79. Word Search

https://leetcode.com/problems/word-search/#/description

 

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.

For example,
Given board =

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

 

 

Sol:

 

Recursively call the dfs method on four directions - up, down, left, right - and check the letter the input word one by one. If certain letter meets the end conditions then False is returned. Otherwise, all letters are checked, and end conditions are not triggered, then True is returned.

 

3 End Conditions:

1) possible answer crosses the border of the board

2)  the board does not have the letter in the word

3) possible answer is visited - We don't trace back

 

class Solution(object):

    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        # dfs, recursion
        # time O(n^2 * m^2), space O(n^2)
        
        row = len(board)
        col = len(board[0])
        visited = {}
        for i in range(row):
            for j in range(col):
                if self.dfs(board, word, i, j, visited):
                    return True
        return False
    
    # pos variable here is to check the letter in word one by one 
    def dfs(self, board, word, x, y, visited, pos = 0):

            # convergence condition, all letters in the word pass 
            if pos == len(word):
                return True
            
            # end condition: cross the line
            if x < 0 or y < 0 or x >= len(board) or y >= len(board[0]):
                return False
            
            # end condition: the board does not have the letter in the word
            if board[x][y] != word[pos]:
                return False
            
            # end condition: the board has been checked, we don't trace back
            if visited.get((x,y)):
                return False

            visited[(x,y)] = True
            res = self.dfs(board, word, x - 1, y, visited, pos + 1) \
                    or  self.dfs(board, word, x + 1, y, visited, pos + 1) \
                    or  self.dfs(board, word, x, y + 1, visited, pos + 1) \
                    or  self.dfs(board, word, x, y - 1, visited, pos + 1)
            visited[(x,y)] = False
            
            return res