Memoization Top Down DP in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Fibonacci with Memoization

What is the output of the following Rust code that calculates Fibonacci numbers using top-down memoization?

DSA Rust

use std::collections::HashMap;

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

fn main() {
    let mut memo = HashMap::new();
    let result = fib(6, &mut memo);
    println!("{}", result);
}

A 21

B 34

C 8

D 13

💡 Hint

Remember Fibonacci sequence starts with 0, 1, 1, 2, 3, 5, 8, 13...

The Fibonacci sequence starts with 0 and 1. fib(6) = fib(5) + fib(4) = 5 + 3 = 8.

❓ Predict Output

intermediate

Output of Climbing Stairs Problem with Memoization

What is the output of this Rust code that uses top-down memoization to count ways to climb stairs?

DSA Rust

use std::collections::HashMap;

fn climb_stairs(n: usize, memo: &mut HashMap<usize, usize>) -> usize {
    if n <= 1 {
        return 1;
    }
    if let Some(&val) = memo.get(&n) {
        return val;
    }
    let ways = climb_stairs(n - 1, memo) + climb_stairs(n - 2, memo);
    memo.insert(n, ways);
    ways
}

fn main() {
    let mut memo = HashMap::new();
    let result = climb_stairs(4, &mut memo);
    println!("{}", result);
}

A 6

B 7

C 8

D 5

💡 Hint

Number of ways to climb 4 stairs with 1 or 2 steps at a time.

Ways to climb 4 stairs: 1+1+1+1, 1+1+2, 1+2+1, 2+1+1, 2+2 = 5 ways.

🔧 Debug

advanced

Identify the Error in Memoized Factorial Code

What error will this Rust code produce when run?

DSA Rust

use std::collections::HashMap;

fn factorial(n: usize, memo: &mut HashMap<usize, usize>) -> usize {
    if n == 0 {
        return 1;
    }
    if let Some(&val) = memo.get(&n) {
        return val;
    }
    let result = n * factorial(n - 1, memo);
    memo.insert(n, result);
    result
}

fn main() {
    let mut memo = HashMap::new();
    let result = factorial(20, &mut memo);
    println!("{}", result);
}

A No error, prints 2432902008176640000

B Compilation error due to missing return type

C Overflow panic at runtime

D Stack overflow due to infinite recursion

💡 Hint

Check the data type used for factorial result and its max value.

usize on 64-bit systems is u64, which can hold 20! = 2432902008176640000 without overflow.

🧠 Conceptual

advanced

Memoization Cache Behavior

In top-down memoization, what happens if the memo cache is not used properly?

A The algorithm runs slower due to repeated calculations

B The algorithm produces incorrect results

C The algorithm uses less memory but runs faster

D The algorithm terminates immediately

💡 Hint

Memoization avoids repeated work by caching results.

Without memoization cache, the function recalculates the same subproblems multiple times, slowing down the algorithm.

🚀 Application

expert

Memoization for Unique Paths in Grid

Given this Rust code for counting unique paths in a grid using memoization, what is the output?

DSA Rust

use std::collections::HashMap;

fn unique_paths(m: usize, n: usize, memo: &mut HashMap<(usize, usize), usize>) -> usize {
    if m == 1 || n == 1 {
        return 1;
    }
    if let Some(&val) = memo.get(&(m, n)) {
        return val;
    }
    let paths = unique_paths(m - 1, n, memo) + unique_paths(m, n - 1, memo);
    memo.insert((m, n), paths);
    paths
}

fn main() {
    let mut memo = HashMap::new();
    let result = unique_paths(3, 3, &mut memo);
    println!("{}", result);
}

A 9

B 6

C 12

D 8

💡 Hint

Number of ways to move right or down in a 3x3 grid.

Unique paths in 3x3 grid = 6 (combinations of moves).

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