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++
// 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; } |
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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