##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```

Source: codewars | Difficulty: EASY
Tags: String algorithms

Fundamentals

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):
	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):
	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)])