## DEV Community

Abhishek Chaudhary

Posted on

# Word Search

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: true

Example 2:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true

Example 3:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false

Constraints:

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board and word consists of only lowercase and uppercase English letters.

Follow up: Could you use search pruning to make your solution faster with a larger board?

SOLUTION:

# class Solution:
#     def isAdjacent(a, b, c, d):
#         if a == c and abs(b - d) == 1:
#             return True
#         if b == d and abs(a - c) == 1:
#             return True
#         return False

#     def exist(self, board: List[List[str]], word: str) -> bool:
#         m = len(board)
#         n = len(board[0])
#         wlen = len(word)
#         pos = [[] for c in word]
#         for i in range(m):
#             for j in range(n):
#                 for k in range(wlen):
#                     if word[k] == board[i][j]:
#                         pos[k].append((i, j))
#         paths = [[i] for i in pos[0]]
#         while len(paths) > 0:
#             curr = paths.pop(0)
#             currlen = len(curr)
#             lastcell = curr[-1]
#             if currlen == wlen:
#                 return True
#             if currlen < wlen:
#                 for x, y in pos[currlen + 1]:
#                     if (x, y) not in curr and self.isAdjacent(lastcell, x, y):
#                         paths.append(curr + [(x, y)])
#         return False

class Solution:
def getNeighbors(self, i, j, m, n):
nb = []
if i > 0:
nb.append((i - 1, j))
if i < m - 1:
nb.append((i + 1, j))
if j > 0:
nb.append((i, j - 1))
if j < n - 1:
nb.append((i, j + 1))
return nb

def isPossible(self, board, m, n, pos, word, i, used):
if i == len(word) - 1:
return True
neighbors = self.getNeighbors(pos[0], pos[1], m, n)
for a, b in neighbors:
if (a, b) not in used and board[a][b] == word[i + 1]:
if self.isPossible(board, m, n, (a, b), word, i + 1, used.union({(a, b)})):
return True
return False

def exist(self, board: List[List[str]], word: str) -> bool:
m = len(board)
n = len(board[0])
wlen = len(word)
for i in range(m):
for j in range(n):
if board[i][j] == word[0]:
if self.isPossible(board, m, n, (i, j), word, 0, {(i, j)}):
return True
return False