### Task:

You have to write a function `pattern` which creates the following pattern (See Examples) upto desired number of rows.

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

### Examples:

`pattern(9)`:

    123456789
    234567891
    345678912
    456789123
    567891234
    678912345
    789123456
    891234567
    912345678

`pattern(5)`:

    12345
    23451
    34512
    45123
    51234

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(''.join(str((x+y)%n+1) for y in range(n)) for x in range(n))

def pattern(n):
    nums = list(range(1,n+1))
    result = ''
    result += ''.join(str(n) for n in nums)
    
    for i in range(1, n):
        result += '\n'
        nums.append(nums.pop(0))
        result += ''.join(str(n) for n in nums)
        
    return result
    
    


from collections import deque

def pattern(n):
    result = []
    line = deque([str(num) for num in range(1, n + 1)])
    for i in range(n):
        result.append("".join(line))
        line.rotate(-1)
    return "\n".join(result)

from collections import deque

def pattern(n):
    A = deque([str(i+1) for i in range(n)])
    result = []
    for i in range(len(A)):
        print(A)
        result.append(''.join(A))
        A.rotate(-1)
    return '\n'.join(result)


def pattern(s):
    return "\n".join(["".join((str(x + y) if x + y <= s else str(x + y - s * ((x + y) // s)) for y in range(s))) for x in range(1, s + 1)])

from itertools import chain

def pattern(n):
    line = list(i + 1 for i in range(n))   
    return "\n".join("".join(str(x) for x in chain(line[i:], line[:i])) for i in range(n))


from itertools import chain


def pattern(n):
    xs = ''.join(str(i) for i in range(1, n+1))
    return '\n'.join(''.join(str(j) for j in chain(range(i, n+1), range(1, i))) for i in range(1, n+1))

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