Counting letters in a number between 1 and 1000

[ergospherical]

Excluding spaces and hyphens, how many letters are there in a given number between 1 and 1000 (inclusive)? How can I make my algorithm more efficient?

dig_lens = [4,3,3,5,4,4,3,5,5,4]
tens_lens = [0,3,6,6,6,5,5,7,6,6]
teens_lens = [0,6,6,8,8,7,7,9,8,8]

def num_letters(n):
    nlist = []
    while(n>=1):
        nlist.append(n%10)
        n//=10
    length = len(nlist)

    if(n==1000):
        return 11

    count = 0
    if(length == 1):
        count = count + dig_lens[n]

    if(length == 2):
        if(nlist[1]==1 and nlist[0]>0):
            count = count + teens_lens[nlist[0]]
        else:
            count = count + tens_lens[nlist[1]] + dig_lens[nlist[0]]

    if(length == 3):
        count = count + dig_lens[nlist[2]] + 7
        if(nlist[1]==0 and nlist[0]>0):
            count = count + 3 + dig_lens[nlist[0]]
        elif(nlist[1]==1 and nlist[0]>0):
            count = 3 + count + teens_lens[nlist[0]]
        else:
            count = 3 + count + tens_lens[nlist[1]] + dig_lens[nlist[0]]

    return count

##e.g.##
print(num_letters(816))

 

[DaveC426913]

Does it ever pronounce "hundred"?

 

[ergospherical]

it does pronounce hundred - that's accounted for in the first line of the length=3 case!

 

[DaveC426913]

OK, I see what you're doing. Hundred is "+7".
Eleven is "6".

 

[pbuk]

If length == 1 then you do not need to test if length == 2 (use elif) Edit: or if you are done then return the answer.

You could use count += instead of count = count + ... but if count is 0 then you don't need to include it in any sums anyway. Edit: you hardly need an intermediate count, if you know the answer just return it.

Edit: use more spaces and use them consistently

dig_lens = [4, 3, 3, 5, 4, 4, 3, 5, 5, 4]
tens_lens = [0, 3, 6, 6, 6, 5, 5, 7, 6, 6]
teens_lens = [0, 6, 6, 8, 8, 7, 7, 9, 8, 8]def num_letters(n):
    nlist = []
    while n >= 1:
        nlist.append(n % 10)
        n //= 10
    length = len(nlist)

    if n == 1000:
        return 11

    if length == 1:
        return dig_lens[n]

    if length == 2:
        if nlist[1] == 1 and nlist[0] > 0:
            return teens_lens[nlist[0]]
        else:
            return tens_lens[nlist[1]] + dig_lens[nlist[0]]

    if length == 3:
        count = dig_lens[nlist[2]] + 7
        if nlist[1] == 0 and nlist[0] > 0:
            return count + 3 + dig_lens[nlist[0]]
        if nlist[1] == 1 and nlist[0] > 0:
            return count + 3 + teens_lens[nlist[0]]
        else:
            return count + 3 + tens_lens[nlist[1]] + dig_lens[nlist[0]]

    raise RuntimeError("Can't handle more than 1,000")##e.g.##
print(num_letters(10000))

 

[DaveC426913]

Notice that the size of the number is an input, which means a user might not cooperate with the constraints. How robust is your program if the input parameter were 1,001? Or 10,000?

See how flexible and universal and bulletproof you can make it.

 

[ergospherical]

Edit: you hardly need an intermediate count, if you know the answer just return it.

Yeah, I started thinking I was going to try and do some clever recursive algorithm, but realized it'd just be easier to split it into cases - but didn't get rid of the intermediate count.

 

[pbuk]

Sorry, I have edited and re-edited my code, bad habit. It is prettier now. Should have just passed it through the Black parser straight off, my typing is a bit haphazard after a Friday night in the pub for the first time in 18 months.

 

[DaveC426913]

Comments are good, for those who are trying to review your code.
I had a bunch of false starts before I got a handle on what it was doing.

 

[Filip Larsen]

if n == 1000:
return 11

Looks like that test should use the original value of n to trigger properly?

 

[ergospherical]

Looks like that test should use the original value of n to trigger properly?

whoopsie, yeah that and the length=1 case ought to be before the loop