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.lengthn == board[i].length1 <= m, n <= 12board[i][j] is a lowercase English letter.1 <= words.length <= 3 * 1041 <= words[i].length <= 10words[i] consists of lowercase English letters.- All the strings of
words are unique.
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class Solution:
def findWords(
self,
board: List[List[str]],
words: List[str]
) -> List[str]:
def prune_shitty_trie(root, word):
prune_candidate = (root, word[0])
current = root
i = 0
while i < len(word):
char = word[i]
if current[char]["is_word"] or len(current[char]) > 2:
# can't prune anything above and including current[char]
if i + 1 >= len(word):
return
prune_candidate = (current[char], word[i + 1])
current = current[char]
i += 1
if len(current) == 1:
del prune_candidate[0][prune_candidate[1]]
def dfs(r, c, node, accumulated):
if r < 0 or r >= m or c < 0 or c >= n:
return
if board[r][c] not in node:
return
char = board[r][c]
board[r][c] = "#"
accumulated += char
node = node[char]
if node["is_word"]:
res.append(accumulated)
node["is_word"] = False
prune_shitty_trie(root, accumulated)
# 4 directions
for (dr, dc) in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
dfs(r + dr, c + dc, node, accumulated)
board[r][c] = char
m, n = len(board), len(board[0])
# Build trie
root = {"is_word": False}
for word in words:
current = root
for char in word:
current = current.setdefault(char, {"is_word": False})
current["is_word"] = True
res = []
for r in range(m):
for c in range(n):
dfs(r, c, root, '')
return res
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