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recursive solution in Clear category for The Secret Room by rafal.pawlowski
def toWord(num):
d = { 0 : 'zero', 1 : 'one', 2 : 'two', 3 : 'three', 4 : 'four', 5 : 'five',
6 : 'six', 7 : 'seven', 8 : 'eight', 9 : 'nine', 10 : 'ten',
11 : 'eleven', 12 : 'twelve', 13 : 'thirteen', 14 : 'fourteen',
15 : 'fifteen', 16 : 'sixteen', 17 : 'seventeen', 18 : 'eighteen',
19 : 'nineteen', 20 : 'twenty',
30 : 'thirty', 40 : 'forty', 50 : 'fifty', 60 : 'sixty',
70 : 'seventy', 80 : 'eighty', 90 : 'ninety' }
if (num < 20):
return d[num]
elif (num < 100):
if num % 10 == 0: return d[num]
else: return d[num // 10 * 10] + ' ' + d[num % 10]
elif (num < 1000):
if num % 100 == 0: return d[num // 100] + ' hundred'
else: return d[num // 100] + ' hundred ' + toWord(num % 100)
elif num == 1000:
return 'one thousand'
def secret_room(num):
lst=[]
for i in range(1,num+1,1):
lst.append(toWord(i))
if i == num:
find = lst[-1]
return sorted(lst).index(find)+1
Feb. 18, 2019
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