# Convert integers to roman numerals equivalent in Java

Converting Integers to Roman Numerals equivalent in Java

Let's understand what is the Input and the expected output.

Input : 99
Output: XCIX

Input : 81
Output: LXXXI

Input : 0
Output: not defined

Algorithm

STEP 1:
Note down all Unique characters where Roman numbers deviated from usual Pattern and put them in map.

Take an example,
Roman equivalent of '1' is 'I'. So we will add this in map,
map.put(1, "I"); Now no need to add Roman equivalent of '2' as it can be formed from equivalent of '1'
(taking 'I' twice),
Same for 3 (taking 'I' thrice).

This is not the case with Roman equivalent of '4', it has different pattern and not the ('IIII'),
so add it in map. map.put(4, "IV");

Note:
Unique patterns noted here are to support integers from 1 to 399 as program only support till 399.

map.put(100, "C");
map.put(90, "XC");
map.put(50, "L");
map.put(40, "XL");
map.put(10, "X");
map.put(9, "IX");
map.put(5, "V");
map.put(4, "IV");
map.put(1, "I");

If we want program to support more integers then identify patterns where Roman numbers have unusual patterns after 399 and add it in map.

STEP 2:
For converting Integer to Roman equivalent, we will start comparing given Integer with largest number in map,

Eg:
map.put(100, "C");
map.put(90, "XC");
map.put(50, "L");
map.put(40, "XL");
map.put(10, "X");
map.put(9, "IX");
map.put(5, "V");
map.put(4, "IV");
map.put(1, "I");

check how many times given number(say 153) has 100 in it(1 time, So pick "C" and remove 100 from 153, remaining number is 53),
100: Divide 153/100 = 1 time in 153 (remaining number is 153 - 100 = 53), remaining number is 153%100 = 53

then check how many time remaining number(53) has 90 in it(0 time),
90: 53/90 = 0 time in 53 (remaining number 53), remaining number is 53%90 = 53

then check how many time remaining number(53) has 50 in it(1 time, So pick "L" and remove 50 from 53, remaining number is 3),
50: 53/50 = 1 time in 53 (remaining number is  53 - 50 = 3), remaining number is 53 % 50 = 3

then check how many time remaining number(3) has 40 in it(0 time),
40: 3/40 = 0 time in 40 (remaining number 3), remaining number is 3%40 = 3

then check how many time remaining number(3) has 10 in it(0 time),
10: 3/10 = 0 time in 10 (remaining number 3), remaining number is 3%10 = 3

then check how many time remaining number(3) has 9 in it(0 time),
9: 3/9 = 0 time in 9 (remaining number 3), remaining number is 3%9 = 3

then check how many time remaining number(3) has 5 in it(0 time),
5: 3/5 = 0 time in 5 (remaining number 3), remaining number is 3%5 = 3

then check how many time remaining number(3) has 4 in it(0 time),
4: 3/4 = 0 time in 4 (remaining number 3), remaining number is 3%4 = 3

then check how many time remaining number has(3) 1 in it(3 time, So pick "I"*3 = "III" and remove 1 thrice from 3, remaining number is 0),
1: 3/1 = 3 time in 1 (remaining number is 3 - (1+1+1) = 0), remaining number is 3%1 = 0

We reach 0, no more number present, Stop here are return "CLIII"

Convert Integer to Roman numerals Java Program

```package miscellaneous;

import java.util.Map;

public class IntegerToRomanNumber {

public static void main(String[] args) {
System.out.println(getRomanEquivalentOfInteger(399));
}

private static String getRomanEquivalentOfInteger(int number){
if(number<=0){
return "not defined";
}

//Noting down all Unique characters where Roman numbers deviated from usual Pattern.
//unique patterns noted here are to support integers from 1 to 399 as program only support till 399.
//if we want program to support more integers then identify patterns where Roman numbers have unusual patterns after 399 and add it in map.
Map<Integer, String> map = new LinkedHashMap<Integer, String>();
map.put(100, "C");
map.put(90, "XC");
map.put(50, "L");
map.put(40, "XL");
map.put(10, "X");
map.put(9, "IX");
map.put(5, "V");
map.put(4, "IV");
map.put(1, "I");

String romanEqui="";

// Iterate map, check how many times given number has 100 in it, then check how many time remaining number has 90 in it and so on.
// or we can also say, is number divisible by 100, remaining number is divisible by 90 and so on.
// if number is 153, then first will see how many time number has 100 in it, which is 1 time.
// 100 - 1 time in 150 (remaining number is 150 - 100 = 53) OR 153/100 = 1 remaining 153%100 = 53
// 90  - 0 time in  53 (remaining number is  53 -  90 = 0)  OR  53/90 = 0 remaining   53 % 90 = 53 (we only need to find perfectly divisible numbers.)
// 50  - 1 time in  53 (remaining number is  53 -  50 = 3)  OR  53/50 = 1 remaining   53 % 50 = 3
// 40  - 0 time in   3 (remaining number is   3 -  40 = 0)  OR   3/40 = 0 remaining    3 % 40 = 3
// 10  - 0 time in   3 (remaining number is   3 -  10 = 0)  OR   3/10 = 0 remaining    3 % 10 = 3
// 9   - 0 time in   3 (remaining number is   3 -   9 = 0)  OR    3/9 = 0 remaining    3 %  9 = 3
// 5   - 0 time in   3 (remaining number is   3 -   5 = 0)  OR    3/5 = 0 remaining    3 %  5 = 3
// 4   - 0 time in   3 (remaining number is   3 -   4 = 0)  OR    3/4 = 0 remaining    3 %  4 = 3
// 1   - 3 time in   3 (remaining number is   3 -   1 = 0)  OR    3/1 = 3 remaining    3 %  1 = 0
for (Map.Entry<Integer, String> entry : map.entrySet()) {
int key = entry.getKey();
if(number/key!=0){
for (int i = 0; i < (number/key); i++) {
romanEqui = romanEqui + map.get(key);
}
number = number % key;
}
}
return romanEqui;
}
}

```

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Enjoy !!!!

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