How do you solve Check if Word Can Be Placed In Crossword on LeetCode?

Check if Word Can Be Placed In Crossword (LeetCode #2018) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(1) space complexity. Key insight: Identify potential starting points for the word.

What is the time complexity of Check if Word Can Be Placed In Crossword?

The optimal solution for Check if Word Can Be Placed In Crossword runs in O(m * n) time and uses O(1) space. We only traverse the board once, checking each cell and its possible placements, leading to O(m * n).

What is the brute force solution for Check if Word Can Be Placed In Crossword?

The brute force approach for Check if Word Can Be Placed In Crossword is Brute Force, with O(m * n * k) time complexity and O(1) space complexity. The brute-force approach checks every possible starting position on the board for the word, both horizontally and vertically. It verifies if the word can fit in that position without violating any rules. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Check if Word Can Be Placed In Crossword?

Check if Word Can Be Placed In Crossword uses the following concepts: Array, Matrix, Enumeration. The recommended patterns to study are: Enumeration, Matrix Traversal.

What are the interview tips for Check if Word Can Be Placed In Crossword?

Clearly explain your thought process while coding. Consider edge cases, such as words longer than available spaces. Use helper functions to keep your code organized and readable.

What are common mistakes when solving Check if Word Can Be Placed In Crossword?

Not checking adjacent cells properly. Overlooking blocked cells when placing the word.

#2018

Check if Word Can Be Placed In Crossword

Medium

ArrayMatrixEnumerationEnumerationMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n * k)	O(m * n)
Space	O(1)	O(1)

💡

Intuition

Time O(m * n)Space O(1)

The optimal solution uses a systematic approach to identify all possible placements for the word in one pass. By checking the boundaries and conditions before attempting to place the word, we minimize unnecessary checks.

⚙️

Algorithm

3 steps

1Step 1: Iterate through each cell in the board to find potential starting points for the word.
2Step 2: For each starting point, check if the word can fit horizontally and vertically using helper functions.
3Step 3: Ensure that the placement does not violate any rules regarding adjacent cells.

solution.py23 lines

1def canPlaceWord(board, word):
2    m, n = len(board), len(board[0])
3    word_len = len(word)
4    def canPlaceHorizontally(i, j):
5        if j > 0 and board[i][j - 1] != '#': return False
6        if j + word_len < n and board[i][j + word_len] != '#': return False
7        for k in range(word_len):
8            if board[i][j + k] != ' ' and board[i][j + k] != word[k]:
9                return False
10        return True
11    def canPlaceVertically(i, j):
12        if i > 0 and board[i - 1][j] != '#': return False
13        if i + word_len < m and board[i + word_len][j] != '#': return False
14        for k in range(word_len):
15            if board[i + k][j] != ' ' and board[i + k][j] != word[k]:
16                return False
17        return True
18    for i in range(m):
19        for j in range(n):
20            if board[i][j] == ' ' or board[i][j] == word[0]:
21                if canPlaceHorizontally(i, j) or canPlaceVertically(i, j):
22                    return True
23    return False

ℹ

Complexity note: We only traverse the board once, checking each cell and its possible placements, leading to O(m * n).

1Identify potential starting points for the word.
2Check boundaries to avoid out-of-bounds errors.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.