Recursion vs Iteration When Each Wins in DSA Rust - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate factorial using recursion.

DSA Rust

fn factorial(n: u32) -> u32 {
    if n == 0 {
        1
    } else {
        n * factorial([1])
    }
}

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

An + 1

Bn - 1

Cn

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using n + 1 causes infinite recursion.

Using n causes no progress towards base case.

Factorial uses recursion by multiplying n with factorial of n-1 until n reaches 0.

2fill in blank

medium

Complete the code to calculate factorial using iteration.

DSA Rust

fn factorial_iter(n: u32) -> u32 {
    let mut result = 1;
    for i in 1..=[1] {
        result *= i;
    }
    result
}

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An

Bn + 1

Cn - 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using n - 1 misses multiplying by n.

Using n + 1 causes out of range iteration.

Iteration multiplies all numbers from 1 up to n to get factorial.

3fill in blank

hard

Fix the error in the recursive Fibonacci function by completing the blank.

DSA Rust

fn fib(n: u32) -> u32 {
    if n <= 1 {
        n
    } else {
        fib(n - 1) + fib([1])
    }
}

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An + 1

Bn - 3

Cn - 2

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using n causes infinite recursion.

Using n + 1 causes invalid calls.

Fibonacci uses fib(n-1) + fib(n-2) to calculate the nth number.

4fill in blank

hard

Fill both blanks to create a recursive function that sums numbers from n down to 1.

DSA Rust

fn sum_to_n(n: u32) -> u32 {
    if n == 0 {
        0
    } else {
        n + sum_to_n([1])
    }
}

fn main() {
    let result = sum_to_n([2]);
    println!("Sum is {}", result);
}

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An - 1

Bn + 1

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n + 1 causes infinite recursion.

Starting main with 1 misses summing from 0.

The recursion reduces n by 1 each call, and main calls sum_to_n with 0.

5fill in blank

hard

Fill all three blanks to create an iterative function that calculates the nth Fibonacci number.

DSA Rust

fn fib_iter(n: u32) -> u32 {
    let (mut a, mut b) = (0, 1);
    for _ in 0..[1] {
        let temp = [2];
        a = b;
        b = temp;
    }
    b
}

fn main() {
    let result = fib_iter([3]);
    println!("Fibonacci number is {}", result);
}

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Aa + b

Bb + a

Cn

Dn - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping a and b incorrectly.

Looping wrong number of times.

The loop runs n times, updating a and b by adding previous values. main calls fib_iter with n.

Practice

1. Which situation is recursion generally better suited for compared to iteration?

easy

A. When the problem can be divided into similar smaller problems

B. When you need to process large data sets with minimal memory

C. When you want to avoid function call overhead

D. When the problem requires simple repeated loops

Solution

1. Step 1: Understand recursion use case

   Recursion works well when a problem can be broken down into smaller similar problems, like tree traversals or factorial calculation.

2. Step 2: Compare with iteration use case

   Iteration is better for simple repeated loops or when memory overhead of function calls is a concern.

3. Final Answer:

   When the problem can be divided into similar smaller problems -> Option A

4. Quick Check:

   Recursion fits divide-and-conquer problems [OK]

Hint: Recursion fits divide-and-conquer problems best [OK]

Common Mistakes:

• Thinking recursion is always faster
• Confusing iteration with recursion use cases
• Ignoring function call overhead in recursion

2. Which of the following is the correct Rust syntax to write a recursive function that calculates factorial?

easy

A. fn factorial(n: u32) -> u32 { loop { return n * factorial(n - 1); } }

B. fn factorial(n: u32) -> u32 { let mut result = 1; for i in 1..=n { result *= i; } result }

C. fn factorial(n: u32) -> u32 { while n > 0 { n *= n - 1; } n }

D. fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } }

Solution

1. Step 1: Identify recursive function structure

   Recursive factorial calls itself with n-1 until base case n==0 returns 1.

2. Step 2: Check each option's correctness

   fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } } correctly implements recursion with base case and recursive call. fn factorial(n: u32) -> u32 { let mut result = 1; for i in 1..=n { result *= i; } result } is iterative, C has incorrect loop logic, D causes infinite loop.

3. Final Answer:

   fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } } -> Option D

4. Quick Check:

   Recursive factorial syntax [OK]

Hint: Look for base case and recursive call in function [OK]

Common Mistakes:

• Missing base case causing infinite recursion
• Using loops instead of recursion for this question
• Incorrect loop or return statements

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

fn sum_iter(n: u32) -> u32 {
    let mut total = 0;
    for i in 1..=n {
        total += i;
    }
    total
}

fn sum_rec(n: u32) -> u32 {
    if n == 0 {
        0
    } else {
        n + sum_rec(n - 1)
    }
}

fn main() {
    println!("{}", sum_iter(3));
    println!("{}", sum_rec(3));
}

medium

A. 6
6

B. 3
3

C. 0
0

D. Error due to stack overflow

Solution

1. Step 1: Trace sum_iter(3)

   Loop adds 1 + 2 + 3 = 6.

2. Step 2: Trace sum_rec(3)

   Recursively adds 3 + 2 + 1 + 0 = 6.

3. Final Answer:

   6 6 -> Option A

4. Quick Check:

   Both sum functions return 6 for input 3 [OK]

Hint: Sum 1 to n = n*(n+1)/2 = 6 for n=3 [OK]

Common Mistakes:

• Confusing output format with separate lines
• Assuming recursion causes error for small n
• Misreading loop range

4. Identify the bug in this recursive Rust function and choose the fix:

fn countdown(n: u32) {
    println!("{}", n);
    countdown(n - 1);
}

medium

A. Change println! to return n instead

B. Replace recursion with a for loop

C. Add a base case to stop recursion when n == 0

D. Use n + 1 instead of n - 1 in recursive call

Solution

1. Step 1: Analyze the recursive function

   Function calls itself with n-1 without stopping condition, causing infinite recursion.

2. Step 2: Determine fix to prevent infinite recursion

   Adding a base case like if n == 0 { return; } stops recursion safely.

3. Final Answer:

   Add a base case to stop recursion when n == 0 -> Option C

4. Quick Check:

   Base case prevents infinite recursion [OK]

Hint: Always add base case to recursive functions [OK]

Common Mistakes:

• Forgetting base case causes stack overflow
• Changing print statement instead of fixing recursion
• Using wrong recursive call direction

5. You want to process a large list of numbers to find the maximum value. Which approach is better in Rust and why? A recursive function that compares head and tail elements or an iterative loop?

hard

A. Recursive function, because Rust optimizes all recursion automatically

B. Iterative loop, because recursion may cause stack overflow on large lists

C. Recursive function, because it is always faster than iteration

D. Either approach is equally good for large lists

Solution

1. Step 1: Consider recursion limits in Rust

   Rust does not guarantee tail call optimization, so deep recursion risks stack overflow on large inputs.

2. Step 2: Compare iteration benefits

   Iteration uses constant stack space and is safer for large lists, avoiding overflow.

3. Final Answer:

   Iterative loop, because recursion may cause stack overflow on large lists -> Option B

4. Quick Check:

   Iteration safer for large data [OK]

Hint: Use iteration for large data to avoid stack overflow [OK]

Common Mistakes:

• Assuming recursion is always faster
• Ignoring stack overflow risk
• Believing Rust always optimizes recursion