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:

Input: {"eggs", "karat", "apple", "snack", "tuna"}
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.

If you ask in Discord or on the Website for a solution you are disqualified

I'll update the question with my implementation within due time.

Inspired by this post

Please do not moderate this Question as it doesn't break any rules

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.

local words = {"apple", "estonia", "wow"}

local function isChained () 
	local copy = {};
	
	for i, word in pairs(words) do
		copy[i] = word:reverse();
	end
	
	table.sort(copy);
	table.sort(words);
	
	for i = 1, #words do
		if (words[i]:sub(1, 1) ~= copy[i]:sub(1, 1) or copy[i] == words[i]:reverse()) then
			return false;
		end
	end
	return true;
end

print(isChained());

Can't be bothered to optimize it or make it shorter

--words = {"eggs", "karat", "apple", "no", "snack", "tuna"}
words = {"monkey","moan","neh","neat","hort","team","gah"}
--words = {"eggs", "sea", "apple"}
--words = {"apple", "earth", "roda"}

function intable(table, value)
	for i, v in pairs(table) do
		if type(v) == "table" and type(value) == "table" then
			for i2, v2 in pairs(v) do
				if value[i2] ~= v2 then return false end
			end
			for i2, v2 in pairs(value) do
				if v[i2] ~= v2 then return false end
			end

			return true
		else
			if v == value then
				return true
			end
		end
	end
	return false
end


function chain()
	foundlinks = {}

	for i, v in pairs(words) do
		for i2, v2 in pairs(words) do
			if i ~= i2 then 
				if string.sub(v,1,1) == string.sub(v2, #v2) and
				(not intable(foundlinks, {v, 1})) and (not intable(foundlinks, {v2, 2})) then
					print("Found link: "..v2.. " and "..v)
					table.insert(foundlinks, {v, 1})
					table.insert(foundlinks, {v2, 2})
				end

				if string.sub(v,#v) == string.sub(v2,1,1) and 
				(not intable(foundlinks, {v2, 1})) and (not intable(foundlinks, {v, 2})) then
					print("Found link: "..v.. " and "..v2)
					table.insert(foundlinks, {v2, 1})
					table.insert(foundlinks, {v, 2})
				end
			end
		end

		if #foundlinks == #words * 2 then return true end
	end

	return false --if we get this far
end

--print( intable({2, 5, 3, {3, 3}}, {3, 3}) )

print(chain(words))