##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.*
   
##Examples:

pattern(4):

    1234
    234
    34
    4
    
pattern(6):
 
    123456
    23456
    3456
    456
    56
    6



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

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

Difficulty: introductory

Starter code:

def pattern(n):
	

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

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

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


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


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

def pattern(n):
    output = ""
    for i in range(1, n+1):
        for j in range(i, n+1):
            output += str(j)
        if i < n:
            output += '\n'
    return output

def pattern(n):
    # Happy Coding ^_^
    a = ""
    lst=[]
    result=""
    for i in range(1,n+1):
        a=str(i)
        
        for m in range(i+1,n+1):
            a += str(m)
        lst.append(a)
    for i in lst:
        result += i+'\n'
    return result[:-1]

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