Leetcode - Word Search Solution
Poor Man
- Published
- Updated
- Reading time
- 2 min read
- Length
- 308 words
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.
Note: There will be some test cases with a board or a word larger than constraints to test if your solution is using pruning.
Example 1:
Text
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: trueExample 2:
Text
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: trueExample 3:
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Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: falseConstraints:
m == board.lengthn = board[i].length1 <= m, n <= 61 <= word.length <= 15boardandwordconsists of only lowercase and uppercase English letters.
Solution in python#
Python
class Solution(object):
def exist(self, board, word):
board = board
word = word
def ExistCharacter(i,j,index,word):
if (i < 0 or i >= len(board) or j < 0 or j >= len(board[i])) :
return False
# when index character does not match
if(board[i][j] != word[index]) :
return False
# when completely matched
if(index == len(word) - 1) :
return True
# mark the current character
board[i][j] = '#'
# check every direction
found = ExistCharacter(i, j - 1, index + 1, word) or ExistCharacter(i, j + 1, index + 1, word) or ExistCharacter(i - 1, j, index + 1, word) or ExistCharacter(i + 1, j, index + 1, word)
# unmark the current character
board[i][j] = word[index]
return found
if (word == ""):
return False
# iterate over the board
for i in range(len(board)):
for j in range(len(board[0])):
if (ExistCharacter(i, j, 0, word)):
return True
return False