##Task:

You have to write a function **pattern** which creates the following pattern upto n number of rows. 

* If the Argument is 0 or a Negative Integer then it should return "" i.e. empty string.

* If any odd number is passed as argument then the pattern should last upto the largest even number which is smaller than the passed odd number.

* If the argument is 1 then also it should return "".

##Examples:

pattern(8):

    22
    4444
    666666
    88888888
    
pattern(5):
 
    22
    4444



```Note: There are no spaces in the pattern```

```Hint: Use \n in string to jump to next line```

Difficulty: introductory

Starter code:

def pattern(n):
	

Solutions:
def pattern(n):
    return '\n'.join([str(i) * i for i in range(2, n + 1, 2)])


def pattern(n):
    return "\n".join(str(i) * i for i in range(2, n + 1, 2))

def pattern(n):
    # Happy Coding ^_^
    result = []
    
    #If it's one return nothing
    if n == 1:
        return ''

    #just go through n printing the not odd numbers 
    for i in range(n+1):
           if i%2 == 0:
               result.append(str(i)*i + '\n')
            
    return ''.join(result).strip()


def pattern(n):
    if n % 2 != 0:
        n = n - 1
    return "" if n <= 1 else "\n".join([(str(i) * i) for i in range(1, n + 1) if i % 2 == 0])

def pattern(n):
    return '\n'.join(str(i) * i for i in range(2, n//2*2 + 1, 2))

pattern = lambda n: "\n".join([str(x) * x for x in range(2, n + 1, 2)])

def pattern(n):
    # Happy Coding ^_^
    return '\n'.join([str(a) * a for a in range(2, n + 1, 2)])

def pattern(n):
    return '\n'.join(str(x) * x for x in range(2, n + 1, 2))

pattern=lambda n:'\n'.join(i*str(i)for i in range(2,n+1,2))