##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.*

##Pattern:

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

pattern(4):

    4
    43
    432
    4321
    
pattern(6):
 
    6
    65
    654
    6543
    65432
    654321



```Note: There are no blank spaces```

```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(i) for i in range(n, j, -1)) for j in range(n - 1, -1, -1))

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

def pattern(n):
    full = [str(x) for x in range(n, 0, -1)]
    take = lambda x: ''.join(full[:x])
    return '\n'.join((take(x) for x in range(1, n + 1)))

pattern = lambda n: '\n'.join(''.join(str(n - j) for j in range(i + 1)) for i in range(n))

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


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

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

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

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