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DSA Cprogramming~5 mins

Roman to Integer Conversion in DSA C - Cheat Sheet & Quick Revision

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Recall & Review
beginner
What is the value of the Roman numeral 'IV'?
The value of 'IV' is 4. 'I' before 'V' means 1 is subtracted from 5.
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beginner
Explain the subtraction rule in Roman numerals.
If a smaller numeral appears before a larger one, subtract the smaller from the larger. For example, 'IX' is 9 (10 - 1).
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intermediate
What is the integer value of 'MCMXCIV'?
'MCMXCIV' equals 1994. Breakdown: M(1000) + CM(900) + XC(90) + IV(4).
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intermediate
In C, how can you map Roman numerals to their integer values efficiently?
Use a function or array to map characters 'I', 'V', 'X', 'L', 'C', 'D', 'M' to 1, 5, 10, 50, 100, 500, 1000 respectively.
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beginner
Why do we check the next Roman numeral character when converting to integer?
To decide if we add or subtract the current numeral's value. If the next numeral is larger, subtract current; else add it.
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What integer does the Roman numeral 'XL' represent?
A40
B60
C90
D50
Which Roman numeral represents the number 100?
AD
BL
CC
DM
In Roman to integer conversion, what do you do if the current numeral is smaller than the next one?
AAdd current numeral's value
BSubtract current numeral's value
CIgnore current numeral
DMultiply current numeral's value
What is the integer value of 'D' in Roman numerals?
A500
B50
C100
D1000
Which of these is NOT a valid Roman numeral?
AMMXV
BCM
CXLII
DIC
Describe the step-by-step process to convert a Roman numeral string to an integer.
Think about how subtraction rule affects addition.
You got /5 concepts.
    Explain why the Roman numeral 'IX' equals 9 and not 11.
    Look at the order of numerals and their values.
    You got /4 concepts.

      Practice

      (1/5)
      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
      Recommended

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