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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

function exist(board: string[][], word: string): boolean {
    if (word.length === 0) return false
    let ret = false
    const marks = makeArray(board)
    for (let i = 0; i < board.length; i++) {
        for (let j = 0; j < board[i].length; j++) {
            if (board[i][j] !== word.charAt(0)) continue
            if (loop(marks, board, i, j, word, 0)) return true
        }
    }

    return ret
}

function makeArray(board: string[][]) {
    const arr = []
    for (let i = 0; i < board.length; i++) {
        arr[i] = []
        for (let j = 0; j < board[i].length; j++) {
            arr[i][j] = false
        }
    }
    return arr
}

function loop(marks: boolean[][], board: string[][], i: number, j: number, word: string, index: number): boolean {
    if (i < 0 || j < 0 || i >= board.length || j >= board[i].length || marks[i][j]) {
        return false
    }

    if (board[i][j] !== word.charAt(index)) {
        return false
    } else if (index === word.length - 1) {
        return true
    }

    marks[i][j] = true
    index++

    let r = loop(marks, board, i - 1, j, word, index)
    if (r) return true

    r = loop(marks, board, i + 1, j, word, index)
    if (r) return true

    r = loop(marks, board, i, j - 1, word, index)
    if (r) return true

    r = loop(marks, board, i, j + 1, word, index)
    if (r) return true

    marks[i][j] = false
    return false
}

export { exist }

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