Project Euler 17 - Number letter counts

'''
If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters
used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?
NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115
(one hundred and fifteen) contains 20 letters.
The use of "and" when writing out numbers is in compliance with British usage.
'''

units = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"]
string2 = ["eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"]
decs = ["ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]
thousand = "onethousand"
AND = 3
HUNDRED = 7

# 1. From 1 to 9
sum1 = sum(len(units[k]) for k in range(len(units)))
print("sum1 = " + str(sum1))

# 2. Only ten
sum2 = len(decs[0])
print("sum2 = " + str(sum2))

# 3. From 11 to 19
sum3 = sum(len(string2[k]) for k in range(len(string2)))
print("sum3 = " + str(sum3))

# 4. From 20 to 99
sum4 = 0
for i in range(1, len(decs)):
    sum4 += 10*len(decs[i])             # I use 10 times the words "twenty", "thirty", "forty", ......
for j in range(len(units)):
    sum4 += 8*len(units[j])             # I use 8 times the words "one", "two", "three" when counting from 20 to 99
print("sum4 = " + str(sum4))            # Example: 21, 31, 41, 51, 61, 71, 81, 91 ----> I use 8 times the "one"

# SUM1TO99
sum1to99 = sum1 + sum2 + sum3 + sum4
print("sum1to99 = " + str(sum1to99))

# 5. From 100 to 999, there are exactly 900 numbers
# 5-a: Counting "and"
counterAND = (900 - 9) * AND
# 900 = all the numbers from 100 to 999
# 9 = the numbers from 100 to 999 where I dont use "the and-word" ---> 100, 200, ...., 800, 900

# 5-b: Counting "hundred"
counterHUNDRED = 900 * HUNDRED

# 5-c: Now, its time to count the real numbers (without "and" and "hundred")
# I will use 9 more times the sum1to99
counter1 = 9 * sum1to99

# 5-d: Last task is to count the "one", "two", .... before the word "HUNDRED"
counter2 = 0
for i in range(len(units)):
    counter2 += 100*len(units[i])       # "One" is used 100 times from 100 to 199, "two" is used 100 times from 200 to 299

sum100to999 = counter1 + counter2 + counterAND + counterHUNDRED
print("sum100to999 = " + str(sum100to999))

# 6. "1000"
thous = len(thousand)
print("thous = " + str(thous))

# MAIN FUNCTION
sumALL = sum1to99 + sum100to999 + thous
print()
print("SUM = " + str(sumALL))