LeetCode in Kotlin

212. Word Search II

Hard

Given an m x n board of characters and a list of strings words, return all words on the board.

Each word must be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

Example 1:

Input: board = [[“o”,”a”,”a”,”n”],[“e”,”t”,”a”,”e”],[“i”,”h”,”k”,”r”],[“i”,”f”,”l”,”v”]], words = [“oath”,”pea”,”eat”,”rain”]

Output: [“eat”,”oath”]

Example 2:

Input: board = [[“a”,”b”],[“c”,”d”]], words = [“abcb”]

Output: []

Constraints:

• m == board.length
• n == board[i].length
• 1 <= m, n <= 12
• board[i][j] is a lowercase English letter.
• 1 <= words.length <= 3 * 104
• 1 <= words[i].length <= 10
• words[i] consists of lowercase English letters.
• All the strings of words are unique.

Solution

@Suppress("NAME_SHADOWING")
class Solution {
    private var root: Tree? = null

    fun findWords(board: Array<CharArray>, words: Array<String>): List<String> {
        if (board.isEmpty() || board[0].isEmpty()) {
            return emptyList()
        }
        root = Tree()
        for (word in words) {
            Tree.addWord(root, word)
        }
        val collected: MutableList<String> = ArrayList()
        for (i in board.indices) {
            for (j in board[0].indices) {
                dfs(board, i, j, root, collected)
            }
        }
        return collected
    }

    private fun dfs(board: Array<CharArray>, i: Int, j: Int, cur: Tree?, collected: MutableList<String>) {
        var cur: Tree? = cur
        val c = board[i][j]
        if (c == '-') {
            return
        }
        cur = cur!!.getChild(c)
        if (cur == null) {
            return
        }
        if (cur.end != null) {
            val s: String = cur.end!!
            collected.add(s)
            cur.end = null
            if (cur.len() == 0) {
                Tree.deleteWord(root, s)
            }
        }
        board[i][j] = '-'
        if (i > 0) {
            dfs(board, i - 1, j, cur, collected)
        }
        if (i + 1 < board.size) {
            dfs(board, i + 1, j, cur, collected)
        }
        if (j > 0) {
            dfs(board, i, j - 1, cur, collected)
        }
        if (j + 1 < board[0].size) {
            dfs(board, i, j + 1, cur, collected)
        }
        board[i][j] = c
    }
}

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