Recursion Base Case and Recursive Case in DSA Rust - Build from Scratch

Recursion Base Case and Recursive Case

📖 Scenario: Imagine you have a stack of books and you want to count how many books are in the stack. You can count one book at a time until there are no books left.

🎯 Goal: Build a recursive function in Rust that counts the number of books in a stack represented by a list of book titles.

📋 What You'll Learn

Create a vector called books with exactly these titles: "Book A", "Book B", "Book C"

Create a variable called count to hold the number of books counted

Write a recursive function called count_books that takes a slice of books and returns the count as usize

The function must have a base case when the slice is empty returning 0

The recursive case must count 1 plus the result of calling itself with the rest of the slice

Print the final count using println!

💡 Why This Matters

🌍 Real World

Recursion helps solve problems that can be broken into smaller similar problems, like counting items, searching, or navigating structures.

💼 Career

Understanding recursion is important for software development roles that involve algorithms, data processing, and problem solving.

1

Create the book stack

Create a vector called books with these exact titles: "Book A", "Book B", "Book C"

DSA Rust

# Your code here

Hint

Use vec![...] to create a vector with the given book titles.

2

Create the count variable

Create a variable called count of type usize and set it to 0 inside the main function after the books vector

DSA Rust

fn main() {
    let books = vec!["Book A", "Book B", "Book C"];
    // Create count variable here
    # Your code here
}

Hint

Use let count: usize = 0; to create the count variable.

3

Write the recursive function

Write a recursive function called count_books that takes a slice of &[&str] called books and returns a usize. The function must return 0 if the slice is empty (base case). Otherwise, return 1 plus the result of calling count_books with the rest of the slice (recursive case). Place this function above main.

DSA Rust

// Write the recursive function count_books here
# Your code here

fn main() {
    let books = vec!["Book A", "Book B", "Book C"];
    let count: usize = 0;
}

Hint

Use if books.is_empty() for the base case and call count_books(&books[1..]) for the recursive case.

4

Call the function and print the result

Inside main, call count_books with a slice of books and assign the result to count. Then print count using println!("{}", count).

DSA Rust

fn count_books(books: &[&str]) -> usize {
    if books.is_empty() {
        0
    } else {
        1 + count_books(&books[1..])
    }
}

fn main() {
    let books = vec!["Book A", "Book B", "Book C"];
    let count: usize = 0;
    // Call count_books and print count
    # Your code here
}

Hint

Assign count = count_books(&books); and then print it with println!("{}", count);.

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