## Task

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 one’s 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

**Example 2:**

**Input:**

s = "IV"

**Output:**

4

**Example 3:**

**Input:**

s = "IX"

**Output:**

9

**Example 4:**

**Input:**

s = "LVIII"

**Output:**

58

**Explanation:**

L = 50, V= 5, III = 3.

**Example 5:**

**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]`

.

This problem was taken from Leetcode Roman To Integer

## Solution

Solution 1: Left to right pass

/** * @param {string} s * @return {number} */ var romanToInt = function(s) { var len = s.length; var i = 0; var map = { 'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000 } var sum = 0; while(i < len) { var currentVal = map[ s[i] ]; var nextVal = map[ s[i + 1] ]; if( currentVal < nextVal) { sum += nextVal - currentVal; i ++; } else { sum += currentVal; } i ++; } return sum; };

Solution 2: Left to right (or right to left) pass improved

/** * @param {string} s * @return {number} */ var romanToInt = function(s) { var len = s.length; var i = 0; var map = { 'I': 1, 'IV': 4, 'V': 5, 'IX': 9, 'X': 10, 'XL': 40, 'L': 50, 'XC': 90, 'C': 100, 'CD': 400, 'D': 500, 'CM': 900, 'M': 1000 } var sum = 0; while(i < len) { var currentVal = map[ s[i] ]; var nextVal = map[ s[i + 1] ]; if( currentVal < nextVal) { var sumbol = s[i] + s[i+1]; sum += map[sumbol]; i ++; } else { sum += currentVal; } i ++; } return sum; };

Solution3: Right to left pass

In the “subtraction” cases, such as `XC`

, we’ve been updating our running `sum`

as follows:

```
sum += value(C) - value(X)
```

However, notice that this is mathematically equivalent to the following:

```
sum += value(C)
sum -= value(X)
```

Utilizing this means that we can process *one* symbol each time we go around the main loop. We still need to determine whether or not our current symbol should be added or subtracted by looking at the neighbour though.

This way we could start from the most right symbol an initialize the sym with it, since every most right symbol will always be added to the sum.

/** * @param {string} s * @return {number} */ var romanToInt = function(s) { var len = s.length; var i = len - 1; var map = { 'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000 } var sum = map[ s[i] ]; i --; while(i > -1) { var currentVal = map[ s[i] ]; var prevVal = map[ s[i + 1] ]; if( currentVal < prevVal) { sum -= currentVal; } else { sum += currentVal; } i --; } return sum; };