LeetCode 12. Integer to Roman

October 28, 2024

I started feeling down about my past interview performance. Rather than continue ruminating, I decided to continue my LeetCode grind tonight. My goal is to complete the Top 150 Interview Questions before the end of the year.

Solution

Copy
1class Solution:
2    def intToRoman(self, num: int) -> str:
3        digitCount = int((math.log10(num) + 1) // 1)
4        def getLeftmostDigit():
5            return num // (10 ** (digitCount - 1)) % 10
6        romanForm = ""
7        if digitCount == 4:
8            digit = getLeftmostDigit()
9            for i in range(digit): romanForm += "M" 
10            digitCount -= 1
11        if digitCount == 3:
12            digit = getLeftmostDigit()
13            if digit in [4, 9]:
14                romanForm += "CD" if digit == 4 else "CM"
15                digit -= digit
16            elif digit >= 5:
17                romanForm += "D"
18                digit -= 5
19            for i in range(digit): romanForm += "C" 
20            digitCount -= 1
21        if digitCount == 2:
22            digit = getLeftmostDigit()
23            if digit in [4, 9]:
24                romanForm += "XL" if digit == 4 else "XC"
25                digit -= digit
26            elif digit >= 5:
27                romanForm += "L"
28                digit -= 5
29            for i in range(digit): romanForm += "X"
30            digitCount -= 1
31        if digitCount == 1:
32            digit = getLeftmostDigit()
33            if digit in [4, 9]:
34                romanForm += "IV" if digit == 4 else "IX"
35                digit -= digit
36            elif digit >= 5:
37                romanForm += "V"
38                digit -= 5
39            for i in range(digit): romanForm += "I" 
40
41        return romanForm

O(n) = 1

Reflection

Next time, I would like to reduce the code complexity by abstracting the strings into a dictionary. With the dictionary, I could loop for digitCount iterations and select the appropriate value from the dictionary. This would remove the repeated code, thereby making it easier to read and maintain.

LeetCode 74. Search a 2D Matrix

LeetCode 6. Zigzag Conversion