LeetCode #2018 — MEDIUM

Check if Word Can Be Placed In Crossword

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[["#"," ","#"],[" "," ","#"],["#","c"," "]]
"abc"

Example 2

[[" ","#","a"],[" ","#", "c"],[" ","#","a"]]
"ac"

Example 3

[["#"," ","#"],[" "," ","#"],["#"," ","c"]]
"ca"

Step 02

Core Insight

What unlocks the optimal approach

Check all possible placements for the word.
There is a limited number of places where a word can start.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2018: Check if Word Can Be Placed In Crossword
class Solution {
    private int m;
    private int n;
    private char[][] board;
    private String word;
    private int k;

    public boolean placeWordInCrossword(char[][] board, String word) {
        m = board.length;
        n = board[0].length;
        this.board = board;
        this.word = word;
        k = word.length();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                boolean leftToRight = (j == 0 || board[i][j - 1] == '#') && check(i, j, 0, 1);
                boolean rightToLeft = (j == n - 1 || board[i][j + 1] == '#') && check(i, j, 0, -1);
                boolean upToDown = (i == 0 || board[i - 1][j] == '#') && check(i, j, 1, 0);
                boolean downToUp = (i == m - 1 || board[i + 1][j] == '#') && check(i, j, -1, 0);
                if (leftToRight || rightToLeft || upToDown || downToUp) {
                    return true;
                }
            }
        }
        return false;
    }

    private boolean check(int i, int j, int a, int b) {
        int x = i + a * k, y = j + b * k;
        if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] != '#') {
            return false;
        }
        for (int p = 0; p < k; ++p) {
            if (i < 0 || i >= m || j < 0 || j >= n
                || (board[i][j] != ' ' && board[i][j] != word.charAt(p))) {
                return false;
            }
            i += a;
            j += b;
        }
        return true;
    }
}

// Accepted solution for LeetCode #2018: Check if Word Can Be Placed In Crossword
func placeWordInCrossword(board [][]byte, word string) bool {
	m, n := len(board), len(board[0])
	k := len(word)
	check := func(i, j, a, b int) bool {
		x, y := i+a*k, j+b*k
		if x >= 0 && x < m && y >= 0 && y < n && board[x][y] != '#' {
			return false
		}
		for _, c := range word {
			if i < 0 || i >= m || j < 0 || j >= n || (board[i][j] != ' ' && board[i][j] != byte(c)) {
				return false
			}
			i, j = i+a, j+b
		}
		return true
	}
	for i := range board {
		for j := range board[i] {
			leftToRight := (j == 0 || board[i][j-1] == '#') && check(i, j, 0, 1)
			rightToLeft := (j == n-1 || board[i][j+1] == '#') && check(i, j, 0, -1)
			upToDown := (i == 0 || board[i-1][j] == '#') && check(i, j, 1, 0)
			downToUp := (i == m-1 || board[i+1][j] == '#') && check(i, j, -1, 0)
			if leftToRight || rightToLeft || upToDown || downToUp {
				return true
			}
		}
	}
	return false
}

# Accepted solution for LeetCode #2018: Check if Word Can Be Placed In Crossword
class Solution:
    def placeWordInCrossword(self, board: List[List[str]], word: str) -> bool:
        def check(i, j, a, b):
            x, y = i + a * k, j + b * k
            if 0 <= x < m and 0 <= y < n and board[x][y] != '#':
                return False
            for c in word:
                if (
                    i < 0
                    or i >= m
                    or j < 0
                    or j >= n
                    or (board[i][j] != ' ' and board[i][j] != c)
                ):
                    return False
                i, j = i + a, j + b
            return True

        m, n = len(board), len(board[0])
        k = len(word)
        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

// Accepted solution for LeetCode #2018: Check if Word Can Be Placed In Crossword
/**
 * [2018] Check if Word Can Be Placed In Crossword
 *
 * 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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/10/04/crossword-ex1-1.png" style="width: 478px; height: 180px;" />
 * Input: board = [["#", " ", "#"], [" ", " ", "#"], ["#", "c", " "]], word = "abc"
 * Output: true
 * Explanation: The word "abc" can be placed as shown above (top to bottom).
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/10/04/crossword-ex2-1.png" style="width: 180px; height: 180px;" />
 * 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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/10/04/crossword-ex3-1.png" style="width: 478px; height: 180px;" />
 * 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 * 10^5
 * 	board[i][j] will be ' ', '#', or a lowercase English letter.
 * 	1 <= word.length <= max(m, n)
 * 	word will contain only lowercase English letters.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/check-if-word-can-be-placed-in-crossword/
// discuss: https://leetcode.com/problems/check-if-word-can-be-placed-in-crossword/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn place_word_in_crossword(board: Vec<Vec<char>>, word: String) -> bool {
        false
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2018_example_1() {
        let board = vec![
            vec!['#', ' ', '#'],
            vec![' ', ' ', '#'],
            vec!['#', 'c', ' '],
        ];
        let word = "abc".to_string();

        let result = true;

        assert_eq!(Solution::place_word_in_crossword(board, word), result);
    }

    #[test]
    #[ignore]
    fn test_2018_example_2() {
        let board = vec![
            vec![' ', '#', 'a'],
            vec![' ', '#', 'c'],
            vec![' ', '#', 'a'],
        ];
        let word = "ac".to_string();

        let result = false;

        assert_eq!(Solution::place_word_in_crossword(board, word), result);
    }

    #[test]
    #[ignore]
    fn test_2018_example_3() {
        let board = vec![
            vec!['#', ' ', '#'],
            vec![' ', ' ', '#'],
            vec!['#', ' ', 'c'],
        ];
        let word = "ca".to_string();

        let result = true;

        assert_eq!(Solution::place_word_in_crossword(board, word), result);
    }
}

// Accepted solution for LeetCode #2018: Check if Word Can Be Placed In Crossword
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2018: Check if Word Can Be Placed In Crossword
// class Solution {
//     private int m;
//     private int n;
//     private char[][] board;
//     private String word;
//     private int k;
// 
//     public boolean placeWordInCrossword(char[][] board, String word) {
//         m = board.length;
//         n = board[0].length;
//         this.board = board;
//         this.word = word;
//         k = word.length();
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 boolean leftToRight = (j == 0 || board[i][j - 1] == '#') && check(i, j, 0, 1);
//                 boolean rightToLeft = (j == n - 1 || board[i][j + 1] == '#') && check(i, j, 0, -1);
//                 boolean upToDown = (i == 0 || board[i - 1][j] == '#') && check(i, j, 1, 0);
//                 boolean downToUp = (i == m - 1 || board[i + 1][j] == '#') && check(i, j, -1, 0);
//                 if (leftToRight || rightToLeft || upToDown || downToUp) {
//                     return true;
//                 }
//             }
//         }
//         return false;
//     }
// 
//     private boolean check(int i, int j, int a, int b) {
//         int x = i + a * k, y = j + b * k;
//         if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] != '#') {
//             return false;
//         }
//         for (int p = 0; p < k; ++p) {
//             if (i < 0 || i >= m || j < 0 || j >= n
//                 || (board[i][j] != ' ' && board[i][j] != word.charAt(p))) {
//                 return false;
//             }
//             i += a;
//             j += b;
//         }
//         return true;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.