0212: Word Search II

Problem Statement

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.

Code Solution

class TrieNode:
    def __init__(self):
        self.children = {}
        self.end = False
     
    def addWord(self, word):
        temp = self
        for cx, c in enumerate(word):
            if c not in temp.children:
                temp.children[c] = TrieNode()
            temp = temp.children[c]
        temp.end = True

class Solution:
    def findWords(self, board, words):
	    
        root = TrieNode()
        for word in words:
            root.addWord(word)
        
        nR = len(board)
        nC = len(board[0])
        
        res = set()
        visit = set()
        def dfs(rx, cx, node, word):
            if 0 <= rx < nR and 0 <= cx < nC and (rx, cx) not in visit and board[rx][cx] in node.children:
                
                c = board[rx][cx]
                
                word.append(c)
                visit.add((rx, cx))
                
                node = node.children[c]
                
                if node.end: res.add("".join(word))
                
                dfs(rx + 1, cx, node, word)
                dfs(rx - 1, cx, node, word)
                dfs(rx, cx + 1, node, word)
                dfs(rx, cx - 1, node, word)
                
                visit.remove((rx, cx))
                word.pop()
        
        for rx in range(nR):
            for cx in range(nC):
                dfs(rx, cx, root, [])
                
        return list(res)