Roman to Integer Conversion in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the integer value of the Roman numeral 'I'.

DSA C

int romanToInt(char * s) {
    if (s[0] == 'I') {
        return [1];
    }
    return 0;
}

Drag options to blanks, or click blank then click option'

A1

B0

C10

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of 1 for 'I'.

Confusing 'I' with 'V' or 'X'.

The Roman numeral 'I' equals 1, so the function should return 1.

2fill in blank

medium

Complete the code to get the integer value of a single Roman numeral character.

DSA C

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

Drag options to blanks, or click blank then click option'

A-1

B1

C0

D1000

Attempts:

3 left

💡 Hint

Common Mistakes

Returning -1 which may cause incorrect calculations.

Returning 1 which is a valid numeral value and may cause confusion.

If the character is not a valid Roman numeral, returning 0 is safe to indicate no value.

3fill in blank

hard

Fix the error in the loop condition to correctly iterate over the string until the end.

DSA C

int romanToInt(char * s) {
    int total = 0;
    int i = 0;
    while (s[i] [1] '\0') {
        total += value(s[i]);
        i++;
    }
    return total;
}

Drag options to blanks, or click blank then click option'

A!=

B==

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' which stops immediately if first char is not '\0'.

Using '<' or '>' which are invalid for char comparison here.

The loop should continue while the current character is not the null character '\0'.

4fill in blank

hard

Fill both blanks to correctly subtract when a smaller numeral precedes a larger one.

DSA C

int romanToInt(char * s) {
    int total = 0;
    int i = 0;
    while (s[i] != '\0') {
        int s1 = value(s[i]);
        if (s[i + 1] != '\0') {
            int s2 = value(s[i + 1]);
            if (s1 [1] s2) {
                total += s2 - s1;
                i += [2];
            } else {
                total += s1;
                i++;
            }
        } else {
            total += s1;
            i++;
        }
    }
    return total;
}

Drag options to blanks, or click blank then click option'

A<

B2

C>

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' for comparison.

Incrementing i by 1 instead of 2 after subtraction.

If the current value is less than the next, subtract current from next and move two steps ahead.

5fill in blank

hard

Fill all three blanks to create a dictionary-like structure mapping Roman numerals to integers using arrays.

DSA C

typedef struct {
    char symbol;
    int value;
} RomanMap;

RomanMap map[] = {
    {'I', [1],
    {'V', [2],
    {'X', [3]
};

Drag options to blanks, or click blank then click option'

A1

B5

C10

D50

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up values for 'V' and 'X'.

Using incorrect integer values.

The Roman numerals 'I', 'V', and 'X' correspond to 1, 5, and 10 respectively.

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