4 ms·

What's the elegant mathematical way to do this? I learned engineering school programming (read: programming with no standards or formal theory whatsoever) and t

What's the elegant mathematical way to do this? I learned engineering school programming (read: programming with no standards or formal theory whatsoever) and the only thing that's coming to mind is num2str/split/multiply/concatenate/str2num, but that seems too easy for a real programming interview.

linuxftw 6y ago
You can detect each digit like this: For example, take 999 (N). Using only math, how can I determine the value of each digit? (be sure to take abs(N) and record value of N/abs(N) for later) res = N - (N - 1). Digit = res + 1; If Digit == 10, Digit = 0. So, given that function to find each digit, you can build a loop using 10^i to detect each digit (subtracting the sum of all preceding digits and dividing by 10^i to trim the lower order digits you've already resolved). We'll need a place to store each digit, such as an array. Of course, this assumes integers in base 10. If they are another base (8, 16), replace the 10^i above with the base. It's also ambiguous as to if the function should just return a concatenated version of the results or some possibly a sum. In any case, using a cast to string would be way easier.
- quacked 6y ago
  Thanks!
pmiller2 6y ago
If you want to be tricky about it, you can do it in one line in Python. Let d be your original integer. Then [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))] will be a list containing the base 10 digits of d as integers. Technically, you need to do from math import log10 to make this work, but you can fit that on the same line with a little hacking around using __import__, such as this quick and very dirty attempt: [ [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))] for log10 in [__import__("math", fromlist=["log10"]).log10]][0] Edit: Just to carry this a little further, if we interpret "a new integer that's the result of each of the original integer's digits squared" as doing something like this: 129 -> 1481 (i.e. 9 gets replaced with 81), then we can do it in one line like this: int(''.join([str(x**2) for x in [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))]]))
- quacked 6y ago
  Ooh, nice.
zadkey 6y ago
Well the mathematical way of doing this involves doing a Mod 10 to pull off a digit and then incrementing a number that multiples the mod 10. I've also seen incrementing an exponent. And I hear from more math oriented people that there are other ways.

linuxftw 6y ago

You can detect each digit like this: For example, take 999 (N). Using only math, how can I determine the value of each digit? (be sure to take abs(N) and record value of N/abs(N) for later) res = N - (N - 1). Digit = res + 1; If Digit == 10, Digit = 0. So, given that function to find each digit, you can build a loop using 10^i to detect each digit (subtracting the sum of all preceding digits and dividing by 10^i to trim the lower order digits you've already resolved). We'll need a place to store each digit, such as an array. Of course, this assumes integers in base 10. If they are another base (8, 16), replace the 10^i above with the base. It's also ambiguous as to if the function should just return a concatenated version of the results or some possibly a sum. In any case, using a cast to string would be way easier.

quacked 6y ago
Thanks!

pmiller2 6y ago

If you want to be tricky about it, you can do it in one line in Python. Let d be your original integer. Then [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))] will be a list containing the base 10 digits of d as integers. Technically, you need to do from math import log10 to make this work, but you can fit that on the same line with a little hacking around using __import__, such as this quick and very dirty attempt: [ [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))] for log10 in [__import__("math", fromlist=["log10"]).log10]][0] Edit: Just to carry this a little further, if we interpret "a new integer that's the result of each of the original integer's digits squared" as doing something like this: 129 -> 1481 (i.e. 9 gets replaced with 81), then we can do it in one line like this: int(''.join([str(x**2) for x in [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))]]))

quacked 6y ago
Ooh, nice.

zadkey 6y ago

Well the mathematical way of doing this involves doing a Mod 10 to pull off a digit and then incrementing a number that multiples the mod 10. I've also seen incrementing an exponent. And I hear from more math oriented people that there are other ways.