Recursion Base Case and Recursive Case in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Recursive Factorial Function

What is the output of this Rust code when calling factorial(4)?

DSA Rust

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

fn main() {
    println!("{}", factorial(4));
}

A 24

B 1

C 120

D 10

💡 Hint

Remember factorial of 0 is 1, and factorial(n) = n * factorial(n-1).

The function calculates factorial recursively. factorial(4) = 4 * 3 * 2 * 1 = 24.

❓ Predict Output

intermediate

Output of Recursive Sum Function

What is printed when the following Rust code runs sum(3)?

DSA Rust

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

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

A 0

B 3

C 6

D 9

💡 Hint

Sum of numbers from 1 to n is n*(n+1)/2.

sum(3) = 3 + 2 + 1 + 0 = 6.

🔧 Debug

advanced

Identify the Error in Recursive Fibonacci Function

What error occurs when running this Rust code?

DSA Rust

fn fibonacci(n: u32) -> u32 {
    if n == 0 {
        0
    } else if n == 1 {
        1
    } else {
        fibonacci(n - 1) + fibonacci(n - 2)
    }
}

fn main() {
    println!("{}", fibonacci(42));
}

A Stack overflow due to deep recursion

B No error, prints 267914296

C Type error: mismatched types

D Syntax error: missing semicolon

💡 Hint

Think about how many times fibonacci calls itself for large n.

The recursive calls grow exponentially causing stack overflow for large n like 42.

🧠 Conceptual

advanced

Base Case Importance in Recursion

Why is the base case essential in a recursive function?

A It speeds up the recursion by caching results

B It stops recursion to prevent infinite calls

C It initializes variables before recursion starts

D It converts recursion into iteration automatically

💡 Hint

Without base case, recursion never ends.

Base case stops recursion by providing a condition to end recursive calls.

🚀 Application

expert

Output of Recursive String Reversal

What is the output of this Rust code when reversing the string "rust"?

DSA Rust

fn reverse(s: &str) -> String {
    if s.is_empty() {
        String::new()
    } else {
        let (first, rest) = s.split_at(1);
        format!("{}{}", reverse(rest), first)
    }
}

fn main() {
    println!("{}", reverse("rust"));
}

A "urts"

B "rust"

C "trus"

D "tsur"

💡 Hint

The function moves the first character to the end recursively.

The function reverses the string by recursively moving first character to the end.

Practice

1. In a recursive function, what is the purpose of the base case?

easy

A. To increase the input size for the next call

B. To call the function again with a smaller input

C. To print the result of the recursion

D. To stop the recursion and prevent infinite calls

Solution

1. Step 1: Understand the base case role

   The base case is the condition that stops the recursion from continuing indefinitely.

2. Step 2: Differentiate from recursive case

   The recursive case calls the function again with smaller input, but the base case stops recursion.

3. Final Answer:

   To stop the recursion and prevent infinite calls -> Option D

4. Quick Check:

   Base case = stop recursion [OK]

Hint: Base case stops recursion to avoid infinite loops [OK]

Common Mistakes:

• Confusing base case with recursive case
• Thinking base case increases input size
• Assuming base case prints output

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

easy

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

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

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

D. fn factorial(n: u32) -> u32 { return n + factorial(n - 1); }

Solution

1. Step 1: Check base case and recursive call

   fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } } correctly uses if n == 0 { 1 } as base case and recursive multiplication.

2. Step 2: Identify syntax errors in other options

   fn factorial(n: u32) -> u32 { while n > 0 { n * factorial(n - 1) } } uses while loop incorrectly, C loops infinitely, D uses addition instead of multiplication.

3. Final Answer:

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

4. Quick Check:

   Correct base case and recursive call = fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } } [OK]

Hint: Look for base case with if and recursive call with smaller input [OK]

Common Mistakes:

• Using loops instead of recursion
• Missing base case
• Wrong operation in recursive call

3. What is the output of this Rust code?

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

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

medium

A. 3

B. 6

C. 0

D. Stack overflow error

Solution

1. Step 1: Trace recursive calls for sum(3)

   sum(3) = 3 + sum(2), sum(2) = 2 + sum(1), sum(1) = 1 + sum(0), sum(0) = 0.

2. Step 2: Calculate total sum

   sum(3) = 3 + 2 + 1 + 0 = 6.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Sum 1 to 3 = 6 [OK]

Hint: Add numbers down to zero using recursion [OK]

Common Mistakes:

• Confusing base case value
• Forgetting to add current n
• Expecting runtime error

4. Identify the error in this recursive Rust function that calculates factorial:

fn factorial(n: u32) -> u32 {
    n * factorial(n - 1)
}

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

medium

A. Missing base case to stop recursion

B. Wrong return type for factorial

C. Incorrect function name

D. Using multiplication instead of addition

Solution

1. Step 1: Check for base case presence

   The function lacks a base case like if n == 0 { 1 } to stop recursion.

2. Step 2: Understand recursion consequences

   Without base case, recursion never stops, causing stack overflow error.

3. Final Answer:

   Missing base case to stop recursion -> Option A

4. Quick Check:

   Base case missing = Missing base case to stop recursion [OK]

Hint: Always include base case to stop recursion [OK]

Common Mistakes:

• Assuming recursion stops automatically
• Ignoring base case importance
• Confusing return types

5. You want to write a recursive function in Rust to find the nth Fibonacci number. Which of the following correctly shows the base case and recursive case?

hard

A. if n == 0 { 0 } else if n == 1 { 1 } else { fib(n - 1) + fib(n - 2) }

B. if n == 1 { 0 } else if n == 2 { 1 } else { fib(n - 1) * fib(n - 2) }

C. if n == 0 { 1 } else { fib(n - 1) + fib(n - 2) }

D. if n == 0 { 0 } else { fib(n - 1) - fib(n - 2) }

Solution

1. Step 1: Identify correct base cases for Fibonacci

   Fibonacci sequence starts with fib(0) = 0 and fib(1) = 1, so base cases must reflect that.

2. Step 2: Check recursive case correctness

   Recursive case sums previous two Fibonacci numbers: fib(n-1) + fib(n-2).

3. Final Answer:

   if n == 0 { 0 } else if n == 1 { 1 } else { fib(n - 1) + fib(n - 2) } -> Option A

4. Quick Check:

   Correct base and recursive case for Fibonacci = if n == 0 { 0 } else if n == 1 { 1 } else { fib(n - 1) + fib(n - 2) } [OK]

Hint: Fibonacci base cases: 0 and 1; recursive case sums previous two [OK]

Common Mistakes:

• Wrong base case values
• Using multiplication or subtraction instead of addition
• Missing one base case