# A Space Efficient Solution

• This solution reflects the rules of roman numbers and uses it to construct a roman number. You don't need a very large dictionary or hash map. All you have to store is 1, 5 for every digit. And it uses a loop to decide which digit it is at, instead of hard coding. Again, this solution might be time consuming(128ms) but space efficient.

``````class Solution(object):
def intToRoman(self, num):
d = {1: {1:'I', 5:'V'},
10: {1:'X', 5:'L'},
100: {1:'C', 5:'D'},
1000: {1:'M',},}
roman = ""
length = len(str(num))
for i in range(length, -1, -1):
digit = 10**i
value = num/digit
if value == 0:
continue
elif value == 4:
roman += d[digit][1] + d[digit][5]
elif value == 9:
roman += d[digit][1] + d[digit*10][1]
elif value >0 and value < 4:
roman += value*d[digit][1]
elif value >= 5 and value<9:
roman += d[digit][5] + (value-5)*d[digit][1]
else:
print 'wrong input'
return
num %= digit
return roman
``````

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