Given an integer n, add a dot (".") as the thousands separator and return it in string format.
 
Example 1:
Input: n = 987
Output: "987"

Example 2:
Input: n = 1234
Output: "1.234"

Example 3:
Input: n = 123456789
Output: "123.456.789"

Example 4:
Input: n = 0
Output: "0"

 
Constraints:

0 <= n < 2^31

Source: leetcode | Difficulty: EASY
Tags: String algorithms

Starter code: class Solution:
    def thousandSeparator(self, n: int) -> str:
        

Solutions:
class Solution:

	def thousandSeparator(self, n: int) -> str:
		arr = []
		(i, count) = (0, 0)
		num = str(n)
		while i < len(num):
			if count != 3:
				arr.append(num[~i])
				i += 1
				count += 1
			else:
				arr.append('.')
				count = 0
		return ''.join(arr[::-1])


class Solution:

	def thousandSeparator(self, n: int) -> str:
		res = ''
		for i in range(1, len(str(n)) + 1):
			res = str(n)[::-1][i - 1] + res
			if i % 3 == 0 and i != len(str(n)):
				res = '.' + res
		return res


class Solution:

	def thousandSeparator(self, n: int) -> str:
		s = str(n)
		arr = []
		for (i, c) in enumerate(reversed(s)):
			if i and (not i % 3):
				arr.append('.')
			arr.append(c)
		return ''.join(reversed(arr))


class Solution:

	def thousandSeparator(self, n: int) -> str:
		x = 3
		aux = list(str(n))
		ans = ''
		while len(aux) != 0:
			i = 0
			temp = ''
			while i < 3 and len(aux) != 0:
				temp = aux.pop() + temp
				i += 1
			if n > 10 ** x:
				ans = '.' + temp + ans
				x += 3
			else:
				ans = temp + ans
		return ans


class Solution:

	def thousandSeparator(self, n: int) -> str:
		n = list(str(n))
		if len(n) > 3:
			for i in range(len(n) - 3, 0, -3):
				n.insert(i, '.')
		return ''.join(n)