    // Given array of strings
    String str[] = { "gee", "geeks", "ape", "apple",
                     "geeksforgeeks" };

    String key = "geek";

    // Initialize a Trie
    TrieNode root = new TrieNode();

    // Insert strings to trie
    for(int i = 0; i < str.length; i++)
    {
        InsertTrie(root, str[i]);
    }

    if (checkPresent(root, key))
    {
        System.out.println("YES");
    }
}
}

// This code is contributed by Dharanendra L V.

Python3

# Python program to implement
# the above approach

# Structure of a Trie node
class TrieNode:
    def __init__(self):

        # Store address of a character
        self.trie = [None] * 256

        # Check if the character is
        # last character of a string or not
        self.isEnd = False

# Function to insert a string into Trie
def insert_trie(root, s):
    temp = root
    # Traverse the string, s
    for i in range(len(s)):
        if not temp.trie[ord(s[i])]:
            # Initialize a node
            temp.trie[ord(s[i])] = TrieNode()
        # Update temp
        temp = temp.trie[ord(s[i])]

    # Mark the last character of
    # the string to true
    temp.isEnd = True

# Function to print suggestions of the string
def print_suggestions(root, res):
    # If current character is
    # the last character of a string
    if root.isEnd:
        print(res,end=" ")

    # Iterate over all possible
    # characters of the string
    for i in range(256):
        # If current character
        # present in the Trie
        if root.trie[i]:
            # Insert current character
            # into Trie
            res_list = list(res)
            res_list.append(chr(i))
            print_suggestions(root.trie[i], "".join(res_list))

# Function to check if the string
# is present in Trie or not
def check_present(root, key):
    # Traverse the string
    for i in range(len(key)):
        # If current character not
        # present in the Trie
        if not root.trie[ord(key[i])]:
            print_suggestions(root, key[:i])
            return False
        # Update root
        root = root.trie[ord(key[i])]
    if root.isEnd:
        return True
    print_suggestions(root, key)
    return False

#  Driver Code

# Given array of strings
strs = ["gee", "geeks", "ape", "apple", "geeksforgeeks"]
key = "geek"

# Initialize a Trie
root = TrieNode()

# Insert strings to trie
for s in strs:
    insert_trie(root, s)

if check_present(root, key):
    print("YES")


#  This code is contributed by Aman Kumar

C#

// C# program to implement
// the above approach
using System;

// Structure of a Trie node
public class TrieNode
{
    // Store address of a character
    public TrieNode[] Trie;

    // Check if the character is
    // last character of a string or not
    public bool isEnd;

    // Constructor function
    public TrieNode()
    {
        Trie = new TrieNode[256];
        for (int i = 0; i < 256; i++)
        {
            Trie[i] = null;
        }
        isEnd = false;
    }
}

public class GFG
{
    // Function to insert a string into Trie
    static void InsertTrie(TrieNode root, string s)
    {
        TrieNode temp = root;

        // Traverse the string, s
        for (int i = 0; i < s.Length; i++)
        {
            if (temp.Trie[s[i]] == null)
            {
                // Initialize a node
                temp.Trie[s[i]] = new TrieNode();
            }
            // Update temp
            temp = temp.Trie[s[i]];
        }
        // Mark the last character of
        // the string to true
        temp.isEnd = true;
    }

    // Function to print suggestions of the string
    static void printSuggestions(TrieNode root, string res)
    {
        // If current character is
        // the last character of a string
        if (root.isEnd == true)
        {
            Console.Write(res + " ");
        }

        // Iterate over all possible
        // characters of the string
        for (int i = 0; i < 256; i++)
        {
            // If current character
            // present in the Trie
            if (root.Trie[i] != null)
            {
                // Insert current character
                // into Trie
                res += (char)i;
                printSuggestions(root.Trie[i], res);
                res = res.Substring(0, res.Length - 1);
            }
        }
    }

    // Function to check if the string
    // is present in Trie or not
    static bool checkPresent(TrieNode root, string key)
    {
        // Traverse the string
        for (int i = 0; i < key.Length; i++)
        {
            // If current character not
            // present in the Trie
            if (root.Trie[key[i]] == null)
            {
                printSuggestions(root, key.Substring(0, i));
                return false;
            }
            // Update root
            root = root.Trie[key[i]];
        }

        if (root.isEnd == true)
        {
            return true;
        }
        printSuggestions(root, key);
        return false;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        // Given array of strings
        string[] str = { "gee", "geeks", "ape", "apple", "geeksforgeeks" };
        string key = "geek";

        // Initialize a Trie
        TrieNode root = new TrieNode();

        // Insert strings to trie
        for (int i = 0; i < str.Length; i++)
        {
            InsertTrie(root, str[i]);
        }
        if (checkPresent(root, key))
        {
            Console.WriteLine("YES");
        }
    }
}

// This code is contributed by Utkarsh Kumar.

Javascript

// javascript program to implement
// the above approach

// Structure of a Trie node
class TrieNode {

    constructor(){

        // Store address of a character
        this.Trie = new Array(256).fill(null);

        // Check if the character is
        // last character of a string or not
        this.isEnd = false;

    }
};

// Function to insert a string into Trie
function InsertTrie(root, s)
{

    let temp = root;

    // Traverse the string, s
    for (let i = 0; i < s.length; i++) {

        if (temp.Trie[s[i].charCodeAt(0)] == null) {

            // Initialize a node
            temp.Trie[s[i].charCodeAt(0)] = new TrieNode();
        }

        // Update temp
        temp = temp.Trie[s[i].charCodeAt(0)];
    }

    // Mark the last character of
    // the string to true
    temp.isEnd = true;

}

// Function to print suggestions of the string
function printSuggestions(root, res)
{

    // If current character is
    // the last character of a string
    if (root.isEnd == true) {
        document.write(res + " ");
    }

    // Iterate over all possible
    // characters of the string
    for (let i = 0; i < 256; i++) {

        // If current character
        // present in the Trie
        if (root.Trie[i] != null) {

            // Insert current character
            // into Trie
            let res_list = res.split('');
            res_list.push(String.fromCharCode(i))

            printSuggestions(root.Trie[i], res_list.join(""));
        }
    }
}

// Function to check if the string
// is present in Trie or not
function checkPresent(root, key)
{

    // Traverse the string
    for (let i = 0; i < key.length; i++) {

        // If current character not
        // present in the Trie
        if (root.Trie[key[i].charCodeAt(0)] == null) {

            printSuggestions(root, key.substr(0, i));

            return false;
        }

        // Update root
        root = root.Trie[key[i].charCodeAt(0)];
    }
    if (root.isEnd == true) {

        return true;
    }

    printSuggestions(root, key);

    return false;
}

// Driver Code

// Given array of strings
let str = ["gee", "geeks", "ape", "apple", "geeksforgeeks"];

let key = "geek";

// Initialize a Trie
let root = new TrieNode();

// Insert strings to trie
for (let i = 0; i < str.length; i++) {
    InsertTrie(root, str[i]);
}

if (checkPresent(root, key)) {
    console.log("YES");
}

// The code is contributed by Nidhi goel.

Output:

geeks geeksforgeeks

Time Complexity: O(N * M), where M is the maximum length of the string
Auxiliary Space: O(N * 256)