Problem
Given a 2D board and a word, find if the word exists in the grid.
The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.
Example:
board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]
Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.
Notes
Backtrack problem.
- when found, mark the point to one.
- Use dfs to go down from this point.
- Can’t go down anymore, mark the point back to zero (backtrack step).
Solution
Solution 1: Backtrack, dfs
class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
for i in range(len(board)):
for j in range(len(board[0])):
if self.dfs(board, word, 0, i, j):
return True
return False
def dfs(self, board, word, w, i, j):
if w == len(word):
return True
if i < 0 or j < 0 or i >= len(board) or j >= len(board[0]):
return False
if board[i][j] == word[w]:
board[i][j] = '-'
for next_i, next_j in [[i + 1, j], [i - 1, j], [i, j - 1],
[i, j + 1]]:
if self.dfs(board, word, w + 1, next_i, next_j):
return True
board[i][j] = word[w]
return False