Date and Time: Sep 11, 2024, 0:03 (EST)
Link: https://leetcode.com/problems/integer-to-roman/
| Date | Stopwatch | Y/N | Feedback |
|---|---|---|---|
| Jul 6, 2025 | 20m 59s | Y | Watched tutorial |
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
1 <= num <= 3999
To handle special subtractive form, we just add them into the Roman-Value lists or hashmap. Then, we want to eliminate from the highest decimal place values to the lowest, so we traverse every pair of Roman-Value table from the biggest to the smallest, we use // to know if this current val can eliminate the highest decimal place values and how many this Roman symbol we need to add to res string, then update value of num by num %= val.
class Solution:
def intToRoman(self, num: int) -> str:
# 1. Map symbol with value
# 2. Always start with the greatest value to divide then update num and append roman to ans""
# TC: O(1), SC: O(1)
roman = [['M', 1000], ['CM', 900], ['D', 500], ['CD', 400], ['C', 100], ['XC', 90], ['L', 50], ['XL', 40], ['X', 10], ['IX', 9], ['V', 5], ['IV', 4], ['I', 1]]
ans = ""
for sym, digit in roman:
if num // digit:
# Append sym with times into ans, and update num
ans += sym * (num // digit)
num %= digit
return ansTime Complexity:
Space Complexity: