LeetCode in Kotlin

211. Design Add and Search Words Data Structure

Medium

Design a data structure that supports adding new words and finding if a string matches any previously added string.

Implement the WordDictionary class:

• WordDictionary() Initializes the object.
• void addWord(word) Adds word to the data structure, it can be matched later.
• bool search(word) Returns true if there is any string in the data structure that matches word or false otherwise. word may contain dots '.' where dots can be matched with any letter.

Example:

Input

["WordDictionary","addWord","addWord","addWord","search","search","search","search"] [[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]

Output

[null,null,null,null,false,true,true,true]

Explanation

WordDictionary wordDictionary = new WordDictionary();
wordDictionary.addWord("bad");
wordDictionary.addWord("dad");
wordDictionary.addWord("mad");
wordDictionary.search("pad"); // return False
wordDictionary.search("bad"); // return True
wordDictionary.search(".ad"); // return True
wordDictionary.search("b.."); // return True 

Constraints:

• 1 <= word.length <= 500
• word in addWord consists lower-case English letters.
• word in search consist of '.' or lower-case English letters.
• At most 50000 calls will be made to addWord and search.

Solution

class WordDictionary {
    class TrieNode() {
        val children = Array<TrieNode?>(26) { null }
        var isWord = false
    }

    val trieTree = TrieNode()

    fun addWord(word: String) {
        var p = trieTree
        for (w in word) {
            val i = w - 'a'
            if (p.children[i] == null) p.children[i] = TrieNode()
            p = p.children[i]!!
        }
        p.isWord = true
    }

    fun search(word: String): Boolean {
        fun dfs(p: TrieNode?, start: Int): Boolean {
            if (p == null) return false
            if (start == word.length) return p.isWord
            return if (word[start] == '.') {
                for (i in 0..25) {
                    if (dfs(p.children[i], start + 1)) {
                        return true
                    }
                }
                false
            } else {
                val i = word[start] - 'a'
                dfs(p.children[i], start + 1)
            }
        }
        return dfs(trieTree, 0)
    }
}

/*
 * Your WordDictionary object will be instantiated and called as such:
 * var obj = WordDictionary()
 * obj.addWord(word)
 * var param_2 = obj.search(word)
 */

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