LeetCode 2018: Check if Word Can Be Placed In Crossword Solution

Master LeetCode problem 2018 (Check if Word Can Be Placed In Crossword), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2018. Check if Word Can Be Placed In Crossword

Medium

Array Matrix Enumeration

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible starting positions on the board and check if the word can be placed horizontally or vertically from that position. We need to ensure that the word can fit in the given direction and that the surrounding cells are either empty or match the letters of the word.

For each starting position (cell), we check if we can place the word horizontally or vertically.
To check if the word can be placed horizontally, we ensure that there is enough space to the right and left of the starting cell for the word length, and that the surrounding cells are either empty or match the word's letters.
To check if the word can be placed vertically, we ensure that there is enough space above and below the starting cell for the word length, and that the surrounding cells are either empty or match the word's letters.
If we find a valid position where the word can be placed, we return true. If no such position is found, we return false.

Time Complexity: O(m * n * max(m, n)) where m is the number of rows, n is the number of columns in the board.

Space Complexity: O(1)

Solution Code

class Solution {
    public boolean placeWordInCrossword(char[][] board, String word) {
        int m = board.length;
        int n = board[0].length;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (canPlaceWord(board, word, i, j, 1, 0) || canPlaceWord(board, word, i, j, 0, 1)) {
                    return true;
                }
            }
        }
        
        return false;
    }
    
    private boolean canPlaceWord(char[][] board, String word, int row, int col, int dx, int dy) {
        int m = board.length;
        int n = board[0].length;
        
        for (int i = 0; i < word.length(); i++) {
            if (row < 0 || row >= m || col < 0 || col >= n || (board[row][col] != ' ' && board[row][col] != word.charAt(i))) {
                return false;
            }
            row += dx;
            col += dy;
        }
        
        return (row == m || row == -1 || col == n || col == -1 || board[row - dx][col - dy] == '#');
    }
}

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