Roman Numeral to Integer - Amazon Top Interview Questions

Problem Statement :

Given a string numeral representing a Roman numeral, convert it to an integer.

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

Roman numerals are typically written largest to smallest, from left to right, and can be computed by summing up 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.


1 ≤ n ≤ 15 where n is the length of numeral

1 ≤ k ≤ 3000 where k is the number numeral represents

Example 1


numeral = "XII"




"XII" = 10 + 1 + 1 = 12

Example 2


numeral = "XIV"




"XIV" = 10 + 4 = 14

Solution :


                        Solution in C++ :

int solve(string numeral) {
    unordered_map<char, int> mp;
    mp['I'] = 1;
    mp['V'] = 5;
    mp['X'] = 10;
    mp['L'] = 50;
    mp['C'] = 100;
    mp['D'] = 500;
    mp['M'] = 1000;

    int n = numeral.length();

    int prev = INT_MAX;
    int curr;
    int ans = 0;

    for (int i = 0; i < n; i++) {
        curr = mp[numeral[i]];  // get the current value
        ans += curr;            // add it to result
        if (curr > prev) {      // subtractive case
            ans -= 2 * prev;
        prev = curr;  // update prev for next iteration

    return ans;

                        Solution in Python : 
class Solution:
    def solve(self, numeral):
        value = {"M": 1000, "D": 500, "C": 100, "L": 50, "X": 10, "V": 5, "I": 1}
        res = 0
        prev = 0
        for i in range(len(numeral) - 1, -1, -1):
            n = value[numeral[i]]
            if n >= prev:
                res += int(n)
                res -= int(n)
            prev = n
        return res

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