# GeetCode Hub

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.

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`

public string IntToRoman(int num) { string result = string.Empty; string[] baseFive = { "V", "L", "D"}; string[] baseNum = { "I", "X", "C", "M" }; int pos = 0; while (num > 0) { int pop = num % 10; num /= 10; if (pop == 5) { result = baseFive[pos] + result; } else if (pop > 5) { if (pop == 6) { result = (baseFive[pos] + baseNum[pos] + result); } else if (pop == 9) { result = (baseNum[pos] + baseNum[pos + 1] + result); } else { var temp = string.Empty; var reminder = pop % 5; for (int i = 0; i < reminder; i++) { temp += baseNum[pos]; } result = (baseFive[pos] + temp + result); } } else { if (pop == 4) { result = (baseNum[pos] + baseFive[pos] + result); } else { var temp = string.Empty; var reminder = pop % 5; for (int i = 0; i < reminder; i++) { temp += baseNum[pos]; } result = (temp + result); } } pos++; } return result; }

This question can easily be solved if we know the additive and subtractive nature of Roman Numbers, and calculate the values of units, tens, hundreds, and thousands place of the digit.

Time Complexity: O(1)

Space Complexity: O(1)

Easy