Roman to Integer Conversion in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the result variable to zero.

DSA Rust

let mut result = [1];

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

A1

BNone

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing result to 1 or None causes wrong calculations.

The result variable should start at 0 to accumulate the integer value.

2fill in blank

medium

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

DSA Rust

let value = match c {
    'I' => 1,
    'V' => 5,
    'X' => 10,
    'L' => 50,
    'C' => 100,
    'D' => 500,
    'M' => 1000,
    _ => [1],
};

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

A0

Bpanic!("Invalid character")

C-1

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 or None silently ignores invalid input.

If the character is invalid, the code should panic to indicate an error.

3fill in blank

hard

Fix the error in the condition to check if the current value is less than the next value.

DSA Rust

if value [1] next_value {
    result -= value;
} else {
    result += value;
}

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

A>

B>=

C<

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' reverses the logic and causes wrong results.

If the current value is less than the next, subtract it according to Roman numeral rules.

4fill in blank

hard

Fill both blanks to iterate over the characters and get the next value safely.

DSA Rust

for i in 0..s.len() {
    let c = s.chars().nth(i).unwrap();
    let value = roman_to_int(c);
    let next_value = if i + 1 < s.len() [1] else [2];
    // rest of logic
}

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

Aroman_to_int(s.chars().nth(i + 1).unwrap())

B0

CNone

Dpanic!("No next char")

Attempts:

3 left

💡 Hint

Common Mistakes

Using panic! causes program to crash unnecessarily.

If there is a next character, get its value; otherwise, use 0 to avoid errors.

5fill in blank

hard

Fill all three blanks to complete the function that converts a Roman numeral string to an integer.

DSA Rust

fn roman_to_integer(s: &str) -> i32 {
    let mut result = [1];
    for i in 0..s.len() {
        let c = s.chars().nth(i).unwrap();
        let value = match c {
            'I' => 1,
            'V' => 5,
            'X' => 10,
            'L' => 50,
            'C' => 100,
            'D' => 500,
            'M' => 1000,
            _ => panic!("Invalid character"),
        };
        let next_value = if i + 1 < s.len() [2] else [3];
        if value < next_value {
            result -= value;
        } else {
            result += value;
        }
    }
    result
}

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

A0

Broman_to_integer(s.chars().nth(i + 1).unwrap())

Droman_to_int(s.chars().nth(i + 1).unwrap())

Attempts:

3 left

💡 Hint

Common Mistakes

Calling roman_to_integer recursively causes infinite loop.

Initialize result to 0; get next value with roman_to_int; use 0 if no next char.

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