Given a dictionary of words and an input string, find the longest prefix of the string which is also a word in dictionary.
Examples:
Let the dictionary contains the following words: {are, area, base, cat, cater, children, basement} Below are some input/output examples: -------------------------------------- Input String Output -------------------------------------- caterer cater basemexy base child < Empty >
Solution
We build a Trie of all dictionary words. Once the Trie is built, traverse through it using characters of input string. If prefix matches a dictionary word, store current length and look for a longer match. Finally, return the longest match.
Following is Java implementation of the above solution based.
import java.util.HashMap;
// Trie Node, which stores a character and the children in a HashMap class TrieNode {
public TrieNode( char ch) {
value = ch;
children = new HashMap<>();
bIsEnd = false ;
}
public HashMap<Character,TrieNode> getChildren() { return children; }
public char getValue() { return value; }
public void setIsEnd( boolean val) { bIsEnd = val; }
public boolean isEnd() { return bIsEnd; }
private char value;
private HashMap<Character,TrieNode> children;
private boolean bIsEnd;
} // Implements the actual Trie class Trie {
// Constructor
public Trie() { root = new TrieNode(( char ) 0 ); }
// Method to insert a new word to Trie
public void insert(String word) {
// Find length of the given word
int length = word.length();
TrieNode crawl = root;
// Traverse through all characters of given word
for ( int level = 0 ; level < length; level++)
{
HashMap<Character,TrieNode> child = crawl.getChildren();
char ch = word.charAt(level);
// If there is already a child for current character of given word
if ( child.containsKey(ch))
crawl = child.get(ch);
else // Else create a child
{
TrieNode temp = new TrieNode(ch);
child.put( ch, temp );
crawl = temp;
}
}
// Set bIsEnd true for last character
crawl.setIsEnd( true );
}
// The main method that finds out the longest string 'input'
public String getMatchingPrefix(String input) {
String result = "" ; // Initialize resultant string
int length = input.length(); // Find length of the input string
// Initialize reference to traverse through Trie
TrieNode crawl = root;
// Iterate through all characters of input string 'str' and traverse
// down the Trie
int level, prevMatch = 0 ;
for ( level = 0 ; level < length; level++ )
{
// Find current character of str
char ch = input.charAt(level);
// HashMap of current Trie node to traverse down
HashMap<Character,TrieNode> child = crawl.getChildren();
// See if there is a Trie edge for the current character
if ( child.containsKey(ch) )
{
result += ch; //Update result
crawl = child.get(ch); //Update crawl to move down in Trie
// If this is end of a word, then update prevMatch
if ( crawl.isEnd() )
prevMatch = level + 1 ;
}
else break ;
}
// If the last processed character did not match end of a word,
// return the previously matching prefix
if ( !crawl.isEnd() )
return result.substring( 0 , prevMatch);
else return result;
}
private TrieNode root;
} // Testing class public class Test {
public static void main(String[] args) {
Trie dict = new Trie();
dict.insert( "are" );
dict.insert( "area" );
dict.insert( "base" );
dict.insert( "cat" );
dict.insert( "cater" );
dict.insert( "basement" );
String input = "caterer" ;
System.out.print(input + ": " );
System.out.println(dict.getMatchingPrefix(input));
input = "basement" ;
System.out.print(input + ": " );
System.out.println(dict.getMatchingPrefix(input));
input = "are" ;
System.out.print(input + ": " );
System.out.println(dict.getMatchingPrefix(input));
input = "arex" ;
System.out.print(input + ": " );
System.out.println(dict.getMatchingPrefix(input));
input = "basemexz" ;
System.out.print(input + ": " );
System.out.println(dict.getMatchingPrefix(input));
input = "xyz" ;
System.out.print(input + ": " );
System.out.println(dict.getMatchingPrefix(input));
}
} |
Output:
caterer: cater basement: basement are: are arex: are basemexz: base xyz:
Time Complexity: Time complexity of finding the longest prefix is O(n) where n is length of the input string. Refer this for time complexity of building the Trie.
This article is compiled by Ravi Chandra Enaganti.