Is String Array Circle of Chained Words in least amount of characters?
The Challenge:
Two words can be 'chained' if the last character of the first word is the same as the first character of the second word.
Given a list of words, determine if there is a way to 'chain' all the words in a circle.
Example:
1 | Input: {"eggs", "karat", "apple", "snack", "tuna"} |
2 | Output: True |
Print statements are not part of your character count. Your code should explicitly state if the array of words are chained or not.
Shortest and most valid answer within a week gets the solution.
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I'll update the question with my implementation within due time.
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2 answers
My solution.
please check to see if it passes through your testcases or not.
EDIT: Found a bug when in an odd number of elements, n-1 elements form a chain but the last one is something like "wow" where if it is flipped, it would still count as chain. This has been fixed now.
01 | local words = {"apple", "estonia", "wow"} |
02 |
03 | local function isChained () |
04 | local copy = {}; |
05 |
06 | for i, word in pairs(words) do |
07 | copy[i] = word:reverse(); |
08 | end |
09 |
10 | table.sort(copy); |
11 | table.sort(words); |
12 |
13 | for i = 1, #words do |
14 | if (words[i]:sub(1, 1) ~= copy[i]:sub(1, 1) or copy[i] == words[i]:reverse()) then |
15 | return false; |
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Can't be bothered to optimize it or make it shorter
01 | --words = {"eggs", "karat", "apple", "no", "snack", "tuna"} |
02 | words = {"monkey","moan","neh","neat","hort","team","gah"} |
03 | --words = {"eggs", "sea", "apple"} |
04 | --words = {"apple", "earth", "roda"} |
05 |
06 | function intable(table, value) |
07 | for i, v in pairs(table) do |
08 | if type(v) == "table" and type(value) == "table" then |
09 | for i2, v2 in pairs(v) do |
10 | if value[i2] ~= v2 then return false end |
11 | end |
12 | for i2, v2 in pairs(value) do |
13 | if v[i2] ~= v2 then return false end |
14 | end |
15 |