This page explains Java solution to problem Concatenated Words
using Dynamic Programming
algorithm.
Given a list of words (without duplicates), please write a program that returns all concatenated words in the given list of words.
A concatenated word is defined as a string that is comprised entirely of at least two shorter words in the given array.
Example 1:Input: ["cat","cats","catsdogcats","dog","dogcatsdog","hippopotamuses","rat","ratcatdogcat"]
Output: ["catsdogcats","dogcatsdog","ratcatdogcat"]
Explanation:
"catsdogcats" can be concatenated by "cats", "dog" and "cats";
"dogcatsdog" can be concatenated by "dog", "cats" and "dog";
"ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat"
package com.vc.hard;
import java.util.*;
class ConcatenatedWords {
public List<String> findAllConcatenatedWordsInADict(String[] words) {
Arrays.sort(words, new Comparator<String>(){
public int compare(String s1, String s2) {
return Integer.compare(s1.length(), s2.length());
}
});
HashSet<String> set = new HashSet<>();
List<String> res = new ArrayList<>();
for(String word: words) {
if(canForm(word, set)) res.add(word);
set.add(word);
}
return res;
}
private boolean canForm(String word, HashSet<String> set) {
if(set.isEmpty()) return false;
int n = word.length();
boolean[] dp = new boolean[n + 1];
dp[0] = true;
for(int i = 1; i <= word.length(); i++) {
for(int j = 0; j < i; j++) {
if(dp[j] && set.contains(word.substring(j, i))) {
dp[i] = true;
break;
}
}
}
return dp[n];
}
}
O(N * K2) Where
N is total number of elements in an input string array words
K is length of longest word in an input array
O(N) Where
N is total number of elements in an input string array words