Format any integer provided into a string with "," (commas) in the correct places.

**Example:**
``` csharp
Kata.NumberFormat(100000); // return "100,000"
Kata.NumberFormat(5678545); // return "5,678,545"
Kata.NumberFormat(-420902); // return "-420,902"
```
``` javascript
numberFormat(100000); // return '100,000'
numberFormat(5678545); // return '5,678,545'
numberFormat(-420902); // return '-420,902'
```
``` cpp
numberFormat(100000); // return '100,000'
numberFormat(5678545); // return '5,678,545'
numberFormat(-420902); // return '-420,902'
```
``` python
number_format(100000); # return '100,000'
number_format(5678545); # return '5,678,545'
number_format(-420902); # return '-420,902'
```
``` ruby
number_format(100000); # return '100,000'
number_format(5678545); # return '5,678,545'
number_format(-420902); # return '-420,902'
```
``` crystal
number_format(100000); # return '100,000'
number_format(5678545); # return '5,678,545'
number_format(-420902); # return '-420,902'
```

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def number_format(n):
	

Solutions:
def number_format(n):
	return f'{n:,}'


def number_format(n):
	return format(n, ',')


number_format = '{:,}'.format


def number_format(n):
	return str('{:,.3f}'.format(n)).rstrip('0').rstrip('.')


import re

def number_format(n):
	return re.sub('\\B(?=(\\d{3})+(?!\\d))', ',', str(n))


number_format = lambda n: ''.join((',' * (i % 3 == 0) + x for (i, x) in enumerate(str(n)[::-1])))[::-1].replace('-,', '-').strip(',')


import re

def number_format(n):
	n = str(n)[::-1]
	g = re.split('(\\d{3})', n)
	g = list(filter(None, g))
	if g[-1] == '-':
		return g[-1] + ','.join(g[:-1])[::-1]
	return ','.join(g)[::-1]


def number_format(n):
	s = f'{n:,}'
	return s


def number_format(n):
	n = [x for x in str(n)]
	if n[0] == '-':
		n = n[1:]
		result = '-'
	else:
		result = ''
	end = []
	while len(n) > 3:
		end.append(''.join([','] + n[-3:]))
		n = n[:-3]
	end.append(''.join(n))
	for x in end:
		result += end[-1]
		end = end[:-1]
	return result


def number_format(n):
	return f'{n:,.3f}'.rstrip('0').rstrip('.')