# Question

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.


# Solution

class Solution:
def exist(self, board, word):
"""
:type board: List[List[str]]
:type word: str
:rtype: bool
"""
if len(word) == 0:
return True
if len(board) == 0 or len(board) == 0:
return False
x, y = len(board), len(board)

char = word
for i in range(x):
for j in range(y):
if board[i][j] == char:
ret = self.backtracking_one_step(board, x, y, word, 0, i, j, char)
if ret:
return True

return False

def backtracking(self, board, x, y, word, index, i, j):
if index == len(word) - 1:
return True

index += 1
char = word[index]
if i > 0 and board[i - 1][j] == char:
if self.backtracking_one_step(board, x, y, word, index, i - 1, j, char):
return True
if i < x - 1 and board[i + 1][j] == char:
if self.backtracking_one_step(board, x, y, word, index, i + 1, j, char):
return True
if j > 0 and board[i][j - 1] == char:
if self.backtracking_one_step(board, x, y, word, index, i, j - 1, char):
return True
if j < y - 1 and board[i][j + 1] == char:
if self.backtracking_one_step(board, x, y, word, index, i, j + 1, char):
return True

return False

def backtracking_one_step(self, board, x, y, word, index, i, j, char):
board[i][j] = ''
ret = self.backtracking(board, x, y, word, index, i, j)
if ret:
return True
else:
board[i][j] = char
return False