Given a number, find its corresponding Roman numeral.
Examples:
Input : 9
Output : IX
Input : 40
Output : XL
Input : 1904
Output : MCMIV
Following is the list of Roman symbols which include subtractive cases also:
SYMBOL VALUE
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
Idea is to convert the units, tens, hundreds, and thousands of places of the given number separately. If the digit is 0, then there’s no corresponding Roman numeral symbol. The conversion of digit’s 4’s and 9’s are little bit different from other digits because these digits follow subtractive notation.
Algorithm to convert decimal number to Roman Numeral
Compare given number with base values in the order 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1. Base value which is just smaller or equal to the given number will be the initial base value (largest base value) .Divide the number by its largest base value, the corresponding base symbol will be repeated quotient times, the remainder will then become the number for future division and repetitions.The process will be repeated until the number becomes zero.
Example to demonstrate above algorithm:
Convert 3549 to its Roman Numerals
Output:
MMMDXLIX
Explanation:
Explanation:
Step 1
- Initially number = 3549
- Since 3549 >= 1000 ; largest base value will be 1000 initially.
- Divide 3549/1000. Quotient = 3, Remainder =549. The corresponding symbol M will be repeated thrice.
- We append the Result value in the 2nd List.
- Now Remainder is not equal to 0 so we call the function again.
Step 2
- Now, number = 549
- 1000 > 549 >= 500 ; largest base value will be 500.
- Divide 549/500. Quotient = 1, Remainder =49. The corresponding symbol D will be repeated once & stop the loop.
- We append the Result value in the 2nd List.
- Now Remainder is not equal to 0 so we call the function again.
Step 3
- Now, number = 49
- 50 > 49 >= 40 ; largest base value is 40.
- Divide 49/40. Quotient = 1, Remainder = 9. The corresponding symbol XL will be repeated once & stop the loop.
- We append the Result value in the 2nd List.
- Now Remainder is not equal to 0 so we call the function again.
Step 4
- Now, number = 9
- Number 9 is present in list ls so we directly fetch the value from dictionary dict and set Remainder=0 & stop the loop.
- Remainder = 0. The corresponding symbol IX will be repeated once and now remainder value is 0 so we will not call the function again.
Step 5
- Finally, we join the 2nd list values.
- The output obtained MMMDXLIX.
Following is the implementation of the above algorithm:
C++
#include <bits/stdc++.h>
using namespace std;
int printRoman(int number)
{
int num[] = {1,4,5,9,10,40,50,90,100,400,500,900,1000};
string sym[] = {"I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"};
int i=12;
while(number>0)
{
int div = number/num[i];
number = number%num[i];
while(div--)
{
cout<<sym[i];
}
i--;
}
}
int main()
{
int number = 3549;
printRoman(number);
return 0;
}
|
Java
class GFG {
static int sub_digit(char num1, char num2, int i, char[] c) {
c[i++] = num1;
c[i++] = num2;
return i;
}
static int digit(char ch, int n, int i, char[] c) {
for (int j = 0; j < n; j++) {
c[i++] = ch;
}
return i;
}
static void printRoman(int number) {
char c[] = new char[10001];
int i = 0;
if (number <= 0) {
System.out.printf("Invalid number");
return;
}
while (number != 0) {
if (number >= 1000) {
i = digit('M', number / 1000, i, c);
number = number % 1000;
}
else if (number >= 500) {
if (number < 900) {
i = digit('D', number / 500, i, c);
number = number % 500;
}
else {
i = sub_digit('C', 'M', i, c);
number = number % 100;
}
}
else if (number >= 100) {
if (number < 400) {
i = digit('C', number / 100, i, c);
number = number % 100;
}
else {
i = sub_digit('C', 'D', i, c);
number = number % 100;
}
}
else if (number >= 50) {
if (number < 90) {
i = digit('L', number / 50, i, c);
number = number % 50;
}
else {
i = sub_digit('X', 'C', i, c);
number = number % 10;
}
}
else if (number >= 10) {
if (number < 40) {
i = digit('X', number / 10, i, c);
number = number % 10;
}
else {
i = sub_digit('X', 'L', i, c);
number = number % 10;
}
}
else if (number >= 5) {
if (number < 9) {
i = digit('V', number / 5, i, c);
number = number % 5;
}
else {
i = sub_digit('I', 'X', i, c);
number = 0;
}
}
else if (number >= 1) {
if (number < 4) {
i = digit('I', number, i, c);
number = 0;
}
else {
i = sub_digit('I', 'V', i, c);
number = 0;
}
}
}
System.out.printf("Roman numeral is: ");
for (int j = 0; j < i; j++) {
System.out.printf("%c", c[j]);
}
}
public static void main(String[] args) {
int number = 3549;
printRoman(number);
}
}
|
Python3
ls=[1000,900,500,400,100,90,50,40,10,9,5,4,1]
dict={1:"I",4:"IV",5:"V",9:"IX",10:"X",40:"XL",50:"L",90:"XC",100:"C",400:"CD",500:"D",900:"CM",1000:"M"}
ls2=[]
def func(no,res):
for i in range(0,len(ls)):
if no in ls:
res=dict[no]
rem=0
break
if ls[i]<no:
quo=no//ls[i]
rem=no%ls[i]
res=res+dict[ls[i]]*quo
break
ls2.append(res)
if rem==0:
pass
else:
func(rem,"")
if __name__ == "__main__":
func(3549, "")
print("".join(ls2))
|
C#
using System;
class GFG
{
static int sub_digit(char num1, char num2,
int i, char[] c)
{
c[i++] = num1;
c[i++] = num2;
return i;
}
static int digit(char ch, int n, int i, char[] c)
{
for (int j = 0; j < n; j++)
{
c[i++] = ch;
}
return i;
}
static void printRoman(int number)
{
char[] c = new char[10001];
int i = 0;
if (number <= 0)
{
Console.WriteLine("Invalid number");
return;
}
while (number != 0)
{
if (number >= 1000)
{
i = digit('M', number / 1000, i, c);
number = number % 1000;
}
else if (number >= 500)
{
if (number < 900)
{
i = digit('D', number / 500, i, c);
number = number % 500;
}
else
{
i = sub_digit('C', 'M', i, c);
number = number % 100;
}
}
else if (number >= 100)
{
if (number < 400)
{
i = digit('C', number / 100, i, c);
number = number % 100;
}
else
{
i = sub_digit('C', 'D', i, c);
number = number % 100;
}
}
else if (number >= 50)
{
if (number < 90)
{
i = digit('L', number / 50, i, c);
number = number % 50;
}
else
{
i = sub_digit('X', 'C', i, c);
number = number % 10;
}
}
else if (number >= 10)
{
if (number < 40)
{
i = digit('X', number / 10, i, c);
number = number % 10;
}
else
{
i = sub_digit('X', 'L', i, c);
number = number % 10;
}
}
else if (number >= 5)
{
if (number < 9)
{
i = digit('V', number / 5, i, c);
number = number % 5;
}
else
{
i = sub_digit('I', 'X', i, c);
number = 0;
}
}
else if (number >= 1)
{
if (number < 4)
{
i = digit('I', number, i, c);
number = 0;
}
else
{
i = sub_digit('I', 'V', i, c);
number = 0;
}
}
}
Console.WriteLine("Roman numeral is: ");
for (int j = 0; j < i; j++)
{
Console.Write("{0}", c[j]);
}
}
public static void Main()
{
int number = 3549;
printRoman(number);
}
}
|
Javascript
<script>
function printRoman(number)
{
let num = [1,4,5,9,10,40,50,90,100,400,500,900,1000];
let sym = ["I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"];
let i=12;
while(number>0)
{
let div = Math.floor(number/num[i]);
number = number%num[i];
while(div--)
{
document.write(sym[i]);
}
i--;
}
}
let number = 3549;
printRoman(number);
</script>
|
Time Complexity: O(N), where N is the length of ans string that stores the conversion.
Auxiliary Space: O(N)
References : http://blog.functionalfun.net/2009/01/project-euler-89-converting-to-and-from.html
Another Approach 1:
In this approach, we have to first observe the problem. The number given in problem statement can be maximum of 4 digits. The idea to solve this problem is:
- Divide the given number into digits at different places like one’s, two’s, hundred’s or thousand’s.
- Starting from the thousand’s place print the corresponding roman value. For example, if the digit at thousand’s place is 3 then print the roman equivalent of 3000.
- Repeat the second step until we reach one’s place.
Example:
Suppose the input number is 3549. So, starting from thousand’s place we will start printing the roman equivalent. In this case we will print in the order as given below:
First: Roman equivalent of 3000
Second: Roman equivalent of 500
Third: Roman equivalent of 40
Fourth: Roman equivalent of 9
So, the output will be: MMMDXLIX
Below is the implementation of above idea.
C++
#include <bits/stdc++.h>
using namespace std;
string intToRoman(int num)
{
string m[] = { "", "M", "MM", "MMM" };
string c[] = { "", "C", "CC", "CCC", "CD",
"D", "DC", "DCC", "DCCC", "CM" };
string x[] = { "", "X", "XX", "XXX", "XL",
"L", "LX", "LXX", "LXXX", "XC" };
string i[] = { "", "I", "II", "III", "IV",
"V", "VI", "VII", "VIII", "IX" };
string thousands = m[num / 1000];
string hundreds = c[(num % 1000) / 100];
string tens = x[(num % 100) / 10];
string ones = i[num % 10];
string ans = thousands + hundreds + tens + ones;
return ans;
}
int main()
{
int number = 3549;
cout << intToRoman(number);
return 0;
}
|
Java
class GFG {
static String intToRoman(int num)
{
String m[] = { "", "M", "MM", "MMM" };
String c[] = { "", "C", "CC", "CCC", "CD",
"D", "DC", "DCC", "DCCC", "CM" };
String x[] = { "", "X", "XX", "XXX", "XL",
"L", "LX", "LXX", "LXXX", "XC" };
String i[] = { "", "I", "II", "III", "IV",
"V", "VI", "VII", "VIII", "IX" };
String thousands = m[num / 1000];
String hundreds = c[(num % 1000) / 100];
String tens = x[(num % 100) / 10];
String ones = i[num % 10];
String ans = thousands + hundreds + tens + ones;
return ans;
}
public static void main(String[] args)
{
int number = 3549;
System.out.println(intToRoman(number));
}
}
|
Python3
def intToRoman(num):
m = ["", "M", "MM", "MMM"]
c = ["", "C", "CC", "CCC", "CD", "D",
"DC", "DCC", "DCCC", "CM "]
x = ["", "X", "XX", "XXX", "XL", "L",
"LX", "LXX", "LXXX", "XC"]
i = ["", "I", "II", "III", "IV", "V",
"VI", "VII", "VIII", "IX"]
thousands = m[num // 1000]
hundreds = c[(num % 1000) // 100]
tens = x[(num % 100) // 10]
ones = i[num % 10]
ans = (thousands + hundreds +
tens + ones)
return ans
if __name__ == "__main__":
number = 3549
print(intToRoman(number))
|
C#
using System;
class GFG {
static String intToRoman(int num)
{
String[] m = { "", "M", "MM", "MMM" };
String[] c = { "", "C", "CC", "CCC", "CD",
"D", "DC", "DCC", "DCCC", "CM" };
String[] x = { "", "X", "XX", "XXX", "XL",
"L", "LX", "LXX", "LXXX", "XC" };
String[] i = { "", "I", "II", "III", "IV",
"V", "VI", "VII", "VIII", "IX" };
String thousands = m[num / 1000];
String hundreds = c[(num % 1000) / 100];
String tens = x[(num % 100) / 10];
String ones = i[num % 10];
String ans = thousands + hundreds + tens + ones;
return ans;
}
public static void Main()
{
int number = 3549;
Console.WriteLine(intToRoman(number));
}
}
|
Javascript
<script>
function intToRoman(num)
{
let m = ["", "M", "MM", "MMM"];
let c = ["", "C", "CC", "CCC", "CD", "D",
"DC", "DCC", "DCCC", "CM"];
let x = ["", "X", "XX", "XXX", "XL", "L",
"LX", "LXX", "LXXX", "XC"];
let i = ["", "I", "II", "III", "IV", "V",
"VI", "VII", "VIII", "IX"];
let a1 = Math.floor(num/1000);
let a2 = Math.floor((num%1000)/100);
let a3 = Math.floor((num%100)/10);
let thousands = m[a1];
let hundreds = c[a2];
let tens = x[a3];
let ones = i[num%10];
let ans = thousands + hundreds + tens + ones;
return ans;
}
let number = 3549;
document.write(intToRoman(number));
</script>
|
PHP
<?php
function intToRoman($num)
{
$m = array("", "M", "MM", "MMM");
$c = array("", "C", "CC", "CCC", "CD", "D",
"DC", "DCC", "DCCC", "CM");
$x = array("", "X", "XX", "XXX", "XL", "L",
"LX", "LXX", "LXXX", "XC");
$i = array("", "I", "II", "III", "IV", "V",
"VI", "VII", "VIII", "IX");
$thousands = $m[$num / 1000];
$hundreds = $c[($num % 1000) / 100];
$tens = $x[($num % 100) / 10];
$ones = $i[$num % 10];
$ans = $thousands . $hundreds . $tens . $ones;
return $ans;
}
$number = 3549;
echo intToRoman($number);
|
Time Complexity: O(N), where N is the length of ans string that stores the conversion.
Auxiliary Space: O(N)
Thanks to Shashwat Jain for providing the above solution approach.
Another Approach 2:
In this approach we consider the main significant digit in the number. Ex: in 1234, main significant digit is 1. Similarly in 345 it is 3.
In order to extract main significant digit out, we need to maintain a divisor (lets call it div) like 1000 for 1234 (since 1234 / 1000 = 1) and 100 for 345 (345 / 100 = 3).
Also, lets maintain a dictionary called romanNumeral = {1 : ‘I’, 5: ‘V’, 10: ‘X’, 50: ‘L’, 100: ‘C’, 500: ‘D’, 1000: ‘M’}
Following is the algorithm:
if main significant digit <= 3
- romanNumeral[div] * mainSignificantDigit
if main significant digit == 4
- romanNumeral[div] + romanNumeral[div * 5]
if 5 <= main significant digit <=8
- romanNumeral[div * 5] + (romanNumeral[div] * ( mainSignificantDigit-5))
if main significant digit ==9
- romanNumeral[div] + romanNumeral[div*10]
Example:
Suppose the input number is 3649.
Iter 1
- Initially number = 3649
- main significant digit is 3. Div = 1000.
- So, romanNumeral[1000] * 3
- gives: MMM
Iter 2
- now, number = 649
- main significant digit is 6. Div = 100.
- So romanNumeral[100*5] + (romanNumeral[div] * ( 6-5))
- gives: DC
Iter 3
- now, number = 49
- main significant digit is 4. Div = 10.
- So romanNumeral[10] + romanNumeral[10 * 5]
- gives: XL
Iter 4
- now, number = 9
- main significant digit is 9. Div = 1.
- So romanNumeral[1] * romanNumeral[1*10]
- gives: IX
Final result by clubbing all the above: MMMDCXLIX
Below is the Python implementation of above idea.
C++
#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
string integerToRoman(int num) {
unordered_map<int, char> roman;
roman[1] = 'I';
roman[5] = 'V';
roman[10] = 'X';
roman[50] = 'L';
roman[100] = 'C';
roman[500] = 'D';
roman[1000] = 'M';
roman[5000] = 'G';
roman[10000] = 'H';
string tmp = to_string(num);
const int numDigits = tmp.length();
string res = "";
for(int i=0;i<numDigits;++i) {
const char src = tmp[i];
const int number = (src - '0');
const int place = (numDigits-1)-i;
const int absolute = pow(10, place);
if (number == 9) {
res.append(1, roman[absolute]);
res.append(1, roman[(number+1) * absolute]);
} else
if (number >= 5) {
res.append(1, roman[5*absolute]);
res.append(number-5, roman[absolute]);
} else
if (number >= 4) {
res.append(1, roman[absolute]);
res.append(1, roman[5*absolute]);
} else {
res.append(number, roman[absolute]);
}
}
return res;
}
int main() {
cout << integerToRoman(3549) << endl;
return 0;
}
|
Java
import java.util.*;
public class Solution {
static String integerToRoman(int num)
{
HashMap<Integer, Character> roman = new HashMap<>();
roman.put(1, 'I');
roman.put(5, 'V');
roman.put(10, 'X');
roman.put(50, 'L');
roman.put(100, 'C');
roman.put(500, 'D');
roman.put(1000, 'M');
roman.put(5000, 'G');
roman.put(10000, 'H');
String tmp = num + "";
int numDigits = tmp.length();
String res = "";
for (int i = 0; i < numDigits; ++i) {
char src = tmp.charAt(i);
int number = (src - '0');
int place = (numDigits - 1) - i;
int absolute = (int)Math.pow(10, place);
if (number == 9) {
res += roman.get(absolute);
res += roman.get(absolute * 10);
}
else if (number >= 5) {
res += roman.get(absolute * 5);
res += (roman.get(absolute) + "")
.repeat(number - 5);
}
else if (number == 4) {
res += roman.get(absolute);
res += roman.get(absolute * 5);
}
else {
res += (roman.get(absolute) + "")
.repeat(number);
}
}
return res;
}
public static void main(String[] args)
{
System.out.println("Roman Numeral of Integer is:"
+ integerToRoman(3549));
}
}
|
Python3
import math
def integerToRoman(A):
romansDict = \
{
1: "I",
5: "V",
10: "X",
50: "L",
100: "C",
500: "D",
1000: "M",
5000: "G",
10000: "H"
}
div = 1
while A >= div:
div *= 10
div //= 10
res = ""
while A:
lastNum = (A // div)
if lastNum <= 3:
res += (romansDict[div] * lastNum)
elif lastNum == 4:
res += (romansDict[div] +
romansDict[div * 5])
elif 5 <= lastNum <= 8:
res += (romansDict[div * 5] +
(romansDict[div] * (lastNum - 5)))
elif lastNum == 9:
res += (romansDict[div] +
romansDict[div * 10])
A = math.floor(A % div)
div //= 10
return res
print("Roman Numeral of Integer is:"
+ str(integerToRoman(3549)))
|
C#
using System;
using System.Collections.Generic;
public class GFG {
static String integerToRoman(int num)
{
Dictionary<int, char> roman
= new Dictionary<int, char>();
roman[1] = 'I';
roman[5] = 'V';
roman[10] = 'X';
roman[50] = 'L';
roman[100] = 'C';
roman[500] = 'D';
roman[1000] = 'M';
roman[5000] = 'G';
roman[10000] = 'H';
string tmp = num + "";
int numDigits = tmp.Length;
String res = "";
for (int i = 0; i < numDigits; ++i) {
char src = tmp[i];
int number = (src - '0');
int place = (numDigits - 1) - i;
int absolute = (int)Math.Pow(10, place);
if (number == 9) {
res += roman[absolute];
res += roman[absolute * 10];
}
else if (number >= 5) {
res += roman[absolute * 5];
res += new string(roman[absolute],
number - 5);
}
else if (number == 4) {
res += roman[absolute];
res += roman[absolute * 5];
}
else {
res += new string(roman[absolute], number);
}
}
return res;
}
public static void Main(string[] args)
{
Console.WriteLine("Roman Numeral of Integer is:"
+ integerToRoman(3549));
}
}
|
Javascript
function integerToRoman(num) {
let roman = new Map();
roman.set(1, 'I');
roman.set(5, 'V');
roman.set(10, 'X');
roman.set(50, 'L');
roman.set(100, 'C');
roman.set(500, 'D');
roman.set(1000, 'M');
roman.set(5000, 'G');
roman.set(10000, 'H');
let tmp = Array.from(String(num));
let numDigits = tmp.length;
let res = [];
for(let i=0;i<numDigits;++i) {
let src = tmp[i];
let number = (src.charCodeAt(0) - 48);
let place = (numDigits-1)-i;
let absolute = Math.pow(10, place);
if (number == 9) {
res.push(roman.get(absolute));
res.push(roman.get((number+1) * absolute));
} else
if (number >= 5) {
res.push(roman.get(5*absolute));
let cnt = number-5;
while(cnt--) res.push(roman.get(absolute));
}
else{
if (number >= 4) {
res.push(roman.get(absolute));
res.push(roman.get(5*absolute));
}
else {
let cnt = number;
while(cnt--) res.push(roman.get(absolute));
}
}
}
return res;
}
let ans = integerToRoman(3549).join('');
console.log(ans);
|
OutputRoman Numeral of Integer is:MMMDXLIX
Time Complexity: O(N), where N is the length of res string that stores the conversion.
Auxiliary Space: O(N)
Thanks to Sriharsha Sammeta for providing the above solution approach.
Approach 3: Using Ladder if-else
The idea is simple, it follows the basic fundamentals of roman numbers as iterating through the greatest to smallest number using if else and check of it exist then add that roman symbol to the answer and decrement that number.
Steps to solve the problem:
- Declare ans variable to store the roman symbol.
- Iterate through all the roman integer value from greatest to smallest until the number is not equal to zero:
- If num>=1000 then ans+=”M” and num-=1000.
- else if num>=900 && num<1000 then ans+=”CM” and num-=900, and so on till num is not zero
- 4. Return the ans
Implementation of the approach:
C++
#include <bits/stdc++.h>
using namespace std;
string intToRoman(int num)
{
string ans = "";
while (num) {
if (num >= 1000) {
ans += "M";
num -= 1000;
}
else if (num >= 900 && num < 1000) {
ans += "CM";
num -= 900;
}
else if (num >= 500 && num < 900) {
ans += "D";
num -= 500;
}
else if (num >= 400 && num < 500) {
ans += "CD";
num -= 400;
}
else if (num >= 100 && num < 400) {
ans += "C";
num -= 100;
}
else if (num >= 90 && num < 100) {
ans += "XC";
num -= 90;
}
else if (num >= 50 && num < 90) {
ans += "L";
num -= 50;
}
else if (num >= 40 && num < 50) {
ans += "XL";
num -= 40;
}
else if (num >= 10 && num < 40) {
ans += "X";
num -= 10;
}
else if (num == 9) {
ans += "IX";
num -= 9;
}
else if (num >= 5 && num < 9) {
ans += "V";
num -= 5;
}
else if (num == 4) {
ans += "IV";
num -= 4;
}
else if (num < 4) {
ans += "I";
num--;
}
}
return ans;
}
int main()
{
int number = 3549;
cout << intToRoman(number);
return 0;
}
|
Java
import java.util.*;
public class Solution {
static String integerToRoman(int num)
{
String ans = "";
while (num>0) {
if (num >= 1000) {
ans += "M";
num -= 1000;
}
else if (num >= 900 && num < 1000) {
ans += "CM";
num -= 900;
}
else if (num >= 500 && num < 900) {
ans += "D";
num -= 500;
}
else if (num >= 400 && num < 500) {
ans += "CD";
num -= 400;
}
else if (num >= 100 && num < 400) {
ans += "C";
num -= 100;
}
else if (num >= 90 && num < 100) {
ans += "XC";
num -= 90;
}
else if (num >= 50 && num < 90) {
ans += "L";
num -= 50;
}
else if (num >= 40 && num < 50) {
ans += "XL";
num -= 40;
}
else if (num >= 10 && num < 40) {
ans += "X";
num -= 10;
}
else if (num == 9) {
ans += "IX";
num -= 9;
}
else if (num >= 5 && num < 9) {
ans += "V";
num -= 5;
}
else if (num == 4) {
ans += "IV";
num -= 4;
}
else if (num < 4) {
ans += "I";
num--;
}
}
return ans;
}
public static void main(String[] args)
{
System.out.println("Roman Numeral of Integer is:"
+ integerToRoman(3549));
}
}
|
Python3
def intToRoman(num):
ans = ""
while(num > 0):
if(num >= 1000):
ans += "M"
num -= 1000
elif(num >= 900 and num < 1000):
ans += "CM"
num -= 900
elif(num >= 500 and num < 900):
ans += "D"
num -= 500
elif(num >= 400 and num < 500):
ans += "CD"
num -= 400
elif(num >= 100 and num < 400):
ans += "C"
num -= 100
elif(num >= 90 and num < 100):
ans += "XC"
num -= 90
elif(num >= 50 and num < 90):
ans += "L"
num -= 50
elif(num >= 40 and num < 50):
ans += "XL"
num -= 40
elif(num >= 10 and num < 40):
ans += "X"
num -= 10
elif(num == 9):
ans += "IX"
num -= 9
elif(num >= 5 and num < 9):
ans += "V"
num -= 5
elif(num == 4):
ans += "IV"
num -= 4
elif(num < 4):
ans += "I"
num = num - 1
return ans
number = 3549
print(intToRoman(number))
|
C#
using System;
using System.Collections.Generic;
public class GFG {
static string intToRoman(int num)
{
string ans = "";
while (num > 0) {
if (num >= 1000) {
ans += "M";
num -= 1000;
}
else if (num >= 900 && num < 1000) {
ans += "CM";
num -= 900;
}
else if (num >= 500 && num < 900) {
ans += "D";
num -= 500;
}
else if (num >= 400 && num < 500) {
ans += "CD";
num -= 400;
}
else if (num >= 100 && num < 400) {
ans += "C";
num -= 100;
}
else if (num >= 90 && num < 100) {
ans += "XC";
num -= 90;
}
else if (num >= 50 && num < 90) {
ans += "L";
num -= 50;
}
else if (num >= 40 && num < 50) {
ans += "XL";
num -= 40;
}
else if (num >= 10 && num < 40) {
ans += "X";
num -= 10;
}
else if (num == 9) {
ans += "IX";
num -= 9;
}
else if (num >= 5 && num < 9) {
ans += "V";
num -= 5;
}
else if (num == 4) {
ans += "IV";
num -= 4;
}
else if (num < 4) {
ans += "I";
num--;
}
}
return ans;
}
public static void Main(string[] args)
{
int number = 3549;
Console.WriteLine(intToRoman(number));
}
}
|
Javascript
function intToRoman(num)
{
let ans = "";
while(num != 0){
if(num >= 1000){
ans += "M";
num -= 1000;
}
else if(num >= 900 && num < 1000){
ans += "CM";
num -= 900;
}
else if (num >= 500 && num < 900) {
ans += "D";
num -= 500;
}
else if (num >= 400 && num < 500) {
ans += "CD";
num -= 400;
}
else if (num >= 100 && num < 400) {
ans += "C";
num -= 100;
}
else if (num >= 90 && num < 100) {
ans += "XC";
num -= 90;
}
else if (num >= 50 && num < 90) {
ans += "L";
num -= 50;
}
else if (num >= 40 && num < 50) {
ans += "XL";
num -= 40;
}
else if (num >= 10 && num < 40) {
ans += "X";
num -= 10;
}
else if (num == 9) {
ans += "IX";
num -= 9;
}
else if (num >= 5 && num < 9) {
ans += "V";
num -= 5;
}
else if (num == 4) {
ans += "IV";
num -= 4;
}
else if (num < 4) {
ans += "I";
num--;
}
}
return ans;
}
let number = 3549;
document.write(intToRoman(number));
|
Time Complexity: O(N), where N is the length of ans string that stores the conversion.
Auxiliary Space: O(N)
Approach 4: Using Recursion
- First, we define a function called printRoman which takes an integer num as input.
- The function checks whether the input integer is greater than or equal to 1000. If it is, it prints the corresponding Roman numeral “M” and recursively calls the function with the updated input by subtracting 1000 from it.
- If the input integer is not greater than or equal to 1000, the function checks whether it is greater than or equal to 900. If it is, it prints the corresponding Roman numeral “CM” and recursively calls the function with the updated input by subtracting 900 from it.
- This process is repeated for each Roman numeral, with the function checking whether the input integer is greater than or equal to the corresponding value for each numeral. If it is, the function prints the corresponding numeral and recursively calls the function with the updated input by subtracting the corresponding value from it.
- Finally, if the input integer is less than 1, the function simply returns and the recursion ends.
- In the main function, we define an integer number and initialize it to 3549. We then call the printRoman function with this integer as input.
- The printRoman function prints the corresponding Roman numeral representation of the input integer (MMMDXLIX) to the console.
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
void printRoman(int num) {
if(num >= 1000) {
cout << "M";
printRoman(num - 1000);
}
else if(num >= 900) {
cout << "CM";
printRoman(num - 900);
}
else if(num >= 500) {
cout << "D";
printRoman(num - 500);
}
else if(num >= 400) {
cout << "CD";
printRoman(num - 400);
}
else if(num >= 100) {
cout << "C";
printRoman(num - 100);
}
else if(num >= 90) {
cout << "XC";
printRoman(num - 90);
}
else if(num >= 50) {
cout << "L";
printRoman(num - 50);
}
else if(num >= 40) {
cout << "XL";
printRoman(num - 40);
}
else if(num >= 10) {
cout << "X";
printRoman(num - 10);
}
else if(num >= 9) {
cout << "IX";
printRoman(num - 9);
}
else if(num >= 5) {
cout << "V";
printRoman(num - 5);
}
else if(num >= 4) {
cout << "IV";
printRoman(num - 4);
}
else if(num >= 1) {
cout << "I";
printRoman(num - 1);
}
}
int main() {
int number = 3549;
printRoman(number);
return 0;
}
|
Java
import java.io.*;
public class GFG {
public static void printRoman(int num) {
if (num >= 1000) {
System.out.print("M");
printRoman(num - 1000);
} else if (num >= 900) {
System.out.print("CM");
printRoman(num - 900);
} else if (num >= 500) {
System.out.print("D");
printRoman(num - 500);
} else if (num >= 400) {
System.out.print("CD");
printRoman(num - 400);
} else if (num >= 100) {
System.out.print("C");
printRoman(num - 100);
} else if (num >= 90) {
System.out.print("XC");
printRoman(num - 90);
} else if (num >= 50) {
System.out.print("L");
printRoman(num - 50);
} else if (num >= 40) {
System.out.print("XL");
printRoman(num - 40);
} else if (num >= 10) {
System.out.print("X");
printRoman(num - 10);
} else if (num >= 9) {
System.out.print("IX");
printRoman(num - 9);
} else if (num >= 5) {
System.out.print("V");
printRoman(num - 5);
} else if (num >= 4) {
System.out.print("IV");
printRoman(num - 4);
} else if (num >= 1) {
System.out.print("I");
printRoman(num - 1);
}
}
public static void main(String[] args) {
int number = 3549;
printRoman(number);
}
}
|
Python3
def print_roman(num):
if num >= 1000:
print("M", end="")
print_roman(num - 1000)
elif num >= 900:
print("CM", end="")
print_roman(num - 900)
elif num >= 500:
print("D", end="")
print_roman(num - 500)
elif num >= 400:
print("CD", end="")
print_roman(num - 400)
elif num >= 100:
print("C", end="")
print_roman(num - 100)
elif num >= 90:
print("XC", end="")
print_roman(num - 90)
elif num >= 50:
print("L", end="")
print_roman(num - 50)
elif num >= 40:
print("XL", end="")
print_roman(num - 40)
elif num >= 10:
print("X", end="")
print_roman(num - 10)
elif num >= 9:
print("IX", end="")
print_roman(num - 9)
elif num >= 5:
print("V", end="")
print_roman(num - 5)
elif num >= 4:
print("IV", end="")
print_roman(num - 4)
elif num >= 1:
print("I", end="")
print_roman(num - 1)
number = 3549
print_roman(number)
|
C#
using System;
class Program {
static void PrintRoman(int num)
{
if (num >= 1000) {
Console.Write("M");
PrintRoman(num - 1000);
}
else if (num >= 900) {
Console.Write("CM");
PrintRoman(num - 900);
}
else if (num >= 500) {
Console.Write("D");
PrintRoman(num - 500);
}
else if (num >= 400) {
Console.Write("CD");
PrintRoman(num - 400);
}
else if (num >= 100) {
Console.Write("C");
PrintRoman(num - 100);
}
else if (num >= 90) {
Console.Write("XC");
PrintRoman(num - 90);
}
else if (num >= 50) {
Console.Write("L");
PrintRoman(num - 50);
}
else if (num >= 40) {
Console.Write("XL");
PrintRoman(num - 40);
}
else if (num >= 10) {
Console.Write("X");
PrintRoman(num - 10);
}
else if (num >= 9) {
Console.Write("IX");
PrintRoman(num - 9);
}
else if (num >= 5) {
Console.Write("V");
PrintRoman(num - 5);
}
else if (num >= 4) {
Console.Write("IV");
PrintRoman(num - 4);
}
else if (num >= 1) {
Console.Write("I");
PrintRoman(num - 1);
}
}
static void Main(string[] args)
{
int number = 3549;
PrintRoman(number);
}
}
|
Time complexity: O(n)
Auxiliary Space: O(n)
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