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
```

Difficulty: introductory

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

import re
def triple_double(num1, num2):
    #code me ^^
    r3 = re.compile(r'(\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