I am new to python and I was practicing my coding skills with Hacker rank. I came across a question where I have to print numbers in a series one after another without adding any new line or white space in between them and most important is I cannot use the string method to get desired output.
For example if input is
3
then output should be
123
How can do it without string?
Please guide me...
Thanks in advance
You can use end keyword in print() method to avoid spaces:
num = int(input("Enter number: "))
for i in range(1, num + 1):
print(i, end='')
You may create a custom function using range() along with the usage of math.log10() as:
import math
def get_my_number(num):
my_num = 0
for i in range(1, num+1):
digits = int(math.log10(i))+1 # count number of digits in number
my_num = my_num * (10 ** digits) + i
return my_num
Sample run:
>>> get_my_number(3)
123
>>> get_my_number(13) # For number greater than 10
12345678910111213
This way you will get the actual value and you won't be just printing it on the console
for statement below takes elements starting from 1 to n.
printing without new line requires and addition in print command.
end = "" would print 123 for example.
end = " " with 1 space between would print 1 2 3 and so on.
if __name__ == '__main__':
n = int(input())
for i in range(1,n+1):
print (i, end = "")
n=int(input())
for i in range(n):
print(i+1, end="")
If you want to print n natural number without using string then you can try this code:
n=int(input())
print(*range(1,n+1),sep="")
simple input: 5
output will be: 12345
The following copy integer values to array ls before writing to a buffer pr and making a single write call (print):
n = int(input())
ls = []
pr = ("")
for i in range(1,n+1):
ls.append(i)
for i in range(len(ls)):
pr += "{}".format(ls[i])
print(pr)
