Converting Decimal Number lying between 1 to 3999 to Roman Numerals

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: 

MMMDXLIX

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: 

1. Divide the given number into digits at different places like one’s, two’s, hundred’s or thousand’s.
2. 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.
3. 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. 

MMMDXLIX

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.

Roman 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:

MMMDXLIX

Time Complexity: O(N), where N is the length of ans string that stores the conversion.
Auxiliary Space: O(N)

Approach 4: Using Recursion 

1. First, we define a function called printRoman which takes an integer num as input.
2. 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.
3. 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.
4. 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.
5. Finally, if the input integer is less than 1, the function simply returns and the recursion ends.
6. 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.
7. 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: