###Task:

You have to write a function `pattern` which creates the following pattern (see examples) up to the desired number of rows.

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

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

###Examples:

pattern(9):

    1
    333
    55555
    7777777
    999999999

pattern(6):

    1
    333
    55555

```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(1,n+1,2))

def pattern(n):
    string = ""
    a = n;
    if a % 2 == 0:
        a -= 1;
    for x in range (1, a + 1):
        if x % 2 != 0:
            string += str(x) * x
            
            if x != a:
                string += "\n"
    return string
        
    # Happy Coding ^_^


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

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

def pattern(n):
    x = []
    if n <= 0:
        return ""
    else:
        for a in range (1, n+1, 2): 
            x.append(str(a)*a)
    return '\n'.join(x)


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

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

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


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