Finish the solution so that it takes an input `n` (integer) and returns a string that is the decimal representation of the number grouped by commas after every 3 digits.

Assume: `0 <= n < 2147483647`

## Examples

```
       1  ->           "1"
      10  ->          "10"
     100  ->         "100"
    1000  ->       "1,000"
   10000  ->      "10,000"
  100000  ->     "100,000"
 1000000  ->   "1,000,000"
35235235  ->  "35,235,235"
```

Difficulty: introductory

Starter code:

def group_by_commas(n):
	

Solutions:
def group_by_commas(n):
    return '{:,}'.format(n)

group_by_commas="{:,}".format


def group_by_commas(n):
    return f'{n:,}'

group_by_commas = lambda n:'{:,}'.format(n)

def group_by_commas(n):
    n = [i for i in str(n)]
    for i in range(len(n) -4, -1, -3):
        n.insert(i+1, ',')
    return ''.join(n)


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

group_by_commas = '{:,}'.format

def group_by_commas(n):
    if n < 0:
        if n >= -999:
            return str(n)
    elif n <= 999:
        return str(n)
    return format(n, ',')

def group_by_commas(n):
    # reverse number as string
    rev = str(n)[::-1]
    # break into groups of 3
    groups = [rev[i:i+3] for i in range(0, len(rev), 3)]
    # insert commas and reverse it
    return ','.join(groups)[::-1]

def group_by_commas(price):
    return '{:,}'.format(price)

import locale
def group_by_commas(n):
    #your code here
    return format(n, ',d')