Published: Oct 20, 2022
Introduction
The problems, “Integer to Roman” and “Roman to Integer,” are well-known pair. Both, tricky conversions are one less to the next order. In this case, the problem is a conversion from integer to string. Defining the mapping is a key to solve the problem. The keys include one less numbers such that 900, 400, 90 … etc as well.
Problem Description
Roman numerals are represented by seven different symbols:
I, V, X, L, C, D and M
.
Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 For example,
2
is written asII
in Roman numeral, just two one’s added together.12
is written asXII
, which is simplyX + II
. The number 27 is written asXXVII
, which isXX + V + II
.Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not
IIII
. Instead, the number four is written asIV
. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written asIX
. There are six instances where subtraction is used:
I
can be placed beforeV
(5) andX
(10) to make 4 and 9.X
can be placed beforeL
(50) andC
(100) to make 40 and 90.C
can be placed beforeD
(500) andM
(1000) to make 400 and 900.Given an integer, convert it to a roman numeral.
Constraints:
1 <= num <= 3999
Examples
Example 1
Input: num = 3
Output: "III"
Explanation: 3 is represented as 3 ones.
Example 2
Input: num = 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.
Example 3
Input: num = 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
Analysis
The solution here defines the integer to string mapping. To handle one less to the next order numbers such as 900 or 400, the mapping includes those mappings as well. Then, divide the number by a key from bigger to smaller. In the next iteration, the number is a reminder. The quotient times mapped string should be added to the result, which gives us the answer.
Solution
class IntegerToRoman:
def intToRoman(self, num: int) -> str:
i2s = {
1000: 'M',
900: 'CM',
500: 'D',
400: 'CD',
100: 'C',
90: 'XC',
50: 'L',
40: 'XL',
10: 'X',
9: 'IX',
5: 'V',
4: 'IV',
1: 'I',
}
result = ''
for o in sorted(i2s.keys(), reverse=True):
q, num = divmod(num, o)
result += i2s[o] * q
return result
Complexities
- Time:
O(n)
– n: the number of keys in the mapping table - Space:
O(1)