###Task:

You have to write a function **pattern** which creates the following Pattern(See Examples) upto n(parameter) number of rows.

####Rules/Note:
* If the Argument is 0 or a Negative Integer then it should return "" i.e. empty string.
* All the lines in the pattern have same length i.e equal to the number of characters in the last line.
* Range of n is (-∞,100]

###Examples:

pattern(5):

        1    
       121   
      12321  
     1234321 
    123454321


pattern(10):

             1         
            121        
           12321       
          1234321      
         123454321     
        12345654321    
       1234567654321   
      123456787654321  
     12345678987654321 
    1234567890987654321

pattern(15):

                  1              
                 121             
                12321            
               1234321           
              123454321          
             12345654321         
            1234567654321        
           123456787654321       
          12345678987654321      
         1234567890987654321     
        123456789010987654321    
       12345678901210987654321   
      1234567890123210987654321  
     123456789012343210987654321 
    12345678901234543210987654321

pattern(20):

                       1                   
                      121                  
                     12321                 
                    1234321                
                   123454321               
                  12345654321              
                 1234567654321             
                123456787654321            
               12345678987654321           
              1234567890987654321          
             123456789010987654321         
            12345678901210987654321        
           1234567890123210987654321       
          123456789012343210987654321      
         12345678901234543210987654321     
        1234567890123456543210987654321    
       123456789012345676543210987654321   
      12345678901234567876543210987654321  
     1234567890123456789876543210987654321 
    123456789012345678909876543210987654321

###Amazing Fact: 


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

Difficulty: introductory

Starter code:

def pattern(n):
	

Solutions:
def pattern(n):
    output = []
    for i in range (1, n + 1):
        wing = ' ' * (n - i) + ''.join(str(d % 10) for d in range(1, i))
        output.append(wing + str(i % 10) + wing[::-1])
    return '\n'.join(output)

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

def pattern(n):
    l = []
    for a, b in zip(range(1, n+2), range(n+1)):
        res = ''.join(str(x%10) for x in range(1,a))
        spaces = n-len(res)
        if res: l.append(spaces * ' ' + res+res[-2::-1] + ' ' * (spaces))
    return '\n'.join(l)

from itertools import cycle, islice

def f(n):
    it = cycle('1234567890')
    xs = ''.join(islice(it, 0, n-1))
    return xs + next(it) + xs[::-1]
    
def pattern(n):
    m = n * 2 - 1
    return '\n'.join(
        '{:^{}}'.format(f(i), m)
        for i in range(1, n+1)
    )

def pattern(n):
    t = []
    for i in range(1,n+1):
        t.append( (n-i)*" " + ls(i)+ +(n-i)*" ")
    return "\n".join(t)
    
def ls(k):
    ls = [str(j%10) for j in range(1,k)]
    return "".join(ls) + str(k%10) + "".join(ls[::-1])

def pattern(n):
  mkP = lambda y: str(y % 10)
  return "\n".join([" " * (n - x - 1) + "".join(map(mkP, range(1, x + 1))) + "".join(map(mkP, range(x + 1, 0, -1))) + " " * (n - x - 1) for x in range(n)])