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.

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Solution

class Solution:
    def intToRoman(self, num: int) -> str:
        digitCount = int((math.log10(num) + 1) // 1)
        def getLeftmostDigit():
            return num // (10 ** (digitCount - 1)) % 10
        romanForm = ""
        if digitCount == 4:
            digit = getLeftmostDigit()
            for i in range(digit): romanForm += "M" 
            digitCount -= 1
        if digitCount == 3:
            digit = getLeftmostDigit()
            if digit in [4, 9]:
                romanForm += "CD" if digit == 4 else "CM"
                digit -= digit
            elif digit >= 5:
                romanForm += "D"
                digit -= 5
            for i in range(digit): romanForm += "C" 
            digitCount -= 1
        if digitCount == 2:
            digit = getLeftmostDigit()
            if digit in [4, 9]:
                romanForm += "XL" if digit == 4 else "XC"
                digit -= digit
            elif digit >= 5:
                romanForm += "L"
                digit -= 5
            for i in range(digit): romanForm += "X"
            digitCount -= 1
        if digitCount == 1:
            digit = getLeftmostDigit()
            if digit in [4, 9]:
                romanForm += "IV" if digit == 4 else "IX"
                digit -= digit
            elif digit >= 5:
                romanForm += "V"
                digit -= 5
            for i in range(digit): romanForm += "I" 

        return romanForm

O(n) = 1

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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.