Convert roman number to arabic using javascript - javascript

Hi I am trying to convert roman numerals to arabic using javascript. I wrote a code but it is failing.
The rules I am trying to follow are :
if Larger number is before smaller number then addition and if smaller number is before larger number then subtraction.
Along with that I have few other rules as well like 'D','L' and 'V' can't be repeated at all and 'M' can be repeated only twice (Not sure how to implement this, can I use regex for it and how?)
Code :
function romanToArabic(roman){
if(roman == null)
return -1;
var value;
for(var i=0;i<roman.length;i++){
current = char_to_int(roman.charAt(i));
next = char_to_int(roman.charAt(i+1));
console.log("Current",current);
console.log("Next",next);
if(current >= next){
value = current + next;
console.log(value);
}
else {
console.log(value);
value = next - current;
}
}
return value;
}
function char_to_int(character) {
switch(character){
case 'I': return 1;
case 'V': return 5;
case 'X': return 10;
case 'L': return 50;
case 'C': return 100;
case 'D': return 500;
case 'M': return 1000;
default: return -1;
}
}
console.log(romanToArabic('IIX'));
Can somebody help? Would appreciate it!
Added screenshots :

To those, who might need to translate conventional roman numbers as opposed to irregular subtractive notation (e.g. 'IIX' instead of 'VIII' for 8), I might suggest my own, slightly shorter method:
const test = ['XIV'/*14*/, 'MXMVI'/*1996*/, 'CII'/*102*/, 'CDI'/*401*/];
const roman2arabic = s => {
const map = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000};
return [...s].reduce((r,c,i,s) => map[s[i+1]] > map[c] ? r-map[c] : r+map[c], 0);
};
console.log(test.map(roman2arabic));
.as-console-wrapper {min-height: 100%}
Though, it can be modified to follow unconventional logic:
const test = ['IIV'/*3*/,'XXMMII'/*1982*/, 'IIIXV'/*12*/, 'XII'/*conventional 12*/];
const roman2arabic = s => {
const map = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000};
return [...s]
.reduceRight(({sum,order},c,i,s) =>
Object.keys(map).indexOf(c) < order ?
{sum: sum-map[c], order} :
{sum: sum+map[c], order: Object.keys(map).indexOf(c)},
{sum:0,order:Object.keys(map).indexOf(s[s.length-1])})
.sum;
};
console.log(test.map(roman2arabic));
.as-console-wrapper {min-height: 100%}

The issue is that your code only subtracts the value corresponding to one character, while in IIX you need to subtract twice (although that kind of representation for the number 8 is quite unconventional -- 8 would normally be represented as VIII).
The solution is to keep collecting a separate sum for when the symbol is the same, so that after reading the first two "I", you have two separate sums:
total: 2
value of all "I": 2
Then when you encounter the "X" and detect that a subtraction is needed, you first undo the addition already done for the grand total, and then perform the subtraction with the value you collected for the "I":
total: -2
After this, you start with a reset value for "X":
total: 10 + -2 = 8
value for all "X": 10
Here is your code adapted for that to happen:
function romanToArabic(roman){
if(roman == null)
return -1;
var totalValue = 0,
value = 0, // Initialise!
prev = 0;
for(var i=0;i<roman.length;i++){
var current = char_to_int(roman.charAt(i));
if (current > prev) {
// Undo the addition that was done, turn it into subtraction
totalValue -= 2 * value;
}
if (current !== prev) { // Different symbol?
value = 0; // reset the sum for the new symbol
}
value += current; // keep adding same symbols
totalValue += current;
prev = current;
}
return totalValue;
}
function char_to_int(character) {
switch(character){
case 'I': return 1;
case 'V': return 5;
case 'X': return 10;
case 'L': return 50;
case 'C': return 100;
case 'D': return 500;
case 'M': return 1000;
default: return -1;
}
}
console.log(romanToArabic('IIX'));
As for your additional question to limit the number of consecutive "I" to at most two, "D" at most one, ... you could use a regular expression test at the start of your function:
if (/III|XXX|CCC|MMM|VV|LL|DD|[^IVXLCDM]/.test(roman))
return -1;
You can just append other invalid sub-sequences separated by |. For instance, if you would not want an "I" to appear directly in front of "L", "C", "D" or "M", then extend to:
if (/III|XXX|CCC|MMM|VV|LL|DD|[^IVXLCDM]|I[LCDM]/.test(roman))
return -1;

const romans = {
'I': 1,
'V': 5,
'X': 10,
'L': 50,
'C': 100,
'D': 500,
'M': 1000
};
// MXMIV
function roman2arabic(nums){
let sum = 0;
const numsArr = nums.split('');
const isSimpleRoman = (num) => num in romans;
const arabicCompare = (current, prev) => romans[current] < romans[prev];
const orderNums = (acc, current) => {
const prev = acc[acc.length - 1] || null;
const arabCurrent = romans[current];
if (prev && isSimpleRoman(prev) && arabicCompare(current, prev)) {
sum -= arabCurrent;
acc.pop() && acc.push(current + prev);
} else {
sum += arabCurrent;
acc.push(current);
}
return acc;
};
return numsArr.reduceRight(orderNums, []) && sum;
}

const romans = {
'I': 1,
'V': 5,
'X': 10,
'L': 50,
'C': 100,
'D': 500,
'M': 1000
};
function roman2arabicRecursion(nums){
const numsArr = nums.split('');
const recursion = (arr, index, sum) => {
const current = arr[index];
const prev = arr[index + 1] || null;
if(prev && romans[current] < romans[prev]){
sum -= romans[current];
} else {
sum += romans[current];
}
if(index === 0) return sum;
return recursion(arr, index - 1, sum);
}
return recursion(numsArr, numsArr.length - 1, 0);
};

Alternatively || operator could be used isntead of ??
const toArabic = (romanNumber) => {
const map = {
M: 1000,
D: 500,
C: 100,
L: 50,
X: 10,
V: 5,
I: 1,
};
const nums = romanNumber.split('');
let result = 0;
for (let i = 0; i < nums.length; i += 1) {
const first = map[nums[i]];
const second = map[nums[i + 1]] ?? 0;
if (first < second) {
result += second - first;
i += 1;
} else {
result += first;
}
}
return result;
};
console.log(toArabic('CMXI')); // 911
console.log(toArabic('MXXIV')); // 1024

We are so used to reading from Left to Right, we overlook the Right to Left alternatives.
The whole point of Roman notation is you want to check if V becomes before X
That is much easier when you reverse the Roman string.
Or in JavaScript, use the hardly ever used reduceRight method
(code optimized for better GZIP/Brotli compression)
const romanToArabic = (input) => [...input].reduceRight((
acc,
letter,
idx,
arr,
value = {m:1000, d:500, c:100, l:50, x:10, v:5, i:1}[letter.toLowerCase()],
doubleSubtraction = letter == arr[idx + 1] // ignore IIX notation
) => {
if (value < acc.high && !doubleSubtraction)
acc.Arabic -= value;
else
acc.Arabic += acc.high = value;
//console.log(idx, letter, acc, 'value:', value, acc.high, arr[idx + 1]);
return acc;
}, { high:0, Arabic:0 }).Arabic; // return Arabic value
//TESTS
Object.entries({
"cxxiv": 124,
"ix": 9,
"iix": 10,
"xL": 40,
"MMMDXLIX": 3549,
"MMMMCMXCIX": 4999}
).map(([roman,value])=>{
let converted = romanToArabic(roman);
console.log(roman, "=", converted);
console.assert(converted == value, "wrong conversion,", roman, "must be", value)
})

I resolved this exercise like this:
function arabic(num) {
let ch;
let sum = 0;
for (let i = 0; i < num.length; i++) {
ch = num[i];
switch (ch) {
case 'I':
if (num[i + 1] === 'V' || num[i + 1] === 'X') {
continue;
}
sum = sum + 1;
break;
case 'V':
if (num[i - 1] === 'I') {
sum = sum + 4;
break;
}
sum = sum + 5;
break;
case 'X':
if (num[i - 1] === 'I') {
sum = sum + 9;
break;
}
if (num[i + 1] === 'C') {
continue;
}
sum = sum + 10;
break;
case 'L':
sum = sum + 50;
break;
case 'C':
if (num[i + 1] === 'D' || num[i + 1] === 'M') {
continue;
}
if (num[i - 1] === 'X') {
sum = sum + 90;
break;
}
sum = sum + 100;
break;
case 'D':
if (num[i - 1] === 'C') {
sum = sum + 400;
break;
}
sum = sum + 500;
break;
case 'M':
if (num[i - 1] === 'C') {
sum = sum + 900;
break;
}
sum = sum + 1000;
break;
}
}
return sum;
}

Related

I am trying to use a if statement to determine if the roman numerals is 4, 9, 40, 90, 400, 900 but I am getting undefined [duplicate]

How can I convert integers into roman numerals?
function romanNumeralGenerator (int) {
}
For example, see the following sample inputs and outputs:
1 = "I"
5 = "V"
10 = "X"
20 = "XX"
3999 = "MMMCMXCIX"
Caveat: Only support numbers between 1 and 3999
There is a nice one here on this blog I found using google:
http://blog.stevenlevithan.com/archives/javascript-roman-numeral-converter
function romanize (num) {
if (isNaN(num))
return NaN;
var digits = String(+num).split(""),
key = ["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM",
"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC",
"","I","II","III","IV","V","VI","VII","VIII","IX"],
roman = "",
i = 3;
while (i--)
roman = (key[+digits.pop() + (i * 10)] || "") + roman;
return Array(+digits.join("") + 1).join("M") + roman;
}
function romanize(num) {
var lookup = {M:1000,CM:900,D:500,CD:400,C:100,XC:90,L:50,XL:40,X:10,IX:9,V:5,IV:4,I:1},roman = '',i;
for ( i in lookup ) {
while ( num >= lookup[i] ) {
roman += i;
num -= lookup[i];
}
}
return roman;
}
Reposted from a 2008 comment located at: http://blog.stevenlevithan.com/archives/javascript-roman-numeral-converter
VIEW DEMO
I don't understand why everyone's solution is so long and uses multiple for loops.
function convertToRoman(num) {
var roman = {
M: 1000,
CM: 900,
D: 500,
CD: 400,
C: 100,
XC: 90,
L: 50,
XL: 40,
X: 10,
IX: 9,
V: 5,
IV: 4,
I: 1
};
var str = '';
for (var i of Object.keys(roman)) {
var q = Math.floor(num / roman[i]);
num -= q * roman[i];
str += i.repeat(q);
}
return str;
}
I've developed the recursive solution below. The function returns one letter and then calls itself to return the next letter. It does it until the number passed to the function is 0 which means that all letters have been found and we can exit the recursion.
var romanMatrix = [
[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']
];
function convertToRoman(num) {
if (num === 0) {
return '';
}
for (var i = 0; i < romanMatrix.length; i++) {
if (num >= romanMatrix[i][0]) {
return romanMatrix[i][1] + convertToRoman(num - romanMatrix[i][0]);
}
}
}
These functions convert any positive whole number to its equivalent Roman Numeral string; and any Roman Numeral to its number.
Number to Roman Numeral:
Number.prototype.toRoman= function () {
var num = Math.floor(this),
val, s= '', i= 0,
v = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1],
r = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
function toBigRoman(n) {
var ret = '', n1 = '', rem = n;
while (rem > 1000) {
var prefix = '', suffix = '', n = rem, s = '' + rem, magnitude = 1;
while (n > 1000) {
n /= 1000;
magnitude *= 1000;
prefix += '(';
suffix += ')';
}
n1 = Math.floor(n);
rem = s - (n1 * magnitude);
ret += prefix + n1.toRoman() + suffix;
}
return ret + rem.toRoman();
}
if (this - num || num < 1) num = 0;
if (num > 3999) return toBigRoman(num);
while (num) {
val = v[i];
while (num >= val) {
num -= val;
s += r[i];
}
++i;
}
return s;
};
Roman Numeral string to Number:
Number.fromRoman = function (roman, accept) {
var s = roman.toUpperCase().replace(/ +/g, ''),
L = s.length, sum = 0, i = 0, next, val,
R = { M: 1000, D: 500, C: 100, L: 50, X: 10, V: 5, I: 1 };
function fromBigRoman(rn) {
var n = 0, x, n1, S, rx =/(\(*)([MDCLXVI]+)/g;
while ((S = rx.exec(rn)) != null) {
x = S[1].length;
n1 = Number.fromRoman(S[2])
if (isNaN(n1)) return NaN;
if (x) n1 *= Math.pow(1000, x);
n += n1;
}
return n;
}
if (/^[MDCLXVI)(]+$/.test(s)) {
if (s.indexOf('(') == 0) return fromBigRoman(s);
while (i < L) {
val = R[s.charAt(i++)];
next = R[s.charAt(i)] || 0;
if (next - val > 0) val *= -1;
sum += val;
}
if (accept || sum.toRoman() === s) return sum;
}
return NaN;
};
I personally think the neatest way (not by any means the fastest) is with recursion.
function convert(num) {
if(num < 1){ return "";}
if(num >= 40){ return "XL" + convert(num - 40);}
if(num >= 10){ return "X" + convert(num - 10);}
if(num >= 9){ return "IX" + convert(num - 9);}
if(num >= 5){ return "V" + convert(num - 5);}
if(num >= 4){ return "IV" + convert(num - 4);}
if(num >= 1){ return "I" + convert(num - 1);}
}
console.log(convert(39));
//Output: XXXIX
This will only support numbers 1-40, but it can easily be extended by following the pattern.
This version does not require any hard coded logic for edge cases such as 4(IV),9(IX),40(XL),900(CM), etc. as the others do.
I have tested this code against a data set from 1-3999 and it works.
TLDR;
This also means this solution can handle numbers greater than the maximum roman scale could (3999).
It appears there is an alternating rule for deciding the next major roman numeral character. Starting with I multiply by 5 to get the next numeral V and then by 2 to get X, then by 5 to get L, and then by 2 to get C, etc to get the next major numeral character in the scale. In this case lets assume "T" gets added to the scale to allow for larger numbers than 3999 which the original roman scale allows. In order to maintain the same algorithm "T" would represent 5000.
I = 1
V = I * 5
X = V * 2
L = X * 5
C = L * 2
D = C * 5
M = D * 2
T = M * 5
This could then allow us to represent numbers from 4000 to 5000; MT = 4000 for example.
Code:
function convertToRoman(num) {
//create key:value pairs
var romanLookup = {M:1000, D:500, C:100, L:50, X:10, V:5, I:1};
var roman = [];
var romanKeys = Object.keys(romanLookup);
var curValue;
var index;
var count = 1;
for(var numeral in romanLookup){
curValue = romanLookup[numeral];
index = romanKeys.indexOf(numeral);
while(num >= curValue){
if(count < 4){
//push up to 3 of the same numeral
roman.push(numeral);
} else {
//else we had to push four, so we need to convert the numerals
//to the next highest denomination "coloring-up in poker speak"
//Note: We need to check previous index because it might be part of the current number.
//Example:(9) would attempt (VIIII) so we would need to remove the V as well as the I's
//otherwise removing just the last three III would be incorrect, because the swap
//would give us (VIX) instead of the correct answer (IX)
if(roman.indexOf(romanKeys[index - 1]) > -1){
//remove the previous numeral we worked with
//and everything after it since we will replace them
roman.splice(roman.indexOf(romanKeys[index - 1]));
//push the current numeral and the one that appeared two iterations ago;
//think (IX) where we skip (V)
roman.push(romanKeys[index], romanKeys[index - 2]);
} else {
//else Example:(4) would attemt (IIII) so remove three I's and replace with a V
//to get the correct answer of (IV)
//remove the last 3 numerals which are all the same
roman.splice(-3);
//push the current numeral and the one that appeared right before it; think (IV)
roman.push(romanKeys[index], romanKeys[index - 1]);
}
}
//reduce our number by the value we already converted to a numeral
num -= curValue;
count++;
}
count = 1;
}
return roman.join("");
}
convertToRoman(36);
I know this is an old question but I'm pretty proud of this solution :) It only handles numbers less than 1000 but could easily be expanded to include however large you'd need by adding on to the 'numeralCodes' 2D array.
var numeralCodes = [["","I","II","III","IV","V","VI","VII","VIII","IX"], // Ones
["","X","XX","XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"], // Tens
["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"]]; // Hundreds
function convert(num) {
var numeral = "";
var digits = num.toString().split('').reverse();
for (var i=0; i < digits.length; i++){
numeral = numeralCodes[i][parseInt(digits[i])] + numeral;
}
return numeral;
}
<input id="text-input" type="text">
<button id="convert-button" onClick="var n = parseInt(document.getElementById('text-input').value);document.getElementById('text-output').value = convert(n);">Convert!</button>
<input id="text-output" style="display:block" type="text">
Loops may be more elegant but I find them hard to read. Came up with a more or less hard coded version that's easy on the eyes. As long as you understand the very first line, the rest is a no-brainer.
function romanNumeralGenerator (int) {
let roman = '';
roman += 'M'.repeat(int / 1000); int %= 1000;
roman += 'CM'.repeat(int / 900); int %= 900;
roman += 'D'.repeat(int / 500); int %= 500;
roman += 'CD'.repeat(int / 400); int %= 400;
roman += 'C'.repeat(int / 100); int %= 100;
roman += 'XC'.repeat(int / 90); int %= 90;
roman += 'L'.repeat(int / 50); int %= 50;
roman += 'XL'.repeat(int / 40); int %= 40;
roman += 'X'.repeat(int / 10); int %= 10;
roman += 'IX'.repeat(int / 9); int %= 9;
roman += 'V'.repeat(int / 5); int %= 5;
roman += 'IV'.repeat(int / 4); int %= 4;
roman += 'I'.repeat(int);
return roman;
}
I created two convert functions.
The first function can convert numbers to roman using reduce.
And the second function is very similar to the first function, the function uses the same way to convert the value.
Everything that you need to change is the _roman property. Because you have to extend this const with scale what you want, I place there max number 1000 but you can put more.
Larger scale with roman numbers you can find here https://www.tuomas.salste.net/doc/roman/numeri-romani.html
const _roman = { M: 1000, CM: 900, D: 500, CD: 400, C: 100, XC: 90, L: 50, XL: 40, X: 10, IX: 9, V: 5, IV: 4, I: 1 };
// 1903 => MCMIII
function toRoman(number = 0) {
return Object.keys(_roman).reduce((acc, key) => {
while (number >= _roman[key]) {
acc += key;
number -= _roman[key];
}
return acc;
}, '');
}
// MCMIII => 1903
function fromRoman(roman = '') {
return Object.keys(_roman).reduce((acc, key) => {
while (roman.indexOf(key) === 0) {
acc += _roman[key];
roman = roman.substr(key.length);
}
return acc;
}, 0);
}
console.log(toRoman(1903)); // should return 'MCMIII
console.log(fromRoman('MCMIII')); // should return 1903
JavaScript
function romanize (num) {
if (!+num)
return false;
var digits = String(+num).split(""),
key = ["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM",
"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC",
"","I","II","III","IV","V","VI","VII","VIII","IX"],
roman = "",
i = 3;
while (i--)
roman = (key[+digits.pop() + (i * 10)] || "") + roman;
return Array(+digits.join("") + 1).join("M") + roman;
}
many other suggestions can be found at http://blog.stevenlevithan.com/archives/javascript-roman-numeral-converter
Here is the solution with recursion, that looks simple:
const toRoman = (num, result = '') => {
const map = {
M: 1000,
CM: 900, D: 500, CD: 400, C: 100,
XC: 90, L: 50, XL: 40, X: 10,
IX: 9, V: 5, IV: 4, I: 1,
};
for (const key in map) {
if (num >= map[key]) {
if (num !== 0) {
return toRoman(num - map[key], result + key);
}
}
}
return result;
};
console.log(toRoman(402)); // CDII
console.log(toRoman(3000)); // MMM
console.log(toRoman(93)); // XCIII
console.log(toRoman(4)); // IV
This function will convert any number smaller than 3,999,999 to roman. Notice that numbers bigger than 3999 will be inside a label with text-decoration set to overline, this will add the overline that is the correct representation for x1000 when the number is bigger than 3999.
Four million (4,000,000) would be IV with two overlines so, you would need to use some trick to represent that, maybe a DIV with border-top, or some background image with those two overlines... Each overline represents x1000.
function convert(num){
num = parseInt(num);
if (num > 3999999) { alert('Number is too big!'); return false; }
if (num < 1) { alert('Number is too small!'); return false; }
var result = '',
ref = ['M','CM','D','CD','C','XC','L','XL','X','IX','V','IV','I'],
xis = [1000,900,500,400,100,90,50,40,10,9,5,4,1];
if (num <= 3999999 && num >= 4000) {
num += ''; // need to convert to string for .substring()
result = '<label style="text-decoration: overline;">'+convert(num.substring(0,num.length-3))+'</label>';
num = num.substring(num.length-3);
}
for (x = 0; x < ref.length; x++){
while(num >= xis[x]){
result += ref[x];
num -= xis[x];
}
}
return result;
}
I created two twin arrays one with arabic numbers the other with the roman characters.
function convert(num) {
var result = '';
var rom = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
var ara = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
Then I added a cycle which scan the roman elements, adding the biggest still comprised in NUM to RESULT, then we decrease NUM of the same amount.
It is like we map a part of NUM in roman numbers and then we decrease it of the same amount.
for (var x = 0; x < rom.length; x++) {
while (num >= ara[x]) {
result += rom[x];
num -= ara[x];
}
}
return result;
}
If you want to convert a big number with more symbols, maybe this algo could help.
The only premise for symbols is that must be odd and follow the same rule (1, 5, 10, 50,100 ...., 10^(N)/2, 10^(N)).
var rnumbers = ["I","V","X","L","C","D","M"];
rnumbers = rnumbers.concat(["V","X","L","C","D","M"].map(function(n) {return '<span style="border-top:1px solid black; padding:1px;">'+n+'</span> '}));
rnumbers = rnumbers.concat(["V","X","L","C","D","M"].map(function(n) {return '<span style="border:1px solid black; border-bottom:1px none black; padding:1px;">'+n+'</span> '}));
rnumbers = rnumbers.concat(["V","X","L","C","D","M"].map(function(n) {return '<span style="border-top:3px double black; padding:1px;">'+n+'</span> '}));
String.prototype.repeat = function( num ) {
return new Array( num + 1 ).join( this );
};
function toRoman(n) {
if(!n) return "";
var strn = new String(n);
var strnlength = strn.length;
var ret = "";
for(var i = 0 ; i < strnlength; i++) {
var index = strnlength*2 -2 - i*2;
var str;
var m = +strn[i];
if(index > rnumbers.length -1) {
str = rnumbers[rnumbers.length-1].repeat(m*Math.pow(10,Math.ceil((index-rnumbers.length)/2)));
}else {
str = rnumbers[index].repeat(m);
if (rnumbers.length >= index + 2) {
var rnregexp = rnumbers[index]
.split("(").join('\\(')
.split(")").join('\\)');
str = str.replace(new RegExp('(' + rnregexp + '){9}'), rnumbers[index] + rnumbers[index + 2])
.replace(new RegExp('(' + rnregexp + '){5}'), rnumbers[index + 1])
.replace(new RegExp('(' + rnregexp + '){4}'), rnumbers[index] + rnumbers[index + 1])
}
}
ret +=str;
}
return ret;
}
<input type="text" value="" onkeyup="document.getElementById('result').innerHTML = toRoman(this.value)"/>
<br/><br/>
<div id="result"></div>
After testing some of the implementations in this post, I have created a new optimized one in order to execute faster. The time execution is really low comparing with the others, but obviously the code is uglier :).
It could be even faster with an indexed array with all the posibilities.
Just in case it helps someone.
function concatNumLetters(letter, num) {
var text = "";
for(var i=0; i<num; i++){
text += letter;
}
return text;
}
function arabicToRomanNumber(arabic) {
arabic = parseInt(arabic);
var roman = "";
if (arabic >= 1000) {
var thousands = ~~(arabic / 1000);
roman = concatNumLetters("M", thousands);
arabic -= thousands * 1000;
}
if (arabic >= 900) {
roman += "CM";
arabic -= 900;
}
if (arabic >= 500) {
roman += "D";
arabic -= 500;
}
if (arabic >= 400) {
roman += "CD";
arabic -= 400;
}
if (arabic >= 100) {
var hundreds = ~~(arabic / 100);
roman += concatNumLetters("C", hundreds);
arabic -= hundreds * 100;
}
if (arabic >= 90) {
roman += "XC";
arabic -= 90;
}
if (arabic >= 50) {
roman += "L";
arabic -= 50;
}
if (arabic >= 40) {
roman += "XL";
arabic -= 40;
}
if (arabic >= 10) {
var dozens = ~~(arabic / 10);
roman += concatNumLetters("X", dozens);
arabic -= dozens * 10;
}
if (arabic >= 9) {
roman += "IX";
arabic -= 9;
}
if (arabic >= 5) {
roman += "V";
arabic -= 5;
}
if (arabic >= 4) {
roman += "IV";
arabic -= 4;
}
if (arabic >= 1) {
roman += concatNumLetters("I", arabic);
}
return roman;
}
function convertToRoman(num) {
var roman = {
M: 1000,
CM: 900,
D: 500,
CD: 400,
C: 100,
XC: 90,
L: 50,
XL: 40,
X: 10,
IX: 9,
V: 5,
IV: 4,
I: 1
}
var result = '';
for (var key in roman) {
if (num == roman[key]) {
return result +=key;
}
var check = num > roman[key];
if(check) {
result = result + key.repeat(parseInt(num/roman[key]));
num = num%roman[key];
}
}
return result;
}
console.log(convertToRoman(36));
I didn't see this posted already so here's an interesting solution using only string manipulation:
var numbers = [1, 4, 5, 7, 9, 14, 15, 19, 20, 44, 50, 94, 100, 444, 500, 659, 999, 1000, 1024];
var romanNumeralGenerator = function (number) {
return 'I'
.repeat(number)
.replace(/I{5}/g, 'V')
.replace(/V{2}/g, 'X')
.replace(/X{5}/g, 'L')
.replace(/L{2}/g, 'C')
.replace(/C{5}/g, 'D')
.replace(/D{2}/g, 'M')
.replace(/DC{4}/g, 'CM')
.replace(/C{4}/g, 'CD')
.replace(/LX{4}/g, 'XC')
.replace(/X{4}/g, 'XL')
.replace(/VI{4}/g, 'IX')
.replace(/I{4}/g, 'IV')
};
console.log(numbers.map(romanNumeralGenerator))
This function works on the the different character sets in each digit. To add another digit add the roman numeral string the 1 place, 5 place and next 1 place. This is nice because you update it with only knowing the next set of characters used.
function toRoman(n){
var d=0,o="",v,k="IVXLCDM".split("");
while(n!=0){
v=n%10,x=k[d],y=k[d+1],z=k[d+2];
o=["",x,x+x,x+x+x,x+y,y,y+x,y+x+x,y+x+x+x,x+z][v]+o;
n=(n-v)/10,d+=2;
}
return o
}
var out = "";
for (var i = 0; i < 100; i++) {
out += toRoman(i) + "\n";
}
document.getElementById("output").innerHTML = out;
<pre id="output"></pre>
function convertToRoman(num) {
var romans = {
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'
};
var popped, rem, roman = '',
keys = Object.keys(romans);
while (num > 0) {
popped = keys.pop();
m = Math.floor(num / popped);
num = num % popped;
console.log('popped:', popped, ' m:', m, ' num:', num, ' roman:', roman);
while (m-- > 0) {
roman += romans[popped];
}
while (num / popped === 0) {
popped = keys.pop();
delete romans[popped];
}
}
return roman;
}
var result = convertToRoman(3999);
console.log(result);
document.getElementById('roman').innerHTML = 'Roman: ' + result;
p {
color: darkblue;
}
<p>Decimal: 3999</p>
<p id="roman">Roman:</p>
I just made this at freecodecamp. It can easily be expanded.
function convertToRoman(num) {
var roman ="";
var values = [1000,900,500,400,100,90,50,40,10,9,5,4,1];
var literals = ["M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"];
for(i=0;i<values.length;i++){
if(num>=values[i]){
if(5<=num && num<=8) num -= 5;
else if(1<=num && num<=3) num -= 1;
else num -= values[i];
roman += literals[i];
i--;
}
}
return roman;
}
Here's a regular expression solution:
function deromanize(roman) {
var r = 0;
// regular expressions to check if valid Roman Number.
if (!/^M*(?:D?C{0,3}|C[MD])(?:L?X{0,3}|X[CL])(?:V?I{0,3}|I[XV])$/.test(roman))
throw new Error('Invalid Roman Numeral.');
roman.replace(/[MDLV]|C[MD]?|X[CL]?|I[XV]?/g, function(i) {
r += {M:1000, CM:900, D:500, CD:400, C:100, XC:90, L:50, XL:40, X:10, IX:9, V:5, IV:4, I:1}[i];
});
return r;
}
I really liked the solution by jaggedsoft but I couldn't reply because my rep is TOO LOW :( :(
I broke it down to explain it a little bit for those that don't understand it. Hopefully it helps someone.
function convertToRoman(num) {
var lookup =
{M:1000,CM:900,D:500,CD:400,C:100,XC:90,L:50,XL:40,X:10,IX:9,V:5,IV:4,I:1},roman = '',i;
for ( i in lookup ) {
while ( num >= lookup[i] ) { //while input is BIGGGER than lookup #..1000, 900, 500, etc.
roman += i; //roman is set to whatever i is (M, CM, D, CD...)
num -= lookup[i]; //takes away the first num it hits that is less than the input
//in this case, it found X:10, added X to roman, then took away 10 from input
//input lowered to 26, X added to roman, repeats and chips away at input number
//repeats until num gets down to 0. This triggers 'while' loop to stop.
}
}
return roman;
}
console.log(convertToRoman(36));
IF this number in HTMLElement (such as span), we recommend to Add HTML attribute data-format :
Number.prototype.toRoman = function() {
var e = Math.floor(this),
t, n = "",
i = 3999,
s = 0;
v = [1e3, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1], r = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"];
if (e < 1 || e > i) return "";
while (s < 13) {
t = v[s];
while (e >= t) {
e -= t;
n += r[s]
}
if (e == 0) return n;
++s
}
return ""
};
var fnrom = function(e) {
if (parseInt(e.innerHTML)) {
e.innerHTML = parseInt(e.innerHTML).toRoman()
}
};
setTimeout(function() {
[].forEach.call(document.querySelectorAll("[data-format=roman]"), fnrom)
}, 10)
Phase <span data-format="roman">4</span> Sales
function convertToRoman(num) {
let roman = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"];
let arabic = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
let index = 0;
let result = "";
while (num > 0) {
if (num >= arabic[index]) {
result += roman[index];
num -= arabic[index];
} else index++;
}
return result;
}
/*my beginner-nooby solution for numbers 1-999 :)*/
function convert(num) {
var RomNumDig = [['','I','II', 'III', 'IV', 'V', 'VI', 'VII', 'VIII', 'IX'],['X','XX', 'XXX', 'XL', 'L', 'LX', 'LXX', 'LXXX', 'XC'], ['C','CC','CCC','CD','D','DC','DCC','DCCC','CM']];
var lastDig = num%10;
var ourNumb1 = RomNumDig[0][lastDig]||'';
if(num>=10) {
var decNum = (num - lastDig)/10;
if(decNum>9)decNum%=10;
var ourNumb2 = RomNumDig[1][decNum-1]||'';}
if(num>=100) {
var hundNum = ((num-num%100)/100);
var ourNumb3 = RomNumDig[2][hundNum-1]||'';}
return ourNumb3+ourNumb2+ourNumb1;
}
console.log(convert(950));//CML
/*2nd my beginner-nooby solution for numbers 1-10, but it can be easy transformed for larger numbers :)*/
function convert(num) {
var ourNumb = '';
var romNumDig = ['I','IV','V','IX','X'];
var decNum = [1,4,5,9,10];
for (var i=decNum.length-1; i>0; i--) {
while(num>=decNum[i]) {
ourNumb += romNumDig[i];
num -= decNum[i];
}
}
return ourNumb;
}
console.log(convert(9));//IX
function toRoman(n) {
var decimals = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
var roman = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
for (var i = 0; i < decimals.length; i++) {
if(n < 1)
return "";
if(n >= decimals[i]) {
return roman[i] + toRoman(n - decimals[i]);
}
}
}
This works for all numbers only in need of roman numerals M and below.
function convert(num) {
var code = [
[1000, "M"], [900, "CM"], [800, "DCCC"], [700, "DCC"], [600, "DC"],
[500, "D"], [400, "CD"], [300, "CCC"], [200, "CC"],
[100, "C"], [90, "XC"], [80, "LXXX"], [70, "LXX"], [60, "LX"],
[50, "L"], [40, "XL"], [30, "XXX"], [20, "XX"],
[10, "X"], [9, "IX"], [8, "VIII"], [7, "VII"], [6, "VI"],
[5, "V"], [4, "IV"], [3, "III"], [2, "II"], [1, "I"],
];
var rom = "";
for(var i=0; i<code.length; i++) {
while(num >= code[i][0]) {
rom += code[i][1];
num -= code[i][0];
}
}
return rom;
}
This is the first time I really got stuck on freecodecamp. I perused through some solutions here and was amazed at how different they all were. Here is what ended up working for me.
function convertToRoman(num) {
var roman = "";
var lookupObj = {
1000:"M",
900:"CM",
500:"D",
400:"CD",
100:"C",
90:"XC",
50:"L",
40:"XL",
10:"X",
9:"IX",
4:"IV",
5:"V",
1:"I",
};
var arrayLen = Object.keys(lookupObj).length;
while(num>0){
for (i=arrayLen-1 ; i>=0 ; i--){
if(num >= Object.keys(lookupObj)[i]){
roman = roman + lookupObj[Object.keys(lookupObj)[i]];
num = num - Object.keys(lookupObj)[i];
break;
}
}
}
return roman;
}
convertToRoman(1231);
function convertToRoman(num) {
var romNumerals = [["M", 1000], ["CM", 900], ["D", 500], ["CD", 400], ["C", 100], ["XC", 90], ["L", 50], ["XL", 40], ["X", 10], ["IX", 9], ["V", 5], ["IV", 4], ["I", 1]];
var runningTotal = 0;
var roman = "";
for (var i = 0; i < romNumerals.length; i++) {
while (runningTotal + romNumerals[i][1] <= num) {
runningTotal += romNumerals[i][1];
roman += romNumerals[i][0];
}
}
return roman;
}

How can I reduce duplication in my JavaScript if syntax?

I want to reduce duplication in my JavaScript syntax.
No matter how much I think about it, it doesn't come to mind.
It doesn't matter if it is for loop or any syntax!
i add some of case and result
if if it is correct answer would be result
// case 1
// const max = [1, 31, 0, 0]
// const min = [1, 31];
// result = [3, 5]
// case 2
// const max = [0, 0, 0, 0, 0, 0]
// const min = [0, 0]
// result = [1, 6]
// case 3
// const max = [45, 4, 35, 20, 3, 9]
// const min = [45, 4, 35, 20, 3, 9]
// result = [1, 1]
if (max.length === 6) {
answer[0] = 1;
} else if (max.length === 5) {
answer[0] = 2;
} else if (max.length === 4) {
answer[0] = 3;
} else if (max.length === 3) {
answer[0] = 4;
} else if (max.length === 2) {
answer[0] = 5;
} else {
answer[0] = 6;
}
if (min.length === 6) {
answer[1] = 1;
} else if (min.length === 5) {
answer[1] = 2;
} else if (min.length === 4) {
answer[1] = 3;
} else if (min.length === 3) {
answer[1] = 4;
} else if (min.length === 2) {
answer[1] = 5;
} else {
answer[1] = 6;
}
is max and min length between 1 and 6? because judging your code, it looks like it.
answer[0] = 7 - max.length
sure looks a bit neater, at the least you could eliminate many if blocks and leave the else block in case max.length is not an integer between 1 and 6
//pseudocode
if( max.length between 1 and 6 inclusive) {
answer[0] = 7- max.length
} else {
answer[0] = some default value, 6?
}
the 7 looks like a magic number, but with more context, you can name it something better
const getAnswers = x => (x < 1 || x > 6) ? 6 : 7 - x;
// case 1
let max = [1, 31, 0, 0]
let min = [1, 31];
// result = [3, 5]
let answer = [getAnswers(max.length), getAnswers(min.length)];
console.log('max.length: ', max.length, ' min.length: ', min.length, ' answer array: ', answer);
// case 2
max = [0, 0, 0, 0, 0, 0]
min = [0]
// result = [1, 6]
answer = [getAnswers(max.length), getAnswers(min.length)];
console.log('max.length: ', max.length, ' min.length: ', min.length, ' answer array: ', answer);
// case 3
max = [45, 4, 35, 20, 3, 9]
min = [45, 4, 35, 20, 3, 9]
// result = [1, 1]
answer = [getAnswers(max.length), getAnswers(min.length)];
console.log('max.length: ', max.length, ' min.length: ', min.length, ' answer array: ', answer);
By reducing duplication, do you mean you want to decrease the length of your code or increase readability?
If you want to increase readability, some people prefer select/case than if/elseif:
switch(max.length) {
case 6:
answer[0] = 1;
break;
case 5:
answer[0] = 2;
break;
case 4:
answer[0] = 3;
break;
case 3:
answer[0] = 4;
break;
case 2:
answer[0] = 5;
break;
default:
answer[0] = 6;
}
If you want to reduce length, you can just do something like #Bergi said in comment:
answer = [7-max.length, 7-min.length];
But if max and min variable is from user input or from external source, unexpected thing may occurs:
max = {length: -5};
min = {length: -99};
answer = [7-max.length, 7-min.length];
console.log(answer)
// outputs [12,106]
The code may outputs a number outside 1-6 integer range.
So you should also add some Math.max and Math.min if you want your code to behave exactly like your if-elseif statement:
max = {length: -5};
min = {length: -99};
answer = [
Math.max(Math.min(7-max.length,1),6),
Math.max(Math.min(7-min.length,1),6)
];
console.log(answer)
// outputs [6,6]
Of course if you take input from external source you should sanitize/validate it first, but if it's an overkill, you can also use the above Math.min and Math.max function
you need to write just these two lines instead :)
answer[0] = 7 - max.length
answer[1] = 7 - min.length

Count function and number

Write a function - countNumbers It should accept string with different symbols and return an object which contains counts of each number.
Tip: consider reusing makeNumber function.
My solution is not full, what should i do the next?
function countNumbers (string) {
let numbers = [];
for (let i = 0; i < string.length; i++) {
if (!isNaN(parseInt(string[i]))){
numbers.push([string[i]])
}
}
return numbers.join('');
};
for example: countNumbers('erer384jj4444666888jfd123');
// => {'1': 1, '2': 1, '3': 2, '4': 5, '6': 3, '8': 4}
countNumbers('jdjjka000466588kkkfs662555');
// => {'0': 3, '2': 1, '4': 1, '5': 4, '6': 4, '8': 2}
You can use an object instead of an array, and check if the property already exists. If it does then add 1, else set it to start with 1.
In your code you return a string with return numbers.join(''); but in this case your can return the object instead.
function countNumbers(string) {
let numbers = {};
for (let i = 0; i < string.length; i++) {
let val = string[i];
if (!isNaN(parseInt(val))) {
val in numbers ? numbers[val]++ : numbers[val] = 1;
}
}
return numbers;
}
console.log(countNumbers('erer384jj4444666888jfd123'));
console.log(countNumbers('jdjjka000466588kkkfs662555'));
Or as pointed out by #secan as an alternative solution you can use a pattern [0-9] with the global flag /g to get the digits from the string.
const countNumbers = str => {
const counterObj = {};
str.match(/[0-9]/g).forEach(
s => counterObj.hasOwnProperty(s) ? counterObj[s] += 1 : counterObj[s] = 1
);
return counterObj;
}
console.log(countNumbers('erer384jj4444666888jfd123'));
console.log(countNumbers('jdjjka000466588kkkfs662555'));

My loop is running twice and not giving the right answer any solution for this [duplicate]

This question already has an answer here:
Linear time algorithm for Maximum contiguous sub-array sum
(1 answer)
Closed 3 years ago.
** The maximum sum subarray problem consists in finding the maximum sum
of a contiguous subsequence in an array or list of integers:
maxSequence([-2, 1, -3, 4, -1, 2, 1, -5, 4]) // should be 6: [4, -1,
2, 1] Easy case is when the list is made up of only positive numbers
and the maximum sum is the sum of the whole array. If the list is made
up of only negative numbers, return 0 instead.
Empty list is considered to have zero greatest sum. Note that the
empty list or array is also a valid sublist/subarray.**
var maxSequence = function(arr) {
// ...
let max = 0;
const sorted = arr.sort((a, b) => {
return a - b;
});
if (arr.length == 0) {
return 0;
} else if (sorted[sorted.length - 1] >= 0 && sorted[0] >= 0) {
let max = 0;
for (let i = 0; i < sorted.length; i++) {
max += sorted[i];
}
return max;
} else if (sorted[0] < 0 && sorted[sorted.length - 1] < 0) {
console.log(0);
} else {
let halfLength = Math.floor(arr.length / 2);
let pivot = 0;
let sequence = 0;
let next = 1;
while (pivot <= arr.length - 1) {
if (arr[next] <= halfLength) {
sequence += arr[next];
next += 1;
} else {
sequence = 0;
halfLength += 1;
pivot += 1;
next = pivot + 1;
}
if (pivot == arr.length - 2) {
sequence += arr[next];
next += 1;
break;
}
if (sequence >= max) {
max = sequence;
}
}
console.log("the answer", max);
}
};
maxSequence([-2, 1, -3, 4, -1, 2, 1, -5, 4]); //, 6)
**The code return 12 instead of 6 any solution i have been trying for an hour now **
You could just test it for all combinations and see which one give you the best score.
function maxSequence(data) {
let result = {
value: null,
seq: null
}
let check = {
pos: true,
neg: true
}
data.forEach(e => {
if (e > 0) check.neg = false;
if (e < 0) check.pos = false;
})
if (check.pos) {
return sum(data)
} else if (check.neg) {
return 0;
}
function sum(seq) {
return seq.reduce((r, e) => r + e, 0)
}
for (let i = 0; i < data.length; i++) {
for (let j = i; j < data.length; j++) {
const seq = data.slice(i, j + 1);
const seqSum = sum(seq);
if (result.value === null || seqSum > result.value) {
result.value = seqSum;
result.seq = seq;
}
}
}
return result;
}
console.log(maxSequence([-2, 1, -3, 4, -1, 2, 1, -5, 4]))
console.log(maxSequence([1, 5, 9, 1]))
console.log(maxSequence([-1, -2, -3, -4]))

Convert a number into a Roman numeral in JavaScript

How can I convert integers into roman numerals?
function romanNumeralGenerator (int) {
}
For example, see the following sample inputs and outputs:
1 = "I"
5 = "V"
10 = "X"
20 = "XX"
3999 = "MMMCMXCIX"
Caveat: Only support numbers between 1 and 3999
There is a nice one here on this blog I found using google:
http://blog.stevenlevithan.com/archives/javascript-roman-numeral-converter
function romanize (num) {
if (isNaN(num))
return NaN;
var digits = String(+num).split(""),
key = ["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM",
"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC",
"","I","II","III","IV","V","VI","VII","VIII","IX"],
roman = "",
i = 3;
while (i--)
roman = (key[+digits.pop() + (i * 10)] || "") + roman;
return Array(+digits.join("") + 1).join("M") + roman;
}
function romanize(num) {
var lookup = {M:1000,CM:900,D:500,CD:400,C:100,XC:90,L:50,XL:40,X:10,IX:9,V:5,IV:4,I:1},roman = '',i;
for ( i in lookup ) {
while ( num >= lookup[i] ) {
roman += i;
num -= lookup[i];
}
}
return roman;
}
Reposted from a 2008 comment located at: http://blog.stevenlevithan.com/archives/javascript-roman-numeral-converter
VIEW DEMO
I don't understand why everyone's solution is so long and uses multiple for loops.
function convertToRoman(num) {
var roman = {
M: 1000,
CM: 900,
D: 500,
CD: 400,
C: 100,
XC: 90,
L: 50,
XL: 40,
X: 10,
IX: 9,
V: 5,
IV: 4,
I: 1
};
var str = '';
for (var i of Object.keys(roman)) {
var q = Math.floor(num / roman[i]);
num -= q * roman[i];
str += i.repeat(q);
}
return str;
}
I've developed the recursive solution below. The function returns one letter and then calls itself to return the next letter. It does it until the number passed to the function is 0 which means that all letters have been found and we can exit the recursion.
var romanMatrix = [
[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']
];
function convertToRoman(num) {
if (num === 0) {
return '';
}
for (var i = 0; i < romanMatrix.length; i++) {
if (num >= romanMatrix[i][0]) {
return romanMatrix[i][1] + convertToRoman(num - romanMatrix[i][0]);
}
}
}
These functions convert any positive whole number to its equivalent Roman Numeral string; and any Roman Numeral to its number.
Number to Roman Numeral:
Number.prototype.toRoman= function () {
var num = Math.floor(this),
val, s= '', i= 0,
v = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1],
r = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
function toBigRoman(n) {
var ret = '', n1 = '', rem = n;
while (rem > 1000) {
var prefix = '', suffix = '', n = rem, s = '' + rem, magnitude = 1;
while (n > 1000) {
n /= 1000;
magnitude *= 1000;
prefix += '(';
suffix += ')';
}
n1 = Math.floor(n);
rem = s - (n1 * magnitude);
ret += prefix + n1.toRoman() + suffix;
}
return ret + rem.toRoman();
}
if (this - num || num < 1) num = 0;
if (num > 3999) return toBigRoman(num);
while (num) {
val = v[i];
while (num >= val) {
num -= val;
s += r[i];
}
++i;
}
return s;
};
Roman Numeral string to Number:
Number.fromRoman = function (roman, accept) {
var s = roman.toUpperCase().replace(/ +/g, ''),
L = s.length, sum = 0, i = 0, next, val,
R = { M: 1000, D: 500, C: 100, L: 50, X: 10, V: 5, I: 1 };
function fromBigRoman(rn) {
var n = 0, x, n1, S, rx =/(\(*)([MDCLXVI]+)/g;
while ((S = rx.exec(rn)) != null) {
x = S[1].length;
n1 = Number.fromRoman(S[2])
if (isNaN(n1)) return NaN;
if (x) n1 *= Math.pow(1000, x);
n += n1;
}
return n;
}
if (/^[MDCLXVI)(]+$/.test(s)) {
if (s.indexOf('(') == 0) return fromBigRoman(s);
while (i < L) {
val = R[s.charAt(i++)];
next = R[s.charAt(i)] || 0;
if (next - val > 0) val *= -1;
sum += val;
}
if (accept || sum.toRoman() === s) return sum;
}
return NaN;
};
I personally think the neatest way (not by any means the fastest) is with recursion.
function convert(num) {
if(num < 1){ return "";}
if(num >= 40){ return "XL" + convert(num - 40);}
if(num >= 10){ return "X" + convert(num - 10);}
if(num >= 9){ return "IX" + convert(num - 9);}
if(num >= 5){ return "V" + convert(num - 5);}
if(num >= 4){ return "IV" + convert(num - 4);}
if(num >= 1){ return "I" + convert(num - 1);}
}
console.log(convert(39));
//Output: XXXIX
This will only support numbers 1-40, but it can easily be extended by following the pattern.
This version does not require any hard coded logic for edge cases such as 4(IV),9(IX),40(XL),900(CM), etc. as the others do.
I have tested this code against a data set from 1-3999 and it works.
TLDR;
This also means this solution can handle numbers greater than the maximum roman scale could (3999).
It appears there is an alternating rule for deciding the next major roman numeral character. Starting with I multiply by 5 to get the next numeral V and then by 2 to get X, then by 5 to get L, and then by 2 to get C, etc to get the next major numeral character in the scale. In this case lets assume "T" gets added to the scale to allow for larger numbers than 3999 which the original roman scale allows. In order to maintain the same algorithm "T" would represent 5000.
I = 1
V = I * 5
X = V * 2
L = X * 5
C = L * 2
D = C * 5
M = D * 2
T = M * 5
This could then allow us to represent numbers from 4000 to 5000; MT = 4000 for example.
Code:
function convertToRoman(num) {
//create key:value pairs
var romanLookup = {M:1000, D:500, C:100, L:50, X:10, V:5, I:1};
var roman = [];
var romanKeys = Object.keys(romanLookup);
var curValue;
var index;
var count = 1;
for(var numeral in romanLookup){
curValue = romanLookup[numeral];
index = romanKeys.indexOf(numeral);
while(num >= curValue){
if(count < 4){
//push up to 3 of the same numeral
roman.push(numeral);
} else {
//else we had to push four, so we need to convert the numerals
//to the next highest denomination "coloring-up in poker speak"
//Note: We need to check previous index because it might be part of the current number.
//Example:(9) would attempt (VIIII) so we would need to remove the V as well as the I's
//otherwise removing just the last three III would be incorrect, because the swap
//would give us (VIX) instead of the correct answer (IX)
if(roman.indexOf(romanKeys[index - 1]) > -1){
//remove the previous numeral we worked with
//and everything after it since we will replace them
roman.splice(roman.indexOf(romanKeys[index - 1]));
//push the current numeral and the one that appeared two iterations ago;
//think (IX) where we skip (V)
roman.push(romanKeys[index], romanKeys[index - 2]);
} else {
//else Example:(4) would attemt (IIII) so remove three I's and replace with a V
//to get the correct answer of (IV)
//remove the last 3 numerals which are all the same
roman.splice(-3);
//push the current numeral and the one that appeared right before it; think (IV)
roman.push(romanKeys[index], romanKeys[index - 1]);
}
}
//reduce our number by the value we already converted to a numeral
num -= curValue;
count++;
}
count = 1;
}
return roman.join("");
}
convertToRoman(36);
I know this is an old question but I'm pretty proud of this solution :) It only handles numbers less than 1000 but could easily be expanded to include however large you'd need by adding on to the 'numeralCodes' 2D array.
var numeralCodes = [["","I","II","III","IV","V","VI","VII","VIII","IX"], // Ones
["","X","XX","XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"], // Tens
["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"]]; // Hundreds
function convert(num) {
var numeral = "";
var digits = num.toString().split('').reverse();
for (var i=0; i < digits.length; i++){
numeral = numeralCodes[i][parseInt(digits[i])] + numeral;
}
return numeral;
}
<input id="text-input" type="text">
<button id="convert-button" onClick="var n = parseInt(document.getElementById('text-input').value);document.getElementById('text-output').value = convert(n);">Convert!</button>
<input id="text-output" style="display:block" type="text">
Loops may be more elegant but I find them hard to read. Came up with a more or less hard coded version that's easy on the eyes. As long as you understand the very first line, the rest is a no-brainer.
function romanNumeralGenerator (int) {
let roman = '';
roman += 'M'.repeat(int / 1000); int %= 1000;
roman += 'CM'.repeat(int / 900); int %= 900;
roman += 'D'.repeat(int / 500); int %= 500;
roman += 'CD'.repeat(int / 400); int %= 400;
roman += 'C'.repeat(int / 100); int %= 100;
roman += 'XC'.repeat(int / 90); int %= 90;
roman += 'L'.repeat(int / 50); int %= 50;
roman += 'XL'.repeat(int / 40); int %= 40;
roman += 'X'.repeat(int / 10); int %= 10;
roman += 'IX'.repeat(int / 9); int %= 9;
roman += 'V'.repeat(int / 5); int %= 5;
roman += 'IV'.repeat(int / 4); int %= 4;
roman += 'I'.repeat(int);
return roman;
}
I created two convert functions.
The first function can convert numbers to roman using reduce.
And the second function is very similar to the first function, the function uses the same way to convert the value.
Everything that you need to change is the _roman property. Because you have to extend this const with scale what you want, I place there max number 1000 but you can put more.
Larger scale with roman numbers you can find here https://www.tuomas.salste.net/doc/roman/numeri-romani.html
const _roman = { M: 1000, CM: 900, D: 500, CD: 400, C: 100, XC: 90, L: 50, XL: 40, X: 10, IX: 9, V: 5, IV: 4, I: 1 };
// 1903 => MCMIII
function toRoman(number = 0) {
return Object.keys(_roman).reduce((acc, key) => {
while (number >= _roman[key]) {
acc += key;
number -= _roman[key];
}
return acc;
}, '');
}
// MCMIII => 1903
function fromRoman(roman = '') {
return Object.keys(_roman).reduce((acc, key) => {
while (roman.indexOf(key) === 0) {
acc += _roman[key];
roman = roman.substr(key.length);
}
return acc;
}, 0);
}
console.log(toRoman(1903)); // should return 'MCMIII
console.log(fromRoman('MCMIII')); // should return 1903
Here is the solution with recursion, that looks simple:
const toRoman = (num, result = '') => {
const map = {
M: 1000,
CM: 900, D: 500, CD: 400, C: 100,
XC: 90, L: 50, XL: 40, X: 10,
IX: 9, V: 5, IV: 4, I: 1,
};
for (const key in map) {
if (num >= map[key]) {
if (num !== 0) {
return toRoman(num - map[key], result + key);
}
}
}
return result;
};
console.log(toRoman(402)); // CDII
console.log(toRoman(3000)); // MMM
console.log(toRoman(93)); // XCIII
console.log(toRoman(4)); // IV
JavaScript
function romanize (num) {
if (!+num)
return false;
var digits = String(+num).split(""),
key = ["","C","CC","CCC","CD","D","DC","DCC","DCCC","CM",
"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC",
"","I","II","III","IV","V","VI","VII","VIII","IX"],
roman = "",
i = 3;
while (i--)
roman = (key[+digits.pop() + (i * 10)] || "") + roman;
return Array(+digits.join("") + 1).join("M") + roman;
}
many other suggestions can be found at http://blog.stevenlevithan.com/archives/javascript-roman-numeral-converter
This function will convert any number smaller than 3,999,999 to roman. Notice that numbers bigger than 3999 will be inside a label with text-decoration set to overline, this will add the overline that is the correct representation for x1000 when the number is bigger than 3999.
Four million (4,000,000) would be IV with two overlines so, you would need to use some trick to represent that, maybe a DIV with border-top, or some background image with those two overlines... Each overline represents x1000.
function convert(num){
num = parseInt(num);
if (num > 3999999) { alert('Number is too big!'); return false; }
if (num < 1) { alert('Number is too small!'); return false; }
var result = '',
ref = ['M','CM','D','CD','C','XC','L','XL','X','IX','V','IV','I'],
xis = [1000,900,500,400,100,90,50,40,10,9,5,4,1];
if (num <= 3999999 && num >= 4000) {
num += ''; // need to convert to string for .substring()
result = '<label style="text-decoration: overline;">'+convert(num.substring(0,num.length-3))+'</label>';
num = num.substring(num.length-3);
}
for (x = 0; x < ref.length; x++){
while(num >= xis[x]){
result += ref[x];
num -= xis[x];
}
}
return result;
}
I created two twin arrays one with arabic numbers the other with the roman characters.
function convert(num) {
var result = '';
var rom = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
var ara = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
Then I added a cycle which scan the roman elements, adding the biggest still comprised in NUM to RESULT, then we decrease NUM of the same amount.
It is like we map a part of NUM in roman numbers and then we decrease it of the same amount.
for (var x = 0; x < rom.length; x++) {
while (num >= ara[x]) {
result += rom[x];
num -= ara[x];
}
}
return result;
}
If you want to convert a big number with more symbols, maybe this algo could help.
The only premise for symbols is that must be odd and follow the same rule (1, 5, 10, 50,100 ...., 10^(N)/2, 10^(N)).
var rnumbers = ["I","V","X","L","C","D","M"];
rnumbers = rnumbers.concat(["V","X","L","C","D","M"].map(function(n) {return '<span style="border-top:1px solid black; padding:1px;">'+n+'</span> '}));
rnumbers = rnumbers.concat(["V","X","L","C","D","M"].map(function(n) {return '<span style="border:1px solid black; border-bottom:1px none black; padding:1px;">'+n+'</span> '}));
rnumbers = rnumbers.concat(["V","X","L","C","D","M"].map(function(n) {return '<span style="border-top:3px double black; padding:1px;">'+n+'</span> '}));
String.prototype.repeat = function( num ) {
return new Array( num + 1 ).join( this );
};
function toRoman(n) {
if(!n) return "";
var strn = new String(n);
var strnlength = strn.length;
var ret = "";
for(var i = 0 ; i < strnlength; i++) {
var index = strnlength*2 -2 - i*2;
var str;
var m = +strn[i];
if(index > rnumbers.length -1) {
str = rnumbers[rnumbers.length-1].repeat(m*Math.pow(10,Math.ceil((index-rnumbers.length)/2)));
}else {
str = rnumbers[index].repeat(m);
if (rnumbers.length >= index + 2) {
var rnregexp = rnumbers[index]
.split("(").join('\\(')
.split(")").join('\\)');
str = str.replace(new RegExp('(' + rnregexp + '){9}'), rnumbers[index] + rnumbers[index + 2])
.replace(new RegExp('(' + rnregexp + '){5}'), rnumbers[index + 1])
.replace(new RegExp('(' + rnregexp + '){4}'), rnumbers[index] + rnumbers[index + 1])
}
}
ret +=str;
}
return ret;
}
<input type="text" value="" onkeyup="document.getElementById('result').innerHTML = toRoman(this.value)"/>
<br/><br/>
<div id="result"></div>
After testing some of the implementations in this post, I have created a new optimized one in order to execute faster. The time execution is really low comparing with the others, but obviously the code is uglier :).
It could be even faster with an indexed array with all the posibilities.
Just in case it helps someone.
function concatNumLetters(letter, num) {
var text = "";
for(var i=0; i<num; i++){
text += letter;
}
return text;
}
function arabicToRomanNumber(arabic) {
arabic = parseInt(arabic);
var roman = "";
if (arabic >= 1000) {
var thousands = ~~(arabic / 1000);
roman = concatNumLetters("M", thousands);
arabic -= thousands * 1000;
}
if (arabic >= 900) {
roman += "CM";
arabic -= 900;
}
if (arabic >= 500) {
roman += "D";
arabic -= 500;
}
if (arabic >= 400) {
roman += "CD";
arabic -= 400;
}
if (arabic >= 100) {
var hundreds = ~~(arabic / 100);
roman += concatNumLetters("C", hundreds);
arabic -= hundreds * 100;
}
if (arabic >= 90) {
roman += "XC";
arabic -= 90;
}
if (arabic >= 50) {
roman += "L";
arabic -= 50;
}
if (arabic >= 40) {
roman += "XL";
arabic -= 40;
}
if (arabic >= 10) {
var dozens = ~~(arabic / 10);
roman += concatNumLetters("X", dozens);
arabic -= dozens * 10;
}
if (arabic >= 9) {
roman += "IX";
arabic -= 9;
}
if (arabic >= 5) {
roman += "V";
arabic -= 5;
}
if (arabic >= 4) {
roman += "IV";
arabic -= 4;
}
if (arabic >= 1) {
roman += concatNumLetters("I", arabic);
}
return roman;
}
function convertToRoman(num) {
var roman = {
M: 1000,
CM: 900,
D: 500,
CD: 400,
C: 100,
XC: 90,
L: 50,
XL: 40,
X: 10,
IX: 9,
V: 5,
IV: 4,
I: 1
}
var result = '';
for (var key in roman) {
if (num == roman[key]) {
return result +=key;
}
var check = num > roman[key];
if(check) {
result = result + key.repeat(parseInt(num/roman[key]));
num = num%roman[key];
}
}
return result;
}
console.log(convertToRoman(36));
I didn't see this posted already so here's an interesting solution using only string manipulation:
var numbers = [1, 4, 5, 7, 9, 14, 15, 19, 20, 44, 50, 94, 100, 444, 500, 659, 999, 1000, 1024];
var romanNumeralGenerator = function (number) {
return 'I'
.repeat(number)
.replace(/I{5}/g, 'V')
.replace(/V{2}/g, 'X')
.replace(/X{5}/g, 'L')
.replace(/L{2}/g, 'C')
.replace(/C{5}/g, 'D')
.replace(/D{2}/g, 'M')
.replace(/DC{4}/g, 'CM')
.replace(/C{4}/g, 'CD')
.replace(/LX{4}/g, 'XC')
.replace(/X{4}/g, 'XL')
.replace(/VI{4}/g, 'IX')
.replace(/I{4}/g, 'IV')
};
console.log(numbers.map(romanNumeralGenerator))
This function works on the the different character sets in each digit. To add another digit add the roman numeral string the 1 place, 5 place and next 1 place. This is nice because you update it with only knowing the next set of characters used.
function toRoman(n){
var d=0,o="",v,k="IVXLCDM".split("");
while(n!=0){
v=n%10,x=k[d],y=k[d+1],z=k[d+2];
o=["",x,x+x,x+x+x,x+y,y,y+x,y+x+x,y+x+x+x,x+z][v]+o;
n=(n-v)/10,d+=2;
}
return o
}
var out = "";
for (var i = 0; i < 100; i++) {
out += toRoman(i) + "\n";
}
document.getElementById("output").innerHTML = out;
<pre id="output"></pre>
function convertToRoman(num) {
var romans = {
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'
};
var popped, rem, roman = '',
keys = Object.keys(romans);
while (num > 0) {
popped = keys.pop();
m = Math.floor(num / popped);
num = num % popped;
console.log('popped:', popped, ' m:', m, ' num:', num, ' roman:', roman);
while (m-- > 0) {
roman += romans[popped];
}
while (num / popped === 0) {
popped = keys.pop();
delete romans[popped];
}
}
return roman;
}
var result = convertToRoman(3999);
console.log(result);
document.getElementById('roman').innerHTML = 'Roman: ' + result;
p {
color: darkblue;
}
<p>Decimal: 3999</p>
<p id="roman">Roman:</p>
I just made this at freecodecamp. It can easily be expanded.
function convertToRoman(num) {
var roman ="";
var values = [1000,900,500,400,100,90,50,40,10,9,5,4,1];
var literals = ["M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"];
for(i=0;i<values.length;i++){
if(num>=values[i]){
if(5<=num && num<=8) num -= 5;
else if(1<=num && num<=3) num -= 1;
else num -= values[i];
roman += literals[i];
i--;
}
}
return roman;
}
Here's a regular expression solution:
function deromanize(roman) {
var r = 0;
// regular expressions to check if valid Roman Number.
if (!/^M*(?:D?C{0,3}|C[MD])(?:L?X{0,3}|X[CL])(?:V?I{0,3}|I[XV])$/.test(roman))
throw new Error('Invalid Roman Numeral.');
roman.replace(/[MDLV]|C[MD]?|X[CL]?|I[XV]?/g, function(i) {
r += {M:1000, CM:900, D:500, CD:400, C:100, XC:90, L:50, XL:40, X:10, IX:9, V:5, IV:4, I:1}[i];
});
return r;
}
I really liked the solution by jaggedsoft but I couldn't reply because my rep is TOO LOW :( :(
I broke it down to explain it a little bit for those that don't understand it. Hopefully it helps someone.
function convertToRoman(num) {
var lookup =
{M:1000,CM:900,D:500,CD:400,C:100,XC:90,L:50,XL:40,X:10,IX:9,V:5,IV:4,I:1},roman = '',i;
for ( i in lookup ) {
while ( num >= lookup[i] ) { //while input is BIGGGER than lookup #..1000, 900, 500, etc.
roman += i; //roman is set to whatever i is (M, CM, D, CD...)
num -= lookup[i]; //takes away the first num it hits that is less than the input
//in this case, it found X:10, added X to roman, then took away 10 from input
//input lowered to 26, X added to roman, repeats and chips away at input number
//repeats until num gets down to 0. This triggers 'while' loop to stop.
}
}
return roman;
}
console.log(convertToRoman(36));
IF this number in HTMLElement (such as span), we recommend to Add HTML attribute data-format :
Number.prototype.toRoman = function() {
var e = Math.floor(this),
t, n = "",
i = 3999,
s = 0;
v = [1e3, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1], r = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"];
if (e < 1 || e > i) return "";
while (s < 13) {
t = v[s];
while (e >= t) {
e -= t;
n += r[s]
}
if (e == 0) return n;
++s
}
return ""
};
var fnrom = function(e) {
if (parseInt(e.innerHTML)) {
e.innerHTML = parseInt(e.innerHTML).toRoman()
}
};
setTimeout(function() {
[].forEach.call(document.querySelectorAll("[data-format=roman]"), fnrom)
}, 10)
Phase <span data-format="roman">4</span> Sales
function convertToRoman(num) {
let roman = ["M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"];
let arabic = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
let index = 0;
let result = "";
while (num > 0) {
if (num >= arabic[index]) {
result += roman[index];
num -= arabic[index];
} else index++;
}
return result;
}
/*my beginner-nooby solution for numbers 1-999 :)*/
function convert(num) {
var RomNumDig = [['','I','II', 'III', 'IV', 'V', 'VI', 'VII', 'VIII', 'IX'],['X','XX', 'XXX', 'XL', 'L', 'LX', 'LXX', 'LXXX', 'XC'], ['C','CC','CCC','CD','D','DC','DCC','DCCC','CM']];
var lastDig = num%10;
var ourNumb1 = RomNumDig[0][lastDig]||'';
if(num>=10) {
var decNum = (num - lastDig)/10;
if(decNum>9)decNum%=10;
var ourNumb2 = RomNumDig[1][decNum-1]||'';}
if(num>=100) {
var hundNum = ((num-num%100)/100);
var ourNumb3 = RomNumDig[2][hundNum-1]||'';}
return ourNumb3+ourNumb2+ourNumb1;
}
console.log(convert(950));//CML
/*2nd my beginner-nooby solution for numbers 1-10, but it can be easy transformed for larger numbers :)*/
function convert(num) {
var ourNumb = '';
var romNumDig = ['I','IV','V','IX','X'];
var decNum = [1,4,5,9,10];
for (var i=decNum.length-1; i>0; i--) {
while(num>=decNum[i]) {
ourNumb += romNumDig[i];
num -= decNum[i];
}
}
return ourNumb;
}
console.log(convert(9));//IX
function toRoman(n) {
var decimals = [1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1];
var roman = ['M', 'CM', 'D', 'CD', 'C', 'XC', 'L', 'XL', 'X', 'IX', 'V', 'IV', 'I'];
for (var i = 0; i < decimals.length; i++) {
if(n < 1)
return "";
if(n >= decimals[i]) {
return roman[i] + toRoman(n - decimals[i]);
}
}
}
This works for all numbers only in need of roman numerals M and below.
function convert(num) {
var code = [
[1000, "M"], [900, "CM"], [800, "DCCC"], [700, "DCC"], [600, "DC"],
[500, "D"], [400, "CD"], [300, "CCC"], [200, "CC"],
[100, "C"], [90, "XC"], [80, "LXXX"], [70, "LXX"], [60, "LX"],
[50, "L"], [40, "XL"], [30, "XXX"], [20, "XX"],
[10, "X"], [9, "IX"], [8, "VIII"], [7, "VII"], [6, "VI"],
[5, "V"], [4, "IV"], [3, "III"], [2, "II"], [1, "I"],
];
var rom = "";
for(var i=0; i<code.length; i++) {
while(num >= code[i][0]) {
rom += code[i][1];
num -= code[i][0];
}
}
return rom;
}
This is the first time I really got stuck on freecodecamp. I perused through some solutions here and was amazed at how different they all were. Here is what ended up working for me.
function convertToRoman(num) {
var roman = "";
var lookupObj = {
1000:"M",
900:"CM",
500:"D",
400:"CD",
100:"C",
90:"XC",
50:"L",
40:"XL",
10:"X",
9:"IX",
4:"IV",
5:"V",
1:"I",
};
var arrayLen = Object.keys(lookupObj).length;
while(num>0){
for (i=arrayLen-1 ; i>=0 ; i--){
if(num >= Object.keys(lookupObj)[i]){
roman = roman + lookupObj[Object.keys(lookupObj)[i]];
num = num - Object.keys(lookupObj)[i];
break;
}
}
}
return roman;
}
convertToRoman(1231);
function convertToRoman(num) {
var romNumerals = [["M", 1000], ["CM", 900], ["D", 500], ["CD", 400], ["C", 100], ["XC", 90], ["L", 50], ["XL", 40], ["X", 10], ["IX", 9], ["V", 5], ["IV", 4], ["I", 1]];
var runningTotal = 0;
var roman = "";
for (var i = 0; i < romNumerals.length; i++) {
while (runningTotal + romNumerals[i][1] <= num) {
runningTotal += romNumerals[i][1];
roman += romNumerals[i][0];
}
}
return roman;
}

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