Given an array of string words, return all strings in words that is a substring of another word. You can return the answer in any order.
A substring is a contiguous sequence of characters within a string
Example 1:
Input: words = ["mass","as","hero","superhero"] Output: ["as","hero"] Explanation: "as" is substring of "mass" and "hero" is substring of "superhero". ["hero","as"] is also a valid answer.
Example 2:
Input: words = ["leetcode","et","code"] Output: ["et","code"] Explanation: "et", "code" are substring of "leetcode".
Example 3:
Input: words = ["blue","green","bu"] Output: [] Explanation: No string of words is substring of another string.
Constraints:
1 <= words.length <= 1001 <= words[i].length <= 30words[i] contains only lowercase English letters.words are unique.
class Solution {
private class TrieNode {
int freq = 0;
TrieNode[] children = new TrieNode[26];
}
private class Trie{
TrieNode root;
Trie(){
root = new TrieNode();
}
public void insertWord(char[] word, int startIdx){
TrieNode curr = root;
for(int i=startIdx; i<word.length; i++){
int c = word[i] - 'a';
if(curr.children[c]==null)
curr.children[c] = new TrieNode();
curr = curr.children[c];
curr.freq += 1;
}
}
public boolean countOccurances(String word){
TrieNode curr = root;
for(char x: word.toCharArray()){
int c = x - 'a';
if(curr.children[c]==null)
return false;
curr = curr.children[c];
}
return curr.freq > 1;
}
}
public List<String> stringMatching(String[] words) {
Trie trie = new Trie();
for(String w: words){
char[] cs = w.toCharArray();
for(int i=0; i<cs.length; i++)
trie.insertWord(cs, i);
}
List<String> ans = new ArrayList<>();
for(String w: words)
if(trie.countOccurances(w))
ans.add(w);
return ans;
}
}