212. Word Search II #

Problem #

Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of sequentially adjacent cell, where “adjacent” cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

Example:

Input: 
board = [
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]
words = ["oath","pea","eat","rain"]

Output: ["eat","oath"]

Note:

1. All inputs are consist of lowercase letters a-z.
2. The values of words are distinct.

Problem Summary #

Given a 2D grid board and a list of words from a dictionary, words, find all words that appear in both the 2D grid and the dictionary.

Each word must be constructed in letter order from letters in adjacent cells, where “adjacent” cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

Solution Ideas #

• This problem is an enhanced version of Problem 79. On top of Problem 79, it adds a word array and requires finding all words that appear in the board. The idea can still follow the DFS search from Problem 79, but the time complexity is extremely high!
• Think about a better solution.

Code #

package leetcode

func findWords(board [][]byte, words []string) []string {
	res := []string{}
	for _, v := range words {
		if exist(board, v) {
			res = append(res, v)
		}
	}
	return res
}

// these is 79 solution
var dir = [][]int{
	{-1, 0},
	{0, 1},
	{1, 0},
	{0, -1},
}

func exist(board [][]byte, word string) bool {
	visited := make([][]bool, len(board))
	for i := 0; i < len(visited); i++ {
		visited[i] = make([]bool, len(board[0]))
	}
	for i, v := range board {
		for j := range v {
			if searchWord(board, visited, word, 0, i, j) {
				return true
			}
		}
	}
	return false
}

func isInBoard(board [][]byte, x, y int) bool {
	return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
}

func searchWord(board [][]byte, visited [][]bool, word string, index, x, y int) bool {
	if index == len(word)-1 {
		return board[x][y] == word[index]
	}
	if board[x][y] == word[index] {
		visited[x][y] = true
		for i := 0; i < 4; i++ {
			nx := x + dir[i][0]
			ny := y + dir[i][1]
			if isInBoard(board, nx, ny) && !visited[nx][ny] && searchWord(board, visited, word, index+1, nx, ny) {
				return true
			}
		}
		visited[x][y] = false
	}
	return false
}