Roman to Integer Conversion in DSA Rust - 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 grows as the input gets longer.

How does the number of steps change when the Roman numeral string is bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn roman_to_int(s: &str) -> i32 {
    let mut total = 0;
    let mut prev = 0;
    for c in s.chars().rev() {
        let value = match c {
            'I' => 1,
            'V' => 5,
            'X' => 10,
            'L' => 50,
            'C' => 100,
            'D' => 500,
            'M' => 1000,
            _ => 0,
        };
        if value < prev {
            total -= value;
        } else {
            total += value;
        }
        prev = value;
    }
    total
}
    

This code converts a Roman numeral string to an integer by reading characters from right to left and adding or subtracting values.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character of the input string once.
• How many times: Exactly once for each character in the string.

How Execution Grows With Input

As the Roman numeral string gets longer, the number of steps grows directly with its length.

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

Pattern observation: The steps increase in a straight line as input size grows.

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: "The code does multiple passes or nested loops, so it must be slower than linear."

[OK] Correct: The code only loops once through the string, doing simple checks each time, so it is linear time.

Interview Connect

Understanding this helps you explain how to efficiently process strings character by character, a common skill in coding interviews.

Self-Check

"What if we changed the code to use recursion instead of a loop? How would the time complexity change?"

Practice

1. What is the main rule to convert Roman numerals to integers?

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 the same

D. Subtract values of all symbols

Solution

1. Step 1: Understand Roman numeral rules

   Roman numerals are added normally, but if a smaller numeral appears before a larger one, it is subtracted.

2. Step 2: Apply the rule to conversion

   When converting, add each symbol's value unless a smaller symbol precedes a larger one, then subtract the smaller.

3. Final Answer:

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

4. Quick Check:

   Roman to integer = add or subtract based on order [OK]

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

Common Mistakes:

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

2. Which Rust code snippet correctly defines a function header for converting a Roman numeral string to an integer?

easy

A. fn roman_to_int(s: &str) -> i32 {

B. fn romanToInt(s: String) -> int {

C. function roman_to_int(s: &str) -> i32 {

D. fn roman_to_int(s: &String) -> i64 {

Solution

1. Step 1: Check Rust function syntax

   Rust functions start with fn, use snake_case, and specify parameter types and return type correctly.

2. Step 2: Validate each option

   fn roman_to_int(s: &str) -> i32 { uses correct Rust syntax: fn, snake_case, &str parameter, and i32 return type. Others have syntax errors or wrong types.

3. Final Answer:

   fn roman_to_int(s: &str) -> i32 { -> Option A

4. Quick Check:

   Rust fn header with &str and i32 return [OK]

Hint: Use fn, snake_case, &str param, i32 return [OK]

Common Mistakes:

• Using camelCase instead of snake_case
• Wrong parameter or return types
• Using 'function' keyword (not Rust)

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

fn main() {
    let s = "IX";
    let result = roman_to_int(s);
    println!("{}", result);
}

fn roman_to_int(s: &str) -> i32 {
    let mut total = 0;
    let mut prev = 0;
    for c in s.chars().rev() {
        let value = match c {
            'I' => 1,
            'V' => 5,
            'X' => 10,
            _ => 0,
        };
        if value < prev {
            total -= value;
        } else {
            total += value;
        }
        prev = value;
    }
    total
}

medium

A. 6

B. 11

C. Error

D. 9

Solution

1. Step 1: Trace characters in reverse

   Input "IX" reversed is 'X' then 'I'. 'X' = 10, 'I' = 1.

2. Step 2: Calculate total with subtraction rule

   Start total=0, prev=0. 'X' (10) >= prev(0), total=10, prev=10. Next 'I' (1) < prev(10), total=10-1=9, prev=1.

3. Final Answer:

   9 -> Option D

4. Quick Check:

   "IX" = 9 [OK]

Hint: Traverse string backward, subtract if smaller before larger [OK]

Common Mistakes:

• Not reversing string for correct subtraction
• Adding all values without subtraction
• Misassigning values to characters

4. Identify the bug in this Rust function for Roman to integer conversion:

fn roman_to_int(s: &str) -> i32 {
    let mut total = 0;
    let mut prev = 0;
    for c in s.chars() {
        let value = match c {
            'I' => 1,
            'V' => 5,
            'X' => 10,
            _ => 0,
        };
        if value > prev {
            total += value - 2 * prev;
        } else {
            total += value;
        }
        prev = value;
    }
    total
}

medium

A. Loop should iterate in reverse order

B. Subtracting twice prev is incorrect

C. Match arms missing some Roman numerals

D. prev variable should be mutable

Solution

1. Step 1: Inspect the match expression

   The match only handles 'I'(1), 'V'(5), 'X'(10). It misses 'L'(50), 'C'(100), 'D'(500), 'M'(1000), causing other symbols to map to 0 and produce wrong results.

2. Step 2: Verify other options

   A: Forward iteration with 'value - 2*prev' is valid for handling subtractive notation. B: The 2*prev subtraction corrects the previously added prev value. D: 'prev' is already mutable.

3. Final Answer:

   Match arms missing some Roman numerals -> Option C

4. Quick Check:

   Match must cover I,V,X,L,C,D,M [OK]

Hint: Include match arms for all Roman numerals [OK]

Common Mistakes:

• Incomplete match arms
• Not handling all Roman numerals
• Forgetting to make prev mutable (already mutable here)

5. Given the Roman numeral string "MCMXCIV", what integer does the following Rust function return?

fn roman_to_int(s: &str) -> i32 {
    let mut total = 0;
    let mut prev = 0;
    for c in s.chars().rev() {
        let value = match c {
            'I' => 1,
            'V' => 5,
            'X' => 10,
            'L' => 50,
            'C' => 100,
            'D' => 500,
            'M' => 1000,
            _ => 0,
        };
        if value < prev {
            total -= value;
        } else {
            total += value;
        }
        prev = value;
    }
    total
}

hard

A. 1994

B. 1904

C. 2004

D. None of the above

Solution

1. Step 1: Reverse and assign values

   Reverse "MCMXCIV" to "VICXMCM". Values: V=5, I=1, C=100, X=10, M=1000, C=100, M=1000.

2. Step 2: Calculate total with subtraction rule

   Start total=0, prev=0. - V(5) >= prev(0): total=5, prev=5 - I(1) < prev(5): total=5-1=4, prev=1 - C(100) >= prev(1): total=4+100=104, prev=100 - X(10) < prev(100): total=104-10=94, prev=10 - M(1000) >= prev(10): total=94+1000=1094, prev=1000 - C(100) < prev(1000): total=1094-100=994, prev=100 - M(1000) >= prev(100): total=994+1000=1994, prev=1000

3. Final Answer:

   1994 -> Option A

4. Quick Check:

   "MCMXCIV" = 1994 [OK]

Hint: Reverse string, subtract if smaller before larger [OK]

Common Mistakes:

• Not reversing string before processing
• Misassigning values to symbols
• Ignoring subtraction rule