## Task:
You have to write a function `pattern` which returns the following Pattern (See Pattern & Examples) upto `n` number of rows. 

* Note: `Returning` the pattern is not the same as `Printing` the pattern.

### Rules/Note:
* If `n < 1` then it should return "" i.e. empty string.
* There are `no whitespaces` in the pattern.

### Pattern:

    (n)(n-1)(n-2)...4321
    (n)(n-1)(n-2)...432
    (n)(n-1)(n-2)...43
    (n)(n-1)(n-2)...4
    ...............
    ..............
    (n)(n-1)(n-2)
    (n)(n-1)
    (n)
    
### Examples:

* pattern(4):

      4321
      432
      43
      4
    
* pattern(11):

      1110987654321
      111098765432
      11109876543
      1110987654
      111098765
      11109876
      1110987
      111098
      11109
      1110
      11


~~~if-not:cfml
* Hint: Use \n in string to jump to next line
~~~
~~~if:cfml
* Hint: Use chr(10) in string to jump to next line
~~~

[List of all my katas]("http://www.codewars.com/users/curious_db97/authored")

Difficulty: introductory

Starter code:

def pattern(n):
	

Solutions:
def pattern(n):
  return "\n".join(["".join([str(y) for y in range(n, x, -1)]) for x in range(n)]);

def pattern(n):
  return '\n'.join([''.join(map(str, list(range(n, x, -1)))) for x in range(n)])


def pattern(n):
    return('\n'.join(''.join(str(i) for i in range(n,j,-1)) for j in range(n)))

def pattern(n):
    r = range(n, 0, -1)
    a = [str(i) for i in r]
    return '\n'.join(''.join(a[:i]) for i in r)


def pattern(n):
    s = ''
    for i in range(n):
        s += str(n)
        for ni in range(n-1,i,-1):
            s += str(ni)
        if n-1 == i:
            pass
        else:
            s += '\n'
    return s

def pattern(n):
    result = []
    for i in range(n, 0, -1):
        string = ""
        for j in range(n, n-i, -1):
             string += str(j)
        result.append(string)
    return "\n".join(result)

def pattern(n):
    return '\n'.join([''.join([str(n-y) for y in range(n-x+1)]) for x in range(1, n + 1)])

def pattern(n):
    return '\n'.join(''.join(map(str, range(n, stop, -1))) for stop in range(n))

pattern = lambda n: "\n".join(["".join([str(n-x) for x in range(0, n-y)]) for y in range(0, n)]);

def pattern(n):
    return '\n'.join("".join(str(n-j) for j in range(0,n-i+1)) for i in range(1, n+1))
    # Happy Coding ^_^