12. Integer to Roman

Hello, reader 👋🏽 ! Welcome to day 20 of the series on Problem Solving. Through this series, I aim to pick up at least one question everyday and share my approach for solving it.

Today, I will be picking up LeetCode's daily challenge problem: 12. Integer to Roman.

🤔 Problem Statement

Given a number, convert it to roman numeral form.
E.g.:
- num = 3 => `"III"
- num = 1994 => "MCMXCIV"

💬 Thought Process

Intuition says that we can convert a number to it's roman numeral by separating the one's, ten's, hundred's and thousand's place.
For e.g. 58 can be split to 50 + 8 and easily be converted to roman form.
This could have been easily done had it not been for the special cases which are also given in the question.
For e.g. 4 in roman is IV and not IIII. Similarly 9 in roman is not VIIII but rather IX.
And the same special cases can be observed in 40, 90, 400, and 900.
Also, we should keep in mind that at numbers 5, 50, and 500, the roman symbol changes in comparison with the previous numbers in 1s, 10s, and 100s places respectively.
So we will also have to check our number against all the possible places where the change happens: 1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, and 1000.
That's what we will do, compare the thousands, hundreds, tens, and ones place with the above numbers.
At any number if the value is less than 0, we move on to the next number and continue until the number is 0.
To better understand, let's walk through an example first. Let's take the number 2947.
Look at the image below to understand the process.
Let's jump into the code for this now.

👩🏽‍💻 Solution

Below is the code for the above discussed approach.

class Solution {
    public String intToRoman(int num) {
        if(num == 0) {
            return "0";
        }

        String s = "";

        String[] mapping = new String[] {
          "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"
        };
        int[] places = new int[] {
            1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1
        };

        for(int i = 0; i<places.length; i++) {
            int place = places[i];
            int digit = num / place;

            if(digit != 0) {
                String symbol = mapping[i];
                s+= getRomanSymbol(digit, symbol);
            }

            num = num % place;
        }

        return s;
    }

    private String getRomanSymbol(int n, String symbol) {
        StringBuilder sb = new StringBuilder("");
        while(n-- > 0) {
            sb.append(symbol);
        }

        return sb.toString();
    }
}

Time Complexity: O(x)
  - x = maximum of number/place
  - while the for loop runs in O(1), the while loop at max runs in `n/place` times, where the symbol at place appears n/place times.
Space Complexity: O(1)
  - We are using constant memory to store the places and their symbol mappings.

There is an O(n) solution to this problem as well, but I won't be sharing it here since it wasn't intuitive for me. You can find that in the discussion forum.

Conclusion

That's a wrap for today's problem. If you liked my explanation then please do drop a like/ comment. Also, please correct me if I've made any mistakes or if you want me to improve something!

Thank you for reading!

12. Integer to Roman

Problem Solving - Day 20

Table of contents

🤔 Problem Statement

💬 Thought Process

👩🏽‍💻 Solution

Conclusion