LeetCode 79. Word Search

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Problem Statement

(Source) 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.

Solution

Tags: Backtracking.

class Solution(object):    def exist(self, board, word):        """        :type board: List[List[str]]        :type word: str        :rtype: bool        """        return any(self.bt(board, word, [word[0]], set([(x, y)]), x, y) for x in xrange(len(board)) for y in xrange(len(board[0])) if board[x][y] == word[0])    def bt(self, board, word, ass, visited, x, y):        if len(ass) == len(word):            return ''.join(ass) == word        else:            m, n = len(board), len(board[0])            if x - 1 >= 0 and board[x-1][y] == word[len(ass)] and (x-1, y) not in visited:                visited.add((x-1, y))                ass.append(board[x-1][y])                if self.bt(board, word, ass, visited, x-1, y):                    return True                else:                    ass.pop()                    visited.remove((x-1, y))            if x + 1 < m and board[x+1][y] == word[len(ass)] and (x+1, y) not in visited:                visited.add((x+1, y))                ass.append(board[x+1][y])                if self.bt(board, word, ass, visited, x+1, y):                    return True                else:                    ass.pop()                    visited.remove((x+1, y))            if y - 1 >= 0 and board[x][y-1] == word[len(ass)] and (x, y-1) not in visited:                visited.add((x, y-1))                ass.append(board[x][y-1])                if self.bt(board, word, ass, visited, x, y-1):                    return True                else:                    ass.pop()                    visited.remove((x, y-1))            if y + 1 < n and board[x][y+1] == word[len(ass)] and (x, y+1) not in visited:                visited.add((x, y+1))                ass.append(board[x][y+1])                if self.bt(board, word, ass, visited, x, y+1):                    return True                else:                    ass.pop()                    visited.remove((x, y+1))            return False

A more concise solution with the same idea:

class Solution(object):    def exist(self, board, word):        """        :type board: List[List[str]]        :type word: str        :rtype: bool        """        return any(self.bt(board, word, [word[0]], set([(x, y)]), x, y) for x in xrange(len(board)) for y in xrange(len(board[0])) if board[x][y] == word[0])    def bt(self, board, word, ass, visited, x, y):        if len(ass) == len(word):            return ''.join(ass) == word        else:            m, n = len(board), len(board[0])            offset = [1, 0, -1, 0, 1]            for index in xrange(4):                xx, yy = x + offset[index], y + offset[index+1]                if 0 <= xx < m and 0 <= yy < n and board[xx][yy] == word[len(ass)] and (xx, yy) not in visited:                    visited.add((xx, yy))                    ass.append(board[xx][yy])                    if self.bt(board, word, ass, visited, xx, yy):                        return True                    else:                        ass.pop()                        visited.remove((xx, yy))            return False

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