word search

word_search.py

#!/usr/bin/env python3
"""
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.
"""


import copy


class Solution:
    def __init__(self):
        self.visited = []
        self.d = [[-1,0],[0,1],[1,0],[0,-1]]
        self.m = 0
        self.n = 0
    
    
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        self.m = len(board)
        self.n = len(board[0])
        
        self.visited = copy.deepcopy(board)
        for i in range(self.m):
            for j in range(self.n):
                self.visited[i][j] = False
        
        for i in range(self.m):
            for j in range(self.n):
                if self.searchWord(board, word, 0, i, j):
                    return True
        return False
            
    def searchWord(self, board, word, index, startx, starty):
        if index == len(word) - 1:
            return board[startx][starty] == word[index]
        
        if board[startx][starty] == word[index]:
            self.visited[startx][starty] = True
            
            for i in range(4):
                newx = startx + self.d[i][0]
                newy = starty + self.d[i][1]
                
                if self.inArea(newx, newy) and not self.visited[newx][newy] and self.searchWord(board, word, index+1, newx, newy):
                    return True
            self.visited[startx][starty] = False
        return False
    
    def inArea(self, x, y):
        return x >= 0 and x < self.m and y >= 0 and y < self.n