# 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?

Python:
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))

berkeman

Gold Member
Does it ever pronounce "hundred"?

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

Last edited:
Gold Member
OK, I see what you're doing. Hundred is "+7".
Eleven is "6".

ergospherical
Homework Helper
Gold Member
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

Python:
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))

Last edited:
ergospherical
Gold Member
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
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.

Homework Helper
Gold Member
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.

Last edited:
Filip Larsen and sysprog
Gold Member
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.

pbuk
Gold Member
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