Seven different symbols represent Roman numerals with the following values:
| Symbol | Value |
|---|---|
| I | 1 |
| V | 5 |
| X | 10 |
| L | 50 |
| C | 100 |
| D | 500 |
| M | 1000 |
Roman numerals are formed by appending the conversions of decimal place values from highest to lowest. Converting a decimal place value into a Roman numeral has the following rules:
- If the value does not start with 4 or 9, select the symbol of the maximal value that can be subtracted from the input, append that symbol to the result, subtract its value, and convert the remainder to a Roman numeral.
- If the value starts with 4 or 9 use the subtractive form representing one symbol subtracted from the following symbol, for example, 4 is 1 (
I) less than 5 (V):IVand 9 is 1 (I) less than 10 (X):IX. Only the following subtractive forms are used: 4 (IV), 9 (IX), 40 (XL), 90 (XC), 400 (CD) and 900 (CM). - Only powers of 10 (
I,X,C,M) can be appended consecutively at most 3 times to represent multiples of 10. You cannot append 5 (V), 50 (L), or 500 (D) multiple times. If you need to append a symbol 4 times use the subtractive form.
Given an integer, convert it to a Roman numeral.
Example 1:
Input: num = 3749
Output: "MMMDCCXLIX"
Explanation:
3000 = MMM as 1000 (M) + 1000 (M) + 1000 (M) 700 = DCC as 500 (D) + 100 (C) + 100 (C) 40 = XL as 10 (X) less of 50 (L) 9 = IX as 1 (I) less of 10 (X) Note: 49 is not 1 (I) less of 50 (L) because the conversion is based on decimal places
Example 2:
Input: num = 58
Output: "LVIII"
Explanation:
50 = L 8 = VIII
Example 3:
Input: num = 1994
Output: "MCMXCIV"
Explanation:
1000 = M 900 = CM 90 = XC 4 = IV
Constraints:
1 <= num <= 3999
Approach 01: Greedy with Symbols
class Solution {
public:
string intToRoman(int num) {
vector> syms = {
{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"}
};
string res = "";
for (auto& sym : syms) {
while (num >= sym.first) {
res += sym.second;
num -= sym.first;
}
}
return res;
}
};
Time Complexity:
- O(1): Since the input range is limited to 3999, the number of iterations is constant.
Space Complexity:
- O(1): Constant space for the symbol list.
class Solution:
def intToRoman(self, num: int) -> str:
syms = [
(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")
]
res = []
for val, sym in syms:
if num == 0: break
count, num = divmod(num, val)
res.append(sym * count)
return "".join(res)
Time Complexity:
- O(1): Since the input range is limited to 3999, the number of iterations is constant.
Space Complexity:
- O(1): Constant space for the symbol list.
class Solution {
public String intToRoman(int num) {
int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
String[] symbols = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"};
StringBuilder res = new StringBuilder();
for (int i = 0; i < values.length && num > 0; i++) {
while (num >= values[i]) {
num -= values[i];
res.append(symbols[i]);
}
}
return res.toString();
}
}
Time Complexity:
- O(1): Since the input range is limited to 3999, the number of iterations is constant.
Space Complexity:
- O(1): Constant space for the symbol list.