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Roman to Integer: Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, 2 is written as II in Roman numeral, just two ones added together. 12 is written as XII, which is simply X + II. The number 27 is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer.

Example 1:

Input: s = "III"
Output: 3
Explanation: III = 3.

Example 2:

Input: s = "LVIII"
Output: 58
Explanation: L = 50, V= 5, III = 3.

Example 3:

Input: s = "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Constraints:

1 <= s.length <= 15
s contains only the characters ('I', 'V', 'X', 'L', 'C', 'D', 'M').
It is guaranteed that s is a valid roman numeral in the range [1, 3999].

Try this Problem on your own or check similar problems:

Integer to Roman

Solution:

public int romanToInt(String s) {
    Map<Character, Set<Character>> validSubtraction = new HashMap<>();
    validSubtraction.put('I', new HashSet<>(List.of('V', 'X')));
    validSubtraction.put('X', new HashSet<>(List.of('L', 'C')));
    validSubtraction.put('C', new HashSet<>(List.of('D', 'M')));

    Map<Character, Integer> romanToIntMap = new HashMap<>();
    romanToIntMap.put('I', 1);
    romanToIntMap.put('V', 5);
    romanToIntMap.put('X', 10);
    romanToIntMap.put('L', 50);
    romanToIntMap.put('C', 100);
    romanToIntMap.put('D', 500);
    romanToIntMap.put('M', 1000);

    int number = 0;
    Character prev = null;
    for(int i = 0; i < s.length(); ++i){
        char c = s.charAt(i);
        number += romanToIntMap.get(c);
        if(prev != null && validSubtraction.containsKey(prev) && validSubtraction.get(prev).contains(c)){
            number -= 2 * romanToIntMap.get(prev);
        }
        prev = c;
    }

    return number;
}

Time/Space Complexity:

Time Complexity: O(n)
Space Complexity: O(1)

Explanation:

Since we traverse the whole string we have linear time complexity proportional to the length of the string, also we keep store of limited number of records in two maps so we don’t spend any additional space leading to contant space complexity. We create two maps, one from mapping the character to integer value romanToIntMap and the other to check if we have valid subtraction validSubtraction. We traverse the string each time adding to our number the corresponding value of the current character once converted from roman to int using romanToIntMap map, we also track the prev character and check if we have a valid subtraction. If we have, we have to subtract the value of prev two times (one for addition of prev in the previous iteration, and the other since we have valid subtraction). We return the number variable as our final result.

Roman to Integer