/**
* This problem was asked by Twitter.
Implement an autocomplete system.
That is, given a query string s and a set of all possible query strings, return all strings in the set that have s as a prefix.
For example, given the query string de and the set of strings [dog, deer, deal], return [deer, deal].
Hint: Try preprocessing the dictionary into a more efficient data structure to speed up queries.
* */
class Problem_949 {
/*
* solution: DFS and Trie Tree(prefix tree), add all word into prefix tree for searching;
* Time and Space complexity are: O(2^L), L is the length of key word
* */
private var root: TrieNode? = null
init {
root = TrieNode()
}
fun getSamePrefix(query: String, strings: ArrayList<String>): ArrayList<String> {
val result = ArrayList<String>()
for (word in strings) {
root?.insert(word)
}
var lastNode = root
val sb = StringBuilder()
//find out the last node for start dfs
for (c in query) {
lastNode = lastNode?.children?.get(c)
if (lastNode != null) {
sb.append(c)
}
}
dfs(lastNode, result, sb)
return result
}
private fun dfs(node: TrieNode?, list: ArrayList<String>, cur: StringBuilder) {
if (node?.isWord!!) {
list.add(cur.toString())
}
if (node.children.isEmpty()) {
return
}
for (child in node.children.values) {
dfs(child, list, cur.append(child.char.toString()))
//reduce the length for backtracking
cur.setLength(cur.length - 1)
}
}
/**
* node for TrieTree
* */
class TrieNode constructor(c: Char?) {
var isWord = false
var char: Char? = null
val children = HashMap<Char, TrieNode>()
constructor() : this(null)
init {
char = c
}
fun insert(word: String?) {
if (word == null || word.isEmpty()) {
return
}
val first = word[0]
var child = children.get(first)
if (child == null) {
child = TrieNode(first)
children.put(first, child)
}
//insert entire word
if (word.length > 1) {
child.insert(word.substring(1))
} else {
child.isWord = true
}
}
}
}
