2018. Check if Word Can Be Placed In Crossword
Description
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 theboard
. - 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:
![](https://spcdn.pages.dev/leetcode/problems/2018.Check%20if%20Word%20Can%20Be%20Placed%20In%20Crossword/images/crossword-ex1-1.png)
Input: board = [["#", " ", "#"], [" ", " ", "#"], ["#", "c", " "]], word = "abc" Output: true Explanation: The word "abc" can be placed as shown above (top to bottom).
Example 2:
![](https://spcdn.pages.dev/leetcode/problems/2018.Check%20if%20Word%20Can%20Be%20Placed%20In%20Crossword/images/crossword-ex2-1.png)
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:
![](https://spcdn.pages.dev/leetcode/problems/2018.Check%20if%20Word%20Can%20Be%20Placed%20In%20Crossword/images/crossword-ex3-1.png)
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.
Solutions
Solution 1: Enumeration
We can enumerate each position $(i, j)$ in the matrix, and judge whether we can place the word word
from left to right or from right to left, or from top to bottom or from bottom to top, starting from this position.
The following conditions must be met for this position to be used as a starting point:
- If the word
word
is to be placed from left to right, then this position must be the left boundary, or the cellboard[i][j - 1]
to the left of this position is'#'
. - If the word
word
is to be placed from right to left, then this position must be the right boundary, or the cellboard[i][j + 1]
to the right of this position is'#'
. - If the word
word
is to be placed from top to bottom, then this position must be the upper boundary, or the cellboard[i - 1][j]
above this position is'#'
. - If the word
word
is to be placed from bottom to top, then this position must be the lower boundary, or the cellboard[i + 1][j]
below this position is'#'
.
Under the above conditions, we can start from this position and judge whether the word word
can be placed. We design a function $check(i, j, a, b)$, which represents whether it is legal to place the word word
from the position $(i, j)$ in the direction $(a, b)$. If it is legal, return true
, otherwise return false
.
The implementation of the function $check(i, j, a, b)$ is as follows:
We first get the other boundary position $(x, y)$ in the current direction, i.e., $(x, y) = (i + a \times k, j + b \times k)$, where $k$ is the length of the word word
. If $(x, y)$ is in the matrix and the cell at $(x, y)$ is not '#'
, it means that the other boundary position in the current direction is not '#'
, so the word word
cannot be placed, and false
is returned.
Otherwise, we start from the position $(i, j)$ and traverse the word word
in the direction $(a, b)$. If we encounter a cell board[i][j]
that is not a space or not the current character of the word word
, it means that the word word
cannot be placed, and false
is returned. If the word word
is traversed, it means that the word word
can be placed, and true
is returned.
The time complexity is $O(m \times n)$, and the space complexity is $O(1)$. Here, $m$ and $n$ are the number of rows and columns in the matrix, respectively.
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