2018. Check if Word Can Be Placed In Crossword

Problem Link

Description

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You are given an m x n matrix board, representing the current state of a crossword puzzle. The crossword contains lowercase English letters (from solved words), ' ' to represent any empty cells, and '#' to represent any blocked cells.

A word can be placed horizontally (left to right or right to left) or vertically (top to bottom or bottom to top) in the board if:

• It does not occupy a cell containing the character '#'.
• The cell each letter is placed in must either be ' ' (empty) or match the letter already on the board.
• There must not be any empty cells ' ' or other lowercase letters directly left or right of the word if the word was placed horizontally.
• There must not be any empty cells ' ' or other lowercase letters directly above or below the word if the word was placed vertically.

Given a string word, return true if word can be placed in board, or false otherwise.

 

Example 1:

Input: board = [["#", " ", "#"], [" ", " ", "#"], ["#", "c", " "]], word = "abc"
Output: true
Explanation: The word "abc" can be placed as shown above (top to bottom).

Example 2:

Input: board = [[" ", "#", "a"], [" ", "#", "c"], [" ", "#", "a"]], word = "ac"
Output: false
Explanation: It is impossible to place the word because there will always be a space/letter above or below it.

Example 3:

Input: board = [["#", " ", "#"], [" ", " ", "#"], ["#", " ", "c"]], word = "ca"
Output: true
Explanation: The word "ca" can be placed as shown above (right to left). 

 

Constraints:

• m == board.length
• n == board[i].length
• 1 <= m * n <= 2 * 105
• board[i][j] will be ' ', '#', or a lowercase English letter.
• 1 <= word.length <= max(m, n)
• word will contain only lowercase English letters.

Solution

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Python3

class Solution:
    def placeWordInCrossword(self, board: List[List[str]], word: str) -> bool:
        words = [word, word[::-1]]
        n = len(word)
        
        for B in board, zip(*board):
            for row in B:
                s = "".join(row).split('#')
                
                for word in words:
                    for z in s:
                        if len(z) == n and all(z[i] == word[i] or z[i] == ' ' for i in range(n)):
                            return True
        
        return False