Memoization Top Down DP in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is memoization in the context of dynamic programming?

Memoization is a technique where we store the results of expensive function calls and reuse them when the same inputs occur again, avoiding repeated calculations.

beginner

How does top-down dynamic programming differ from bottom-up?

Top-down DP solves the problem by breaking it into subproblems recursively and uses memoization to save results. Bottom-up solves smaller subproblems first and builds up to the solution.

intermediate

In Rust, which data structure is commonly used to store memoized results for top-down DP?

A Vec or a HashMap is commonly used to store memoized results, where Option indicates if a result is computed or not.

beginner

Why is memoization important in recursive solutions?

Memoization prevents repeated work by caching results of subproblems, which reduces exponential time complexity to polynomial time in many cases.

beginner

What is a key step to implement memoization in a top-down DP function?

Before computing a subproblem, check if its result is already stored in the memo. If yes, return it immediately; if no, compute and store it.

What does memoization primarily help to avoid in recursive algorithms?

A Repeated calculations of the same subproblems

B Using too much memory

C Writing iterative loops

D Sorting data before processing

Memoization stores results of subproblems to avoid recalculating them multiple times.

Which Rust type is best suited to represent a memo table that may or may not have computed values?

A String

B Vec<T>

C Vec<Option<T>>

D HashSet<T>

Vec> allows storing computed results or None if not computed yet.

In top-down DP, when do you store a computed result in the memo?

A Only at the end of the program

B After computing the subproblem result

C Before computing the subproblem

D Never, memo is read-only

You store the result after computing it to reuse later.

What is the main difference between top-down and bottom-up DP?

A Top-down uses recursion with memoization; bottom-up uses iteration

B Top-down is slower than bottom-up

C Bottom-up uses recursion; top-down uses loops

D They are the same

Top-down solves problems recursively with memoization; bottom-up builds solutions iteratively.

Which problem is a classic example to apply memoization top-down DP?

A Printing a string

B Sorting an array

C Binary search

D Fibonacci sequence calculation

Fibonacci calculation has overlapping subproblems and benefits from memoization.

Explain how memoization works in a top-down dynamic programming approach.

Think about how you remember answers to smaller questions to avoid re-solving them.

Describe how you would implement memoization in Rust for a recursive function.

Consider how Rust handles optional values and indexing.

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