Problem Description

This problem involves a matrix representing a crossword puzzle, which includes lowercase English letters, spaces (represented by ' ') for empty cells, and the '#' character for blocked cells. The goal is to determine if a given word can be placed in the puzzle following certain conditions. The word can be placed either horizontally or vertically and must adhere to the following rules:

1. The word cannot be placed in cells that contain the '#' character (blocked cells).
2. Each letter of the word must either fill an empty cell (designated by spaces ' ') or match an existing letter on the board.
3. If the word is placed horizontally, there should not be any empty cells or other letters immediately to the left or right of the word.
4. If the word is placed vertically, there should not be any empty cells or other letters immediately above or below the word.

The task is to return true if the word can be placed on the board according to the rules, or false otherwise.

Intuition

To decide whether the given word can be positioned within the board, we should check each cell of the matrix where the first letter of the word could potentially be placed. This checking has to take into account the orientation (horizontal and vertical) and also the direction (from left to right, right to left, top to bottom, and bottom to top).

The intuition behind the provided solution is to systematically iterate over every cell in the matrix and try to match the word considering all possible starting positions and directions that adhere to the crossword rules. For every potential starting position, we examine whether the word would fit without violating any constraints such as running into blocked cells or mismatching existing letters. This involves:

• Checking horizontally to the right (left_to_right) and to the left (right_to_left).
• Checking vertically downwards (up_to_down) and upwards (down_to_up).

If any of these checks succeed, indicating that the word fits without issue, the function will return true. If no such fitting place is found across the entire board, the function will return false.

The solution efficiently prunes the search by ensuring the word's placement does not start or end next to an empty cell or a different letter when placing words horizontally or vertically. As such, it starts the placement from the border of the board or next to a blocked cell and checks if every letter of the word can be placed in a suitable position.

Solution Approach

The solution uses a nested loop to iterate through every cell in the board. For each cell, it checks if this cell could be a potential starting point for the word by following these steps:

1. It checks if the current cell and its immediate neighbor in the opposite direction of the check are either on the edge of the board or blocked by a '#'. This ensures that we only start at valid positions according to the rules of the puzzle.
2. If the starting position is valid, it then invokes the check function which will attempt to place the word starting from that position, moving either horizontally or vertically and either forwards or backwards depending on the check being performed.

The check function is designed to validate the placement of the word by iterating over each letter of the word and checking the following conditions:

• The current position is within the bounds of the board.
• The current board cell is not blocked (i.e., not '#').
• The current board cell is either empty (i.e., ' ') or matches the corresponding letter in the word.

The check function also ensures that the letter after the last one of the word (calculated by x, y = i + a * k, j + b * k) is either out of bounds or blocked. This ensures that the word does not end next to a cell that could violate the horizontal or vertical placement rules.

The data structures used in the solution are:

• The board matrix which stores the characters as a 2D list.
• Variables m and n which represent the number of rows and columns in the board, respectively.
• Variable k which is the length of the word.

The solution approach does not use any additional complex algorithms or patterns. It simply leverages careful iteration and checking of board states to determine if the word can be placed.

Here's a code snippet encapsulating that logic:

for i in range(m):
    for j in range(n):
        left_to_right = (j == 0 or board[i][j - 1] == '#') and check(i, j, 0, 1)
        right_to_left = (j == n - 1 or board[i][j + 1] == '#') and check(i, j, 0, -1)
        up_to_down = (i == 0 or board[i - 1][j] == '#') and check(i, j, 1, 0)
        down_to_up = (i == m - 1 or board[i + 1][j] == '#') and check(i, j, -1, 0)
        if left_to_right or right_to_left or up_to_down or down_to_up:
            return True
return False

The and operator in left_to_right, right_to_left, up_to_down, and down_to_up checks combines the start position validation and the check function call. If any of these conditions return True, it means the word can be placed in the board following the puzzle rules.

Example Walkthrough

Consider a 3 x 3 crossword puzzle board and the word "cat":

board = [
    ['#', 'c', '#'],
    [' ', 'a', ' '],
    ['#', 't', '#']
]

We want to check if we can place the word "cat" on this board.

Using our solution approach, we will check each cell starting from (0,0) to (2,2) to find a valid placement for "cat".

First, we check horizontally to the right (left_to_right). Starting from (0,0) we find it's a blocked cell ('#'), so we move to (0,1). The cell (0,1) contains the first letter of the word "cat", 'c', and cell (0,0) is blocked, which is a valid start position. However, we cannot place "cat" horizontally to the right because there is no space to fit the entire word "cat".

Next, we check horizontally to the left (right_to_left). This direction is not applicable in this case as we are looking for starting positions.

Then, we check vertically downwards (up_to_down). From (0,1), we realize (0,0) is blocked, providing a potential starting position for "cat". We can successfully match 'c' with 'c', then move to the next cell (1,1) and match 'a' with 'a', and finally move to (2,1) and see that 't' can match 't'. Thus, "cat" can be placed vertically from (0,1) to (2,1).

Since we found a valid placement for "cat", we do not need to check vertically upwards (down_to_up) from (0,1). The check function would return true, and the algorithm would confirm that "cat" can be placed on the board.

So in our algorithm, as soon as it runs the up_to_down check starting at (0,1), it will return true, indicating that the word "cat" can indeed be placed on the board vertically. The result is obtained without having to check other cells or directions, as the solution is efficiently designed to stop once a valid position is found.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is O(m * n * k), where m is the number of rows in the board, n is the number of columns in the board, and k is the length of the word to be placed. This complexity arises because the code iterates over all cells of the board (m * n) and for each cell, it attempts to place the word in all four directions. The check function, which is called for each direction, runs a loop up to the length of the word (k).

Additionally, when evaluating the board for possible placements, the algorithm checks the perpendicular cells to ensure placement is at the beginning or end of a word sequence (which is a constant time check). Due to these operations being constant in time, they do not impact the linear relationship between the time complexity and the number of cells times the length of the word.

Space Complexity

The space complexity of the code is O(1) (constant space complexity). This is because the algorithm only uses a fixed amount of extra space for variables that store the dimensions of the board and indices during the checks regardless of the input size. No additional space proportional to the input size is required beyond what is used to store the board and word, which are inputs and not counted towards the space complexity.

Learn more about how to find time and space complexity quickly using problem constraints.