Search for Word II
Given a 2-D grid of characters board and a list of strings words, return all words that are present in the grid.
For a word to be present it must be possible to form the word with a path in the board with horizontally or vertically neighboring cells. The same cell may not be used more than once in a word.
Example 1:
Input:
board = [
["a","b","c","d"],
["s","a","a","t"],
["a","c","k","e"],
["a","c","d","n"]
],
words = ["bat","cat","back","backend","stack"]
Output: ["cat","back","backend"]Example 2:
Input:
board = [
["x","o"],
["x","o"]
],
words = ["xoxo"]
Output: []Constraints:
1 <= board.length, board[i].length <= 10
board[i] consists only of lowercase English letter.
1 <= words.length <= 100
1 <= words[i].length <= 10words[i] consists only of lowercase English letters.
All strings within words are distinct.
Solution
Compared with basic Search for Word, this problem involves multiple queries. For one query, we have to go through the whole matrix and apply DFS, which consume O ( n × m × Len ( w o r d ) ) O(n\times m\times \text{Len}(word)) O(n×m×Len(word)) time complexity. If we simply repeat this process on every query, the time complexity is O ( n × m × ∑ w ∈ w o r d s Len ( w ) ) O(n\times m\times \sum_{w\in words}\text{Len}(w)) O(n×m×∑w∈wordsLen(w)), which is too time-consuming.
In fact, we can go through the whole matrix and apply DFS at each position only once, because in just one going through, we can meet all possible words that can be produced by the matrix. There is no need to repeat it.
One possible solution is to record all strings while we go through the matrix, using a set or hash algorithm, but the number of all possible strings might be large and we only need few among them. The overall time and space complexity will be O ( n 2 m 2 ) O(n^2m^2) O(n2m2).
Actually, the DFS can be much more efficient if we search according to the given word list. To achieve this, we can build a Trie tree on all the words we want to find in the matrix. Then, we simultaneously move on both the matrix and Trie tree. Under the guidance of Trie tree, we only search the position that can compose a prefix of target words. Although in the worst case, the time complexity will still be O ( n 2 m 2 ) O(n^2m^2) O(n2m2), but in general cases, the time complexity will be about O ( ∑ w ∈ w o r d s Len ( w ) ) O( \sum_{w\in words}\text{Len}(w)) O(∑w∈wordsLen(w)) and the space complexity will be constantly O ( n × m + ∑ w ∈ w o r d s Len ( w ) ) O(n\times m+ \sum_{w\in words}\text{Len}(w)) O(n×m+∑w∈wordsLen(w)).
Code
py
class Trie:
def __init__(self):
self.is_word = [False]
self.tree = [{}]
self.id_cnt = 0
self.word_cnt = [0]
def insert(self, word):
node = 0
self.word_cnt[0] += 1
for c in word:
if c not in self.tree[node]:
self.id_cnt += 1
self.tree[node][c] = self.id_cnt
self.is_word.append(False)
self.tree.append({})
self.word_cnt.append(0)
node = self.tree[node][c]
self.word_cnt[node] += 1
self.is_word[node] = True
def remove(self, word):
node = 0
self.word_cnt[0] -= 1
for c in word:
if c not in self.tree[node]:
return
node = self.tree[node][c]
if self.word_cnt[node] <= 0:
return
self.word_cnt[node] -= 1
self.is_word[node] = False
class Solution:
def findWords(self, board: List[List[str]], words: List[str]) -> List[str]:
trie = Trie()
for word in words:
trie.insert(word)
ans = []
vis = set()
X = [ 1,-1, 0, 0]
Y = [ 0, 0, 1,-1]
def dfs(x, y, node, string):
print(x, y, node, string)
if trie.word_cnt[node] <= 0:
return
if trie.is_word[node]:
ans.append(string)
trie.remove(string)
vis.add((x, y))
for i in range(4):
nxt_x = x + X[i]
nxt_y = y + Y[i]
if 0 <= nxt_x < len(board) and 0 <= nxt_y < len(board[0]) and (nxt_x, nxt_y) not in vis:
nxt_char = board[nxt_x][nxt_y]
if nxt_char in trie.tree[node]:
dfs(nxt_x, nxt_y, trie.tree[node][nxt_char], string+nxt_char)
vis.remove((x,y))
return False
for i in range(len(board)):
for j in range(len(board[0])):
if board[i][j] in trie.tree[0]:
dfs(i, j, trie.tree[0][board[i][j]], board[i][j])
return ans