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 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.
Example 1:
Input: num = 3
Output: "III"
Example 2:
Input: num = 4
Output: "IV"
Example 3:
Input: num = 9
Output: "IX"
Example 4:
Input: num = 58
Output: "LVIII"
Explanation: L = 50, V = 5, III = 3.
Example 5:
Input: num = 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
Constraints:
1 <= num <= 3999
Solution
First of all, we're going to create a map
with all possible RomanInteger pairs.
The algorithm is going through this map
from biggest to smallest starting from M
to I
and keeping track of a result
variable.
The main algorithm is using the division num
by map[key]
and modulus operator. The division tells us how many particular symbols do we need to repeat. And the modulus operator helps us to change the num
.
The easiest way to understand this solution is to look at an example :) Let's imagine we need to convert 2439
:
In this solution, we can do some improvements:

We can return the result if the num equals
0

We can check whether the division equals
0
or not, if not we don't need to repeat anything
The time complexity is going to be O(n)
, there is n
is the number of Roman numeral characters. And the space complexity is O(1)
.
Hope it was useful for you!
Thanks for reading!