On 21/11/15 12:05, Aadesh Shrestha wrote:
> is there a better solution than this which requires less code?
Yes, several. But less lines of code is not necessarily better.
it might take more memory, or run slower. So better is a relative term
and you always need to think about what you mean by better.
That having been said...
> def square_digits(num):
> arr = []
> list = []
Never use the name of a built in type as a variable name.
You will lose the ability to access the original type name.
> while(num !=0):
> temp = num%10
> num = num /10
You can use the divmod() function to combine these two lines:
num,temp = divmod(num,10)
> arr.insert(0, temp)
> for i in arr:
Rather than inserting each digit into the array then looping
over the array squaring everything, why not insert the
squared number directly into the array and avoid the
second loop?
> list.append( i*i)
> str1 = ''.join(str(e) for e in list)
> return int(str1)
I assume this does what you want? For example if num is 47
The return value is 1649.
If so you could again save a step by converting the square into a string
before inserting into the array:
arr.insert(0, str(temp**2))
return int(''.join(arr) )
So the final code would be:
def square_digits(num):
arr = []
while(num !=0):
num,temp = divmod(num,10)
arr.insert(0, str(temp**2))
return int(''.join(arr))
HTH
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos