Integer to Roman Numeral - Amazon Top Interview Questions

Problem Statement :

Given an integer n, return its corresponding Roman numeral.

Roman numerals contain the symbols representing values in the following list:

"I" = 1
"V" = 5
"X" = 10
"L" = 50
"C" = 100
"D" = 500
"M" = 1000

Symbols are typically written largest to smallest, from left to right, and can be computed by summing the values of all the symbols. However, in some cases, when a symbol of lower value is to the left of a symbol of higher value, then the lower value is subtracted from the higher one.

There are 6 cases where this is possible:

When "I" is before "V", we get 4.
When "I" is before "X", we get 9.
When "X" is before "L", we get 40.
When "X" is before "C", we get 90.
When "C" is before "D", we get 400.
When "C" is before "M", we get 900.

Roman numerals must also follow these rules:

No symbol is repeated more than 3 times.

The symbols "V", "L", and "D" are not repeated.


1 ≤ n ≤ 3000

Example 1


n = 12




"XII" = 10 + 1 + 1 = 12

Example 2


n = 14




"XIV" = 10 + 4 = 14

Solution :


                        Solution in C++ :

string solve(int num) {
    string thousands[] = {"", "M", "MM", "MMM"};  // as num is lesser than 3000
    string hundreds[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
    string tens[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
    string ones[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};

    string str = thousands[num / 1000] + hundreds[(num % 1000) / 100] + tens[(num % 100) / 10] +
                 ones[num % 10];
    return str;

                        Solution in Python : 
class Solution:
    def solve(self, n):
        res = ""
        table = [
            (1000, "M"),
            (900, "CM"),
            (500, "D"),
            (400, "CD"),
            (100, "C"),
            (90, "XC"),
            (50, "L"),
            (40, "XL"),
            (10, "X"),
            (9, "IX"),
            (5, "V"),
            (4, "IV"),
            (1, "I"),
        for cap, roman in table:
            d, m = divmod(n, cap)
            res += roman * d
            n = m

        return res

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