Write a function 
```javascript 
tripledouble(num1,num2)
```
```python
triple_double(num1, num2)
```
which takes numbers `num1` and `num2` and returns `1` if there is a straight triple of a number at any place in `num1` and also a straight double of the **same** number in `num2`.

If this isn't the case, return `0`

## Examples

```python
triple_double(451999277, 41177722899) == 1
# num1 has straight triple 999s and num2 has straight double 99s

triple_double(1222345, 12345) == 0
# num1 has straight triple 2s but num2 has only a single 2

triple_double(12345, 12345) == 0

triple_double(666789, 12345667) == 1
```

Source: codewars | Difficulty: EASY

Starter code: def triple_double(num1, num2):
	

Solutions:
def triple_double(num1, num2):
	return any([i * 3 in str(num1) and i * 2 in str(num2) for i in '0123456789'])


def triple_double(num1, num2):
	for x in range(10):
		if str(x) * 3 in str(num1):
			if str(x) * 2 in str(num2):
				return 1
	return 0


def triple_double(num1, num2):
	(num1, num2) = (str(num1), str(num2))
	for num in '0123456789':
		if num * 3 in num1 and num * 2 in num2:
			return 1
	return 0


def triple_double(num1, num2):
	for i in range(10):
		if str(i) * 3 in str(num1):
			if str(i) * 2 in str(num2):
				return 1
			else:
				return 0
	return 0


import re

def triple_double(num1, num2):
	str = '{0}|{1}'.format(num1, num2)
	regex = re.compile('(\\d)(\\1){2}.*\\|.*(\\1){2}')
	match = regex.search(str)
	return True if match else False


def triple_double(num1, num2):
	num1 = str(num1)
	num2 = str(num2)
	nums = ['1', '0', '2', '3', '4', '5', '6', '7', '8', '9', '0']
	for i in nums:
		if i * 3 in num1 and i * 2 in num2:
			return 1
	return 0


import re

def triple_double(*args):
	return bool(re.search('(\\d)(\\1){2}.*\\ .*(\\1){2}', '%s %s' % args))


def triple_double(num1, num2):
	DECIMAL_BASE = 10
	tripples = []
	(previous, previous_previous) = (None, None)
	while num1:
		(num1, rem) = divmod(num1, DECIMAL_BASE)
		if rem == previous and previous == previous_previous:
			tripples.append(rem)
		previous_previous = previous
		previous = rem
	while num2:
		(num2, rem) = divmod(num2, DECIMAL_BASE)
		if rem == previous and rem in tripples:
			return 1
		previous = rem
	return 0


import re

def triple_double(num1, num2):
	triples = re.findall('(\\d)\\1{2}', str(num1))
	doubles = re.findall('(\\d)\\1{1}', str(num2))
	return len(set(triples) & set(doubles))


import re

def triple_double(num1, num2):
	r3 = re.compile('(\\d)\\1\\1')
	f = r3.findall(str(num1))
	if len(f) > 0:
		r2 = re.compile(f[0] + '{2}')
		if len(r2.findall(str(num2))):
			return 1
	return 0