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DSA Rustprogramming~10 mins

Fibonacci Using DP in DSA Rust - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to initialize the DP array with base Fibonacci values.

DSA Rust
let mut dp = vec![0; n + 1];
dp[0] = 0;
dp[1] = [1];
Drag options to blanks, or click blank then click option'
An
B1
C2
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Setting dp[1] to 0 instead of 1.
Using n instead of 1 for dp[1].
2fill in blank
medium

Complete the for loop to fill the DP array with Fibonacci numbers.

DSA Rust
for i in 2..=[1] {
    dp[i] = dp[i - 1] + dp[i - 2];
}
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An
Bn - 1
Cn + 1
Dn * 2
Attempts:
3 left
💡 Hint
Common Mistakes
Using n - 1 which misses the last Fibonacci number.
Using n + 1 which causes out-of-bound error.
3fill in blank
hard

Fix the error in the return statement to return the nth Fibonacci number.

DSA Rust
return [1];
Drag options to blanks, or click blank then click option'
Adp[0]
Bdp[n - 1]
Cdp[n + 1]
Ddp[n]
Attempts:
3 left
💡 Hint
Common Mistakes
Returning dp[n - 1] which is the (n-1)th Fibonacci number.
Returning dp[n + 1] which is out of bounds.
4fill in blank
hard

Fill both blanks to complete the function signature and return type.

DSA Rust
fn fibonacci([1]: usize) -> [2] {
    // function body
}
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An
Bi32
Cusize
Dfib
Attempts:
3 left
💡 Hint
Common Mistakes
Using i32 for input or output which can cause negative values.
Using variable names other than n for input.
5fill in blank
hard

Fill all three blanks to complete the DP Fibonacci function body.

DSA Rust
fn fibonacci(n: usize) -> usize {
    if n == 0 || n == 1 {
        return [1];
    }
    let mut dp = vec![0; n + 1];
    dp[0] = 0;
    dp[1] = 1;
    for i in 2..=[2] {
        dp[i] = dp[i - 1] + [3];
    }
    dp[n]
}
Drag options to blanks, or click blank then click option'
An
Cdp[i - 2]
Ddp[i - 1]
Attempts:
3 left
💡 Hint
Common Mistakes
Returning 0 or 1 directly without checking n.
Using wrong indices in the loop sum.

Practice

(1/5)
1. What is the main advantage of using dynamic programming (DP) to calculate Fibonacci numbers compared to a simple recursive approach?
easy
A. It avoids repeated calculations by storing intermediate results.
B. It uses less memory by not storing any results.
C. It calculates Fibonacci numbers using only recursion without loops.
D. It always runs in constant time regardless of input size.

Solution

  1. Step 1: Understand recursive Fibonacci inefficiency

    Simple recursion recalculates the same Fibonacci values many times, causing exponential time complexity.
  2. Step 2: How DP improves calculation

    DP stores intermediate Fibonacci values in a table (memoization or tabulation), so each value is computed once.
  3. Final Answer:

    It avoids repeated calculations by storing intermediate results. -> Option A
  4. Quick Check:

    DP stores results to avoid repeats [OK]
Hint: DP saves previous results to speed up Fibonacci [OK]
Common Mistakes:
  • Thinking DP uses no extra memory
  • Believing DP always runs in constant time
  • Confusing recursion with DP
2. Which of the following Rust code snippets correctly initializes a vector to store Fibonacci numbers for dynamic programming?
easy
A. let fib = vec![0; n];
B. let mut fib = vec![0; n+1];
C. let mut fib = Vec::new();
D. let fib = [0; n+1];

Solution

  1. Step 1: Check vector size for DP storage

    We need a vector of size n+1 to store Fibonacci numbers from 0 to n.
  2. Step 2: Verify mutability and initialization

    Vector must be mutable to update values; initialized with zeros for all positions.
  3. Final Answer:

    let mut fib = vec![0; n+1]; -> Option B
  4. Quick Check:

    Mutable vector with n+1 zeros [OK]
Hint: Use mutable vector with size n+1 for DP array [OK]
Common Mistakes:
  • Using immutable vector when updates needed
  • Incorrect vector size causing index errors
  • Using fixed-size array without mutability
3. What will be the output of the following Rust code that uses bottom-up DP to compute Fibonacci numbers?
fn fib(n: usize) -> usize {
    let mut dp = vec![0; n+1];
    dp[1] = 1;
    for i in 2..=n {
        dp[i] = dp[i-1] + dp[i-2];
    }
    dp[n]
}

fn main() {
    println!("{}", fib(5));
}
medium
A. 13
B. 8
C. 5
D. 3

Solution

  1. Step 1: Trace DP array filling for n=5

    dp[0]=0, dp[1]=1; then dp[2]=1, dp[3]=2, dp[4]=3, dp[5]=5.
  2. Step 2: Return dp[5]

    Function returns dp[5], which is 5.
  3. Final Answer:

    5 -> Option C
  4. Quick Check:

    Fibonacci(5) = 5 [OK]
Hint: Bottom-up DP sums previous two values [OK]
Common Mistakes:
  • Confusing index starting points
  • Off-by-one errors in loop range
  • Misreading dp array values
4. Identify the error in this Rust code for Fibonacci using DP and select the correct fix:
fn fib(n: usize) -> usize {
    let mut dp = vec![0; n];
    dp[0] = 0;
    dp[1] = 1;
    for i in 2..=n {
        dp[i] = dp[i-1] + dp[i-2];
    }
    dp[n]
}
medium
A. Change vector size to n+1 and return dp[n]
B. Change vector size to n and return dp[n-1]
C. Initialize dp with zeros and ones alternatively
D. Remove dp[0] assignment since it's zero by default

Solution

  1. Step 1: Check vector size and indexing

    Vector size is n (indices 0 to n-1), but loop 2..=n and dp[n] access index n, causing out of bounds panic.
  2. Step 2: Fix vector size

    Change to vec![0; n+1] to store values up to index n.
  3. Final Answer:

    Change vector size to n+1 and return dp[n] -> Option A
  4. Quick Check:

    Size n+1 for indices 0 to n [OK]
Hint: Vector of size n+1 stores fib[0..n] [OK]
Common Mistakes:
  • Accessing dp[n] causing panic
  • Incorrect vector size for indexing
  • Ignoring zero-based indexing
5. You want to compute the 10th Fibonacci number using top-down DP (memoization) in Rust. Which code snippet correctly implements memoization to avoid repeated calculations?
fn fib_memo(n: usize, memo: &mut Vec) -> usize {
    if n <= 1 {
        return n;
    }
    if let Some(val) = memo[n] {
        return val;
    }
    let result = fib_memo(n-1, memo) + fib_memo(n-2, memo);
    memo[n] = Some(result);
    result
}

fn main() {
    let n = 10;
    let mut memo = vec![None; n+1];
    println!("{}", fib_memo(n, &mut memo));
}
hard
A. Incorrect because base case should return 0 for n=1
B. Incorrect because memo vector should store usize, not Option<usize>
C. Incorrect because memo is not passed as mutable reference
D. Correct implementation of top-down DP with memoization

Solution

  1. Step 1: Check memo vector type and initialization

    Memo uses Option<usize> to track computed values; initialized with None for all indices.
  2. Step 2: Verify memoization logic

    Function returns stored value if present; otherwise computes recursively and stores result.
  3. Step 3: Confirm base case correctness

    Returns n for n=0 or 1, which matches Fibonacci definition.
  4. Final Answer:

    Correct implementation of top-down DP with memoization -> Option D
  5. Quick Check:

    Memo stores Option, uses recursion with caching [OK]
Hint: Use Option to track computed Fibonacci values [OK]
Common Mistakes:
  • Using usize vector without Option causes wrong caching
  • Wrong base case returns
  • Not passing memo as mutable reference
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