# Word Search II

BacktrackingTrie

## # Solution

Let $n$ be the number of rows and $m$ be the number of columns of the board. Let $w$ be the number of words and $l$ be the average length of word to be matched.

### # Trie

B/c search is done in backtrack, no need to implement search in Trie.

See more about Trie in Interview Algorithms.

Complexity

time: $O(wl + nm \cdot 3^wl)$
space: $O(wl + l) = O(wl)$

from collections import defaultdict
class TrieNode:
def __init__(self):
self.children = defaultdict(TrieNode)
self.is_word = False

class Trie:
def __init__(self):
self.root = TrieNode()

def insert(self, word: str) -> None:
curr = self.root
for char in word:
curr = curr.children[char]
curr.is_word = True

class Solution:
def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
n, m = len(board), len(board[0])
res = []
trie = Trie()
root = trie.root
for word in words:
trie.insert(word)

def backtrack(i, j, curr, path):
if curr.is_word:
res.append(path)
curr.is_word = False
if i<0 or i>=n or j<0 or j>=m:
return
temp = board[i][j]
curr = curr.children.get(temp)
if not curr:
return
board[i][j] = '#'
backtrack(i+1, j, curr, path+temp)
backtrack(i-1, j, curr, path+temp)
backtrack(i, j+1, curr, path+temp)
backtrack(i, j-1, curr, path+temp)
board[i][j] = temp

for i in range(n):
for j in range(m):
backtrack(i, j, root, '')

return res

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47