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12. Integer to Roman

Description

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): IV and 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

 

Constraints:

  • 1 <= num <= 3999

Solutions

Solution 1: Greedy

We can first list all possible symbols $cs$ and their corresponding values $vs$, then enumerate each value $vs[i]$ from large to small. Each time, we use as many symbols $cs[i]$ corresponding to this value as possible, until the number $num$ becomes $0$.

The time complexity is $O(m)$, and the space complexity is $O(m)$. Here, $m$ is the number of symbols.

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class Solution:
    def intToRoman(self, num: int) -> str:
        cs = ('M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I')
        vs = (1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1)
        ans = []
        for c, v in zip(cs, vs):
            while num >= v:
                num -= v
                ans.append(c)
        return ''.join(ans)
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class Solution {
    public String intToRoman(int num) {
        List<String> cs
            = List.of("M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I");
        List<Integer> vs = List.of(1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1);
        StringBuilder ans = new StringBuilder();
        for (int i = 0, n = cs.size(); i < n; ++i) {
            while (num >= vs.get(i)) {
                num -= vs.get(i);
                ans.append(cs.get(i));
            }
        }
        return ans.toString();
    }
}
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class Solution {
public:
    string intToRoman(int num) {
        vector<string> cs = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};
        vector<int> vs = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};
        string ans;
        for (int i = 0; i < cs.size(); ++i) {
            while (num >= vs[i]) {
                num -= vs[i];
                ans += cs[i];
            }
        }
        return ans;
    }
};
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func intToRoman(num int) string {
    cs := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
    vs := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
    ans := &strings.Builder{}
    for i, v := range vs {
        for num >= v {
            num -= v
            ans.WriteString(cs[i])
        }
    }
    return ans.String()
}
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function intToRoman(num: number): strin