Integer to Roman

Published: Oct 20, 2022

Medium Hash Table String


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 as II in Roman numeral, just two one’s added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + 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 as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

  • I can be placed before V (5) and X (10) to make 4 and 9.
  • X can be placed before L (50) and C (100) to make 40 and 90.
  • C can be placed before D (500) and M (1000) to make 400 and 900.

Given an integer, convert it to a roman numeral.


  • 1 <= num <= 3999


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.


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.


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


  • Time: O(n) – n: the number of keys in the mapping table
  • Space: O(1)