Roman to Integer Conversion in DSA C - Build from Scratch

Roman to Integer Conversion

📖 Scenario: You are building a simple tool to convert Roman numerals into regular numbers. Roman numerals are used in clocks, books, and old movies. Your program will help translate these symbols into numbers we use every day.

🎯 Goal: Build a program that takes a Roman numeral string and converts it into its integer value using basic C programming concepts.

📋 What You'll Learn

Create a function to map Roman numeral characters to their integer values

Use a loop to process each character in the Roman numeral string

Handle the subtraction rule where a smaller numeral before a larger numeral means subtraction

Print the final integer result

💡 Why This Matters

🌍 Real World

Roman numerals are used in clocks, book chapters, movie titles, and historical documents. Converting them to integers helps in calculations and comparisons.

💼 Career

Understanding string processing, conditional logic, and loops in C is fundamental for software development, embedded systems, and working with legacy data formats.

1

Create the Roman numeral string

Create a character array called roman and initialize it with the string "MCMXCIV" which represents the Roman numeral for 1994.

DSA C

char roman[] = "MCMXCIV";  // Your code here

Hint

Use double quotes to create a string in C, and assign it to a character array.

2

Create a function to get integer value of Roman characters

Write a function called value that takes a char parameter r and returns the integer value of the Roman numeral character. Use switch to return 1000 for 'M', 500 for 'D', 100 for 'C', 50 for 'L', 10 for 'X', 5 for 'V', and 1 for 'I'. Return 0 for any other character.

DSA C

char roman[] = "MCMXCIV";

int value(char r) {
    // Your code here
}

Hint

Use a switch statement to return the correct integer for each Roman numeral character.

3

Convert the Roman numeral to integer

Write a function called romanToInt that takes the character array roman and returns the integer value. Use a for loop to iterate through the string. For each character, compare its value with the next character's value. If the current value is less than the next, subtract it from the result; otherwise, add it. Use the value function to get integer values.

DSA C

char roman[] = "MCMXCIV";

int value(char r) {
    switch(r) {
        case 'M': return 1000;
        case 'D': return 500;
        case 'C': return 100;
        case 'L': return 50;
        case 'X': return 10;
        case 'V': return 5;
        case 'I': return 1;
        default: return 0;
    }
}

int romanToInt(char* roman) {
    // Your code here
}

Hint

Use a while loop to check each character and compare with the next. Add or subtract accordingly.

4

Print the converted integer

In the main function, call romanToInt with the roman string and print the returned integer using printf.

DSA C

char roman[] = "MCMXCIV";

int value(char r) {
    switch(r) {
        case 'M': return 1000;
        case 'D': return 500;
        case 'C': return 100;
        case 'L': return 50;
        case 'X': return 10;
        case 'V': return 5;
        case 'I': return 1;
        default: return 0;
    }
}

int romanToInt(char* roman) {
    int result = 0;
    int i = 0;
    while (roman[i]) {
        int s1 = value(roman[i]);
        int s2 = value(roman[i + 1]);
        if (s1 < s2) {
            result += s2 - s1;
            i += 2;
        } else {
            result += s1;
            i++;
        }
    }
    return result;
}

int main() {
    // Your code here
}

Hint

Call romanToInt with roman and print the result using printf.

Practice

1. What is the main rule to convert Roman numerals to integers?
Example: IV = 4, VI = 6

easy

A. Multiply values of all symbols

B. Add values of symbols, subtract if a smaller symbol is before a larger one

C. Add values only if symbols are in alphabetical order

D. Subtract values of all symbols

Solution

1. Step 1: Understand Roman numeral rules

   Roman numerals add values of symbols, but if a smaller symbol appears before a larger one, it is subtracted.

2. Step 2: Apply rule to example

   For IV, I (1) is before V (5), so subtract 1 from 5 = 4. For VI, V (5) before I (1), so add 5 + 1 = 6.

3. Final Answer:

   Add values of symbols, subtract if a smaller symbol is before a larger one -> Option B

4. Quick Check:

   IV = 4, VI = 6 [OK]

Hint: Subtract when smaller symbol precedes larger one [OK]

Common Mistakes:

• Always adding without subtraction
• Multiplying symbols instead of adding
• Ignoring order of symbols

2. Which of the following C code snippets correctly initializes a map for Roman numeral values?

easy

A. int values = {'I':1, 'V':5, 'X':10};

B. int values[] = {I=1, V=5, X=10};

C. int values[26] = {['I'-'A']=1, ['V'-'A']=5, ['X'-'A']=10};

D. int values[3] = {1, 5, 10};

Solution

1. Step 1: Understand C array initialization for character keys

   In C, to map characters to values, an array indexed by character offset is used, e.g., values['I'-'A'] = 1.

2. Step 2: Check each option

   int values[26] = {['I'-'A']=1, ['V'-'A']=5, ['X'-'A']=10}; uses correct syntax for sparse initialization. Options B and C use invalid syntax for C. int values[3] = {1, 5, 10}; initializes array without mapping to characters.

3. Final Answer:

   int values[26] = {['I'-'A']=1, ['V'-'A']=5, ['X'-'A']=10}; -> Option C

4. Quick Check:

   Correct C syntax for char-indexed array [OK]

Hint: Use array with char offset for mapping in C [OK]

Common Mistakes:

• Using dictionary syntax not valid in C
• Not indexing array by character offset
• Initializing array without specifying indices

3. What is the output of this C code snippet?

char *s = "MCMXCIV";
int values[26] = {['I'-'A']=1, ['V'-'A']=5, ['X'-'A']=10, ['L'-'A']=50, ['C'-'A']=100, ['D'-'A']=500, ['M'-'A']=1000};
int result = 0, i = 0;
while (s[i]) {
  int curr = values[s[i]-'A'];
  int next = values[s[i+1]-'A'];
  if (next > curr) result -= curr;
  else result += curr;
  i++;
}
printf("%d", result);

medium

A. Error

B. 1904

C. 1494

D. 1994

Solution

1. Step 1: Trace each character and apply addition/subtraction

   String: M(1000), C(100), M(1000), X(10), C(100), I(1), V(5). Check pairs:
   M(1000) next C(100) -> 100 < 1000? No, add 1000
   C(100) next M(1000) -> 1000 > 100, subtract 100
   M(1000) next X(10) -> 10 < 1000, add 1000
   X(10) next C(100) -> 100 > 10, subtract 10
   C(100) next I(1) -> 1 < 100, add 100
   I(1) next V(5) -> 5 > 1, subtract 1
   V(5) next 0, add 5

2. Step 2: Calculate total

   1000 - 100 + 1000 - 10 + 100 - 1 + 5 = 1994

3. Final Answer:

   1994 -> Option D

4. Quick Check:

   MCMXCIV = 1994 [OK]

Hint: Subtract when next symbol is larger [OK]

Common Mistakes:

• Adding all values without subtraction
• Off-by-one error accessing next character
• Incorrect mapping of Roman symbols

4. Find the bug in this C code for Roman to Integer conversion:

int romanToInt(char *s) {
  int values[26] = {['I'-'A']=1, ['V'-'A']=5, ['X'-'A']=10, ['L'-'A']=50, ['C'-'A']=100, ['D'-'A']=500, ['M'-'A']=1000};
  int result = 0, i = 0;
  while (s[i]) {
    int curr = values[s[i]-'A'];
    int next = values[s[i+1]-'A'];
    if (curr < next) result += curr;
    else result -= curr;
    i++;
  }
  return result;
}

medium

A. The addition and subtraction logic is reversed

B. Array indexing causes out-of-bounds error

C. Values array is not initialized correctly

D. Function does not handle empty string

Solution

1. Step 1: Analyze addition and subtraction conditions

   Code adds current value if curr < next, else subtracts. This is opposite of Roman rules where smaller before larger means subtract.

2. Step 2: Correct logic

   It should subtract curr if curr < next, else add curr.

3. Final Answer:

   The addition and subtraction logic is reversed -> Option A

4. Quick Check:

   Subtract when smaller before larger [OK]

Hint: Subtract if current < next, else add [OK]

Common Mistakes:

• Swapping add and subtract conditions
• Ignoring last character's next value
• Not initializing values array properly

5. Given the Roman numeral string "XLIX", which of the following C functions correctly returns its integer value?

hard

A. int romanToInt(char *s) { int values[26] = {['I'-'A']=1, ['V'-'A']=5, ['X'-'A']=10, ['L'-'A']=50}; int res=0,i=0; while(s[i]) { int curr=values[s[i]-'A'], next=values[s[i+1]-'A']; if(curr

B. int romanToInt(char *s) { int values[26] = {['I'-'A']=1, ['V'-'A']=5, ['X'-'A']=10, ['L'-'A']=50}; int res=0,i=0; while(s[i]) { int curr=values[s[i]-'A'], next=values[s[i+1]-'A']; if(curr>next) res-=curr; else res+=curr; i++; } return res; }

C. int romanToInt(char *s) { int values[26] = {['I'-'A']=1, ['V'-'A']=5, ['X'-'A']=10, ['L'-'A']=50}; int res=0,i=0; while(s[i]) { int curr=values[s[i]-'A']; res+=curr; i++; } return res; }

D. int romanToInt(char *s) { int values[26] = {['I'-'A']=1, ['V'-'A']=5, ['X'-'A']=10, ['L'-'A']=50}; int res=0,i=0; while(s[i]) { int curr=values[s[i]-'A'], next=values[s[i+1]-'A']; if(curr<=next) res-=curr; else res+=curr; i++; } return res; }

Solution

1. Step 1: Check logic for subtraction and addition

   int romanToInt(char *s) { int values[26] = {['I'-'A']=1, ['V'-'A']=5, ['X'-'A']=10, ['L'-'A']=50}; int res=0,i=0; while(s[i]) { int curr=values[s[i]-'A'], next=values[s[i+1]-'A']; if(curr

2. Step 2: Verify with example "XLIX"

   X(10) < L(50) subtract 10
   L(50) > I(1) add 50
   I(1) < X(10) subtract 1
   X(10) no next add 10
   Total = -10 + 50 - 1 + 10 = 49

3. Final Answer:

   Option A correctly returns 49 for "XLIX" -> Option A

4. Quick Check:

   Subtract if smaller before larger [OK]

Hint: Subtract if current < next, else add [OK]

Common Mistakes:

• Using <= instead of < for subtraction
• Adding all values without subtraction
• Incorrect comparison operators