Memoization Top Down DP in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the memoization cache as a mutable HashMap.

DSA Rust

use std::collections::HashMap;

fn fib(n: u32, memo: &mut HashMap<u32, u64>) -> u64 {
    if n <= 1 {
        return n as u64;
    }
    if let Some(&result) = memo.get(&n) {
        return result;
    }
    let result = fib(n - 1, memo) + fib(n - 2, memo);
    memo.[1](n, result);
    result
}

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

Aremove

Bget

Cinsert

Dclear

Attempts:

3 left

💡 Hint

Common Mistakes

Using get instead of insert to add to the cache.

Trying to remove or clear the cache instead of inserting.

We use the insert method to add the computed result to the memoization cache.

2fill in blank

medium

Complete the code to check if the result for n is already computed in the memo cache.

DSA Rust

if let Some(&result) = memo.[1](&n) {
    return result;
}

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

Ainsert

Bget

Ccontains_key

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using insert instead of get to check the cache.

Using contains_key but not retrieving the value.

The get method returns an Option with the value if the key exists, which we can use to check the cache.

3fill in blank

hard

Fix the error in the recursive call to fib by correctly passing the memo parameter.

DSA Rust

let result = fib(n - 1, [1]) + fib(n - 2, memo);

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

Amemo

Bmemo.clone()

C&memo

D&mut memo

Attempts:

3 left

💡 Hint

Common Mistakes

Passing `&mut memo` causing type mismatch (double mutable reference).

Passing `&memo` causing immutable reference error.

Cloning `memo` unnecessarily causing performance issues.

The memo parameter is already a mutable reference (&mut HashMap), so pass `memo` directly to recursive calls to allow updates.

4fill in blank

hard

Fill both blanks to complete the memoized Fibonacci function with correct base case and recursive calls.

DSA Rust

fn fib(n: u32, memo: &mut HashMap<u32, u64>) -> u64 {
    if n [1] 1 {
        return n as u64;
    }
    if let Some(&result) = memo.get(&n) {
        return result;
    }
    let result = fib(n - 1, [2]) + fib(n - 2, memo);
    memo.insert(n, result);
    result
}

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

A<=

B<

C&memo

Dmemo

Attempts:

3 left

💡 Hint

Common Mistakes

Using `<` instead of `<=` in base case causing wrong results.

Passing `&memo` or `&mut memo` instead of `memo` causing type or borrow errors.

The base case checks if n is less than or equal to 1. Recursive calls pass `memo` directly to allow cache updates.

5fill in blank

hard

Fill all three blanks to create a memoized function that computes factorial with caching.

DSA Rust

fn factorial(n: u32, memo: &mut HashMap<u32, u64>) -> u64 {
    if n [1] 1 {
        return 1;
    }
    if let Some(&result) = memo.get(&n) {
        return result;
    }
    let result = n as u64 * factorial(n - 1, [2]);
    memo.[3](n, result);
    result
}

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

A<=

Bmemo

Cinsert

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of <= in base case.

Passing wrong form like &memo or &mut memo instead of memo.

Using get or remove instead of insert to update cache.

Base case uses <= 1. Recursive call passes `memo`. Insert adds the computed factorial to the cache.

Practice

1. What is the main purpose of memoization in top-down dynamic programming?

easy

A. To store results of subproblems to avoid repeated calculations

B. To sort the input data before processing

C. To convert recursive code into iterative code

D. To reduce memory usage by deleting variables

Solution

1. Step 1: Understand memoization role

   Memoization saves results of subproblems so they are not recalculated multiple times.

2. Step 2: Compare options

   Only To store results of subproblems to avoid repeated calculations describes this purpose correctly. Sorting, converting recursion, or deleting variables are unrelated.

3. Final Answer:

   To store results of subproblems to avoid repeated calculations -> Option A

4. Quick Check:

   Memoization = store subproblem results [OK]

Hint: Memoization means saving answers to reuse later [OK]

Common Mistakes:

• Confusing memoization with sorting
• Thinking memoization reduces memory by deleting data
• Believing memoization changes recursion to iteration

2. Which of the following Rust code snippets correctly declares a memoization cache for a function with input type usize and output type usize?

easy

A. let memo: Vec<usize> = Vec::new();

B. let memo = HashMap::new();

C. let mut memo: Vec = vec![None; n + 1];

D. let mut memo: Vec<usize> = vec![0; n];

Solution

1. Step 1: Identify correct memo type

   Memoization cache should store Option<usize> to mark computed or not, initialized with None.

2. Step 2: Check each option

   let mut memo: Vec = vec![None; n + 1]; uses Vec with None values, correct for memoization. let memo = HashMap::new(); is a HashMap but missing mut and initialization syntax. let memo: Vec<usize> = Vec::new(); lacks Option type. let mut memo: Vec<usize> = vec![0; n]; uses usize with zeros, which can't distinguish computed or not.

3. Final Answer:

   let mut memo: Vec = vec![None; n + 1]; -> Option C

4. Quick Check:

   Memo cache = Vec with None [OK]

Hint: Use Option type to mark uncomputed states [OK]

Common Mistakes:

• Using Vec<usize> without Option to track computed
• Not making memo mutable
• Incorrect initialization of HashMap

3. Consider this Rust function using memoization to compute Fibonacci numbers:

fn fib(n: usize, memo: &mut Vec) -> usize {
    if n <= 1 {
        return n;
    }
    if let Some(val) = memo[n] {
        return val;
    }
    let result = fib(n - 1, memo) + fib(n - 2, memo);
    memo[n] = Some(result);
    result
}

fn main() {
    let n = 5;
    let mut memo = vec![None; n + 1];
    println!("{}", fib(n, &mut memo));
}

What is the output when this code runs?

medium

A. 8

B. 5

C. 3

D. 13

Solution

1. Step 1: Understand Fibonacci calculation

   fib(5) = fib(4) + fib(3). Calculating recursively with memoization avoids repeated work.

2. Step 2: Calculate fib(5)

   fib(0)=0, fib(1)=1, fib(2)=1, fib(3)=2, fib(4)=3, fib(5)=5.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   fib(5) = 5 [OK]

Hint: Recall Fibonacci sequence: 0,1,1,2,3,5... [OK]

Common Mistakes:

• Confusing fib(5) with fib(6)
• Forgetting zero-based indexing
• Ignoring memoization effect

4. This Rust code tries to memoize the factorial function but has a bug:

fn fact(n: usize, memo: &mut Vec) -> usize {
    if n == 0 {
        return 1;
    }
    if memo[n].is_some() {
        return memo[n].unwrap();
    }
    let result = n * fact(n - 1, memo);
    memo[n] = Some(result);
    result
}

fn main() {
    let n = 5;
    let mut memo = vec![None; n];
    println!("{}", fact(n, &mut memo));
}

What is the bug causing a runtime error?

medium

A. The memo vector size is too small causing index out of bounds

B. The base case should return 0 instead of 1

C. The function does not update memo correctly

D. The function should not use recursion

Solution

1. Step 1: Check memo vector size

   Memo is created with size n (5), but fact accesses memo[n], which is memo[5], out of bounds since indices go 0 to 4.

2. Step 2: Identify cause of runtime error

   Accessing memo[5] causes panic due to index out of bounds. The vector should be size n+1 to include index 5.

3. Final Answer:

   The memo vector size is too small causing index out of bounds -> Option A

4. Quick Check:

   Vector size < needed index = runtime error [OK]

Hint: Memo size must cover all indices used [OK]

Common Mistakes:

• Ignoring vector size when indexing
• Changing base case incorrectly
• Thinking recursion is the problem

5. You want to use top-down memoization in Rust to count the number of ways to climb a staircase of n steps, taking 1 or 2 steps at a time. Which of these implementations correctly applies memoization to avoid repeated calculations?

fn climb(n: usize, memo: &mut Vec) -> usize {
    if n == 0 {
        return 1;
    }
    if n < 0 {
        return 0;
    }
    if let Some(val) = memo[n] {
        return val;
    }
    let ways = climb(n - 1, memo) + climb(n - 2, memo);
    memo[n] = Some(ways);
    ways
}

fn main() {
    let n = 4;
    let mut memo = vec![None; n + 1];
    println!("{}", climb(n, &mut memo));
}

hard

A. Incorrect because it does not handle negative indices safely

B. Incorrect because it uses iteration instead of recursion

C. Incorrect because memo vector is not mutable

D. Correct implementation with proper base cases and memo usage

Solution

1. Step 1: Check base cases and recursion

   Base cases handle n == 0 (1 way) and n < 0 (0 ways), which is correct for counting ways.

2. Step 2: Verify memoization usage

   Memo vector is mutable and stores Option<usize>. It checks memo[n] before recursion and stores result after calculation.

3. Step 3: Confirm no unsafe indexing

   Negative n returns 0 before indexing memo, so no out-of-bounds access.

4. Final Answer:

   Correct implementation with proper base cases and memo usage -> Option D

5. Quick Check:

   Base cases + memo check = correct memoization [OK]

Hint: Check base cases and memo before recursion [OK]

Common Mistakes:

• Not handling negative n before memo access
• Forgetting to make memo mutable
• Using iteration instead of recursion in top-down DP