Why Roman to Integer Conversion in DSA C?

The Big Idea

What if you could turn ancient Roman numbers into modern digits with just a few lines of code?

The Scenario

Imagine you have a list of Roman numerals written on old scrolls, and you need to add their values to find the total. Doing this by hand means remembering all the rules and converting each symbol carefully.

The Problem

Manually converting Roman numerals is slow and easy to mess up. You might forget when to subtract instead of add, or mix up symbols, leading to wrong totals and frustration.

The Solution

Roman to Integer Conversion automates this process. It uses simple rules to scan the numeral string and calculate the correct number quickly and without mistakes.

Before vs After

✗ Before

int value = 0;
// Check each symbol and add or subtract manually
if (s[0] == 'I') value += 1;
if (s[1] == 'V') value += 5; // but what if IV means 4?

✓ After

int romanToInt(char* s) {
  int total = 0;
  for (int i = 0; s[i]; i++) {
    // Add or subtract based on next symbol
  }
  return total;
}

What It Enables

This lets you quickly and accurately convert any Roman numeral into a number, enabling calculations, comparisons, and data processing.

Real Life Example

Historians digitizing ancient texts can convert Roman dates into numbers automatically, saving hours of manual work and avoiding errors.

Key Takeaways

Manual conversion is slow and error-prone.

Automated conversion uses rules to add or subtract values correctly.

This makes working with Roman numerals fast and reliable.

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