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Spell Checker using Trie
  • Difficulty Level : Medium
  • Last Updated : 04 Jan, 2021

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 ith 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++

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// 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;
}

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Output:

geeks geeksforgeeks

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

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