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
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