Why Roman to Integer Conversion in DSA Rust?

The Big Idea

What if you could turn ancient Roman numerals into numbers instantly without mistakes?

The Scenario

Imagine you have a list of Roman numerals written on old scrolls, and you need to add them up or compare their values. Doing this by hand means remembering all the rules and subtracting or adding values carefully every time.

The Problem

Manually converting Roman numerals is slow and easy to mess up. You might forget when to subtract or add, especially with tricky cases like IV (4) or IX (9). This leads to mistakes and wastes time.

The Solution

Roman to Integer Conversion automates this process. It uses a simple set of rules to quickly and correctly turn Roman numerals into numbers, so you can do math or comparisons easily without errors.

Before vs After

✗ Before

let mut total = 0;
for ch in roman.chars() {
    if ch == 'I' { total += 1; }
    else if ch == 'V' { total += 5; }
    // ... more rules here
}

✓ After

let integer = roman_to_int(roman);
println!("{}", integer);

What It Enables

This lets you quickly and reliably work with Roman numerals in programs, unlocking tasks like calculations, sorting, and data processing.

Real Life Example

Think of a museum cataloging ancient artifacts labeled with Roman numerals. Converting these labels to numbers helps organize and search the collection efficiently.

Key Takeaways

Manual conversion is slow and error-prone.

Automated conversion uses clear rules to simplify the process.

It enables fast, accurate number handling from Roman numerals.

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