0079-word-search.rb

# frozen_string_literal: true

# 79. Word Search
# https://leetcode.com/problems/word-search
# Medium

=begin
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.
=end

# @param {Character[][]} board
# @param {String} word
# @return {Boolean}
defdfs(board,word,i,j)
returntrueifword == ""
returnfalseifi < 0 || j < 0 || i >= board.length || j >= board[0].length || board[i][j] != word[0]
pom=board[i][j]
board[i][j]="#"
word=word[1..-1]
returntrueifdfs(board,word,i - 1,j)
returntrueifdfs(board,word,i,j - 1)
returntrueifdfs(board,word,i + 1,j)
returntrueifdfs(board,word,i,j + 1)
board[i][j]=pom
false
end

defexist(board,word)
set1=board.flatten.to_set
set2=word.split("").to_set
returnfalseif !(set1 >= set2)
(0...board.length).eachdo |i|
(0...board[0].length).eachdo |j|
returntrueifdfs(board,word,i,j)
end
end
false
end

# **************** #
# TEST #
# **************** #

require"test/unit"
classTest_exist < Test::Unit::TestCase
deftest_
assert_equaltrue,exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]],"ABCCED")
assert_equaltrue,exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]],"SEE")
assert_equalfalse,exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]],"ABCB")
end
end