LeetCode in Kotlin

79. Word Search

Medium

Given an m x n grid of characters board and a string word, return true if word exists in the grid.

The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example 1:

Input: board = [[“A”,”B”,”C”,”E”],[“S”,”F”,”C”,”S”],[“A”,”D”,”E”,”E”]], word = “ABCCED”

Output: true

Example 2:

Input: board = [[“A”,”B”,”C”,”E”],[“S”,”F”,”C”,”S”],[“A”,”D”,”E”,”E”]], word = “SEE”

Output: true

Example 3:

Input: board = [[“A”,”B”,”C”,”E”],[“S”,”F”,”C”,”S”],[“A”,”D”,”E”,”E”]], word = “ABCB”

Output: false

Constraints:

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board and word consists of only lowercase and uppercase English letters.

Follow up: Could you use search pruning to make your solution faster with a larger board?

Solution

class Solution {
    private fun backtrace(
        board: Array<CharArray>,
        visited: Array<BooleanArray>,
        word: String,
        index: Int,
        x: Int,
        y: Int,
    ): Boolean {
        if (index == word.length) {
            return true
        }
        if (x < 0 || x >= board.size || y < 0 || y >= board[0].size || visited[x][y]) {
            return false
        }
        visited[x][y] = true
        return if (word[index] == board[x][y]) {
            val res = (
                backtrace(board, visited, word, index + 1, x, y + 1) ||
                    backtrace(board, visited, word, index + 1, x, y - 1) ||
                    backtrace(board, visited, word, index + 1, x + 1, y) ||
                    backtrace(board, visited, word, index + 1, x - 1, y)
                )
            if (!res) {
                visited[x][y] = false
            }
            res
        } else {
            visited[x][y] = false
            false
        }
    }

    fun exist(board: Array<CharArray>, word: String): Boolean {
        val visited = Array(board.size) {
            BooleanArray(
                board[0].size,
            )
        }
        for (i in board.indices) {
            for (j in board[0].indices) {
                if (backtrace(board, visited, word, 0, i, j)) {
                    return true
                }
            }
        }
        return false
    }
}

This site is open source. Improve this page.