Memoization Top Down DP in DSA Rust - Time & Space Complexity

Time Complexity: Memoization Top Down DP

O(n)

Understanding Time Complexity

We analyze how memoization in top-down dynamic programming affects the time taken by recursive solutions.

We want to know how saving results changes the number of repeated calculations.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn fib_memo(n: usize, memo: &mut Vec<Option<usize>>) -> 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
}

This code calculates the nth Fibonacci number using recursion with memoization to save results.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to fib_memo for smaller n values.
• How many times: Each unique n from 0 to n is computed once and then reused from memo.

How Execution Grows With Input

Each number from 0 up to n is calculated once, then saved to avoid repeats.

Input Size (n)	Approx. Operations
10	About 10 calculations
100	About 100 calculations
1000	About 1000 calculations

Pattern observation: The number of operations grows roughly in a straight line with n.

Final Time Complexity

Time Complexity: O(n)

This means the time grows directly with the input size because each subproblem is solved once.

Common Mistake

[X] Wrong: "Memoization still causes exponential time because of recursion."

[OK] Correct: Memoization stores results, so each subproblem runs once, avoiding repeated work.

Interview Connect

Understanding memoization shows you how to turn slow recursive solutions into efficient ones, a key skill in problem solving.

Self-Check

"What if we removed memoization and only used plain recursion? How would the time complexity change?"

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