Given an m x n grid of characters board and a string
word, return true if word exists in the grid.
The word can 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.
Example 1:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: trueExample 2:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: trueExample 3:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: falseThis requires no copies, but you can make a copy if needed. Otherwise, make a copy of the board and mutate that.
class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
m = len(board)
n = len(board[0])
word_len = len(word)
def inbounds(y, x):
return 0 <= y < m and 0 <= x < n
def dfs(y, x, index):
if index == word_len:
return True
if not inbounds(y, x) or board[y][x] != word[index]:
return False
tmp = board[y][x]
board[y][x] = "#"
res = any([
dfs(y, x + 1, index + 1),
dfs(y, x - 1, index + 1),
dfs(y + 1, x, index + 1),
dfs(y - 1, x, index + 1)
])
board[y][x] = tmp
return res
for y in range(m):
for x in range(n):
if dfs(y, x, 0):
return True
return False