Roman to Integer Conversion in DSA C - Time & Space Complexity

Time Complexity: Roman to Integer Conversion

O(n)

Understanding Time Complexity

We want to know how the time to convert a Roman numeral to an integer changes as the input gets longer.

How does the number of steps grow when the Roman numeral string grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int romanToInt(char * s) {
    int total = 0;
    int i = 0;
    while (s[i]) {
        int current = value(s[i]);
        int next = value(s[i + 1]);
        if (next > current) {
            total += (next - current);
            i += 2;
        } else {
            total += current;
            i++;
        }
    }
    return total;
}

int value(char c) {
    switch(c) {
        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 0;
    }
}
    

This code converts a Roman numeral string to an integer by checking each character and sometimes the next one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that goes through each character in the string.
• How many times: It runs once for each character, sometimes skipping one extra character when two are processed together.

How Execution Grows With Input

As the Roman numeral string gets longer, the number of steps grows roughly the same as the string length.

Input Size (n)	Approx. Operations
10	About 10 checks
100	About 100 checks
1000	About 1000 checks

Pattern observation: The steps increase linearly with the input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to convert grows directly with the length of the Roman numeral string.

Common Mistake

[X] Wrong: "Because we sometimes skip characters, the time is less than linear or constant."

[OK] Correct: Even when skipping, each character is still checked once, so the total work grows with the input size.

Interview Connect

Understanding how to analyze simple loops like this helps you explain your code clearly and shows you know how input size affects performance.

Self-Check

"What if the input was given as a linked list of characters instead of a string? How would the time complexity change?"

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