Overlapping Subproblems and Optimal Substructure 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 the 5th Fibonacci number using memoization?

DSA Rust

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

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

A 3

B 8

C 5

D 13

💡 Hint

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

The 5th Fibonacci number (0-based index) is 5. The code uses memoization to avoid repeated calculations.

🧠 Conceptual

intermediate

Identifying Overlapping Subproblems

Which of the following problems best demonstrates the concept of overlapping subproblems?

A Finding the shortest path in a graph using Dijkstra's algorithm

B Computing Fibonacci numbers using recursion without memoization

C Sorting an array using quicksort

D Calculating factorial of a number using recursion

💡 Hint

Overlapping subproblems means the same subproblems are solved multiple times.

Computing Fibonacci numbers recursively without memoization solves the same subproblems repeatedly, demonstrating overlapping subproblems.

🔧 Debug

advanced

Debugging Memoization Implementation

What error will the following Rust code produce when run, and why?

DSA Rust

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

fn main() {
    let n = 5usize;
    let mut memo = Vec::new();
    let result = fib(n, &mut memo);
    println!("{}", result);
}

A Compilation error due to missing return type

B Runtime error due to unwrap on None value

C Correct output: 5

D Index out of bounds panic because memo vector is empty

💡 Hint

Check how the memo vector is initialized before use.

The memo vector is empty, so accessing memo[n] causes an index out of bounds panic.

🚀 Application

advanced

Optimal Substructure in Shortest Path

Which statement best describes the optimal substructure property in the context of shortest path algorithms?

A Any subpath of a shortest path is itself a shortest path between its endpoints

B The shortest path between two nodes contains cycles to reduce distance

C The shortest path is always the path with the fewest edges

D Optimal substructure means the problem can be solved without recursion

💡 Hint

Think about what happens if a subpath is not shortest.

Optimal substructure means that the best solution contains best solutions to subproblems. For shortest path, any part of the path is also shortest between those points.

❓ Predict Output

expert

Output of Dynamic Programming Table for Coin Change

What is the final state of the dp vector after running the following Rust code to find the number of ways to make change for 5 using coins [1, 2, 5]?

DSA Rust

fn main() {
    let amount = 5usize;
    let coins = vec![1usize, 2usize, 5usize];
    let mut dp = vec![0usize; amount + 1];
    dp[0] = 1;
    for coin in coins {
        for x in coin..=amount {
            dp[x] += dp[x - coin];
        }
    }
    println!("{:?}", dp);
}

A [1, 1, 2, 2, 3, 4]

B [1, 1, 2, 3, 5, 10]

C [1, 1, 1, 1, 1, 1]

D [1, 0, 1, 1, 2, 2]

💡 Hint

dp[x] counts ways to make amount x using given coins.

The dp vector after processing coins [1,2,5] for amount 5 is [1,1,2,2,3,4].

Practice

1. Which of the following best describes the overlapping subproblems property in dynamic programming?

easy

A. The problem can be broken down into smaller subproblems that are solved multiple times.

B. The problem has a unique solution without repeating any subproblems.

C. The problem requires sorting before solving.

D. The problem can only be solved using recursion without memoization.

Solution

1. Step 1: Understand overlapping subproblems

   Overlapping subproblems means the same smaller problems appear multiple times during solving.

2. Step 2: Match definition to options

   The problem can be broken down into smaller subproblems that are solved multiple times. correctly states that subproblems repeat and can be reused to save time.

3. Final Answer:

   The problem can be broken down into smaller subproblems that are solved multiple times. -> Option A

4. Quick Check:

   Overlapping subproblems = repeated smaller problems [OK]

Hint: Look for repeated smaller problems solved multiple times [OK]

Common Mistakes:

• Confusing overlapping subproblems with unique subproblems
• Thinking sorting is required for overlapping subproblems
• Assuming recursion alone solves overlapping subproblems

2. Which Rust syntax correctly declares a function to compute Fibonacci numbers using memoization?

easy

A. fn fib(n: usize, memo: &Vec<usize>) -> usize { ... }

B. fn fib(n: usize) -> usize { let memo = Vec::new(); ... }

C. fn fib(n: usize, memo: Vec<usize>) -> usize { ... }

D. fn fib(n: usize, memo: &mut Vec) -> usize { ... }

Solution

1. Step 1: Identify correct parameter passing for memo

   Memo must be mutable reference to allow updates during recursion.

2. Step 2: Check options for mutable reference

   fn fib(n: usize, memo: &mut Vec) -> usize { ... } uses &mut Vec, which allows updating memo inside the function.

3. Final Answer:

   fn fib(n: usize, memo: &mut Vec) -> usize { ... } -> Option D

4. Quick Check:

   Mutable reference needed for memo updates [OK]

Hint: Use &mut for mutable references in Rust function parameters [OK]

Common Mistakes:

• Passing memo by value causing copies
• Using immutable reference &Vec which disallows updates
• Initializing memo inside function losing previous results

3. What is the output of this Rust code that uses memoization for Fibonacci?

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

fn main() {
    let mut memo = vec![None; 6];
    println!("{}", fib(5, &mut memo));
}

medium

A. 8

B. 5

C. 3

D. 13

Solution

1. Step 1: Trace fib(5) calls with memo

   fib(5) = fib(4) + fib(3), fib(4) = 3, fib(3) = 2, so fib(5) = 5.

2. Step 2: Confirm memo stores results to avoid recomputation

   Memoization stores fib(3)=2, fib(4)=3, so final fib(5) = 5 is returned and printed.

3. Final Answer:

   5 -> Option B

4. Quick Check:

   fib(5) = 5 [OK]

Hint: Recall Fibonacci sequence: 0,1,1,2,3,5,8... [OK]

Common Mistakes:

• Confusing fib(5) with fib(6)
• Ignoring memoization effect
• Miscounting base cases

4. Identify the bug in this Rust code for Fibonacci with memoization:

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

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

medium

A. Vector size is too small causing index out of bounds

B. Memo is not mutable

C. Base case condition is incorrect

D. Using unwrap() causes panic always

Solution

1. Step 1: Check vector size for memo

   Memo vector has size 5 but fib(5) accesses memo[5], which is out of bounds.

2. Step 2: Understand Rust vector indexing

   Indexing beyond vector length causes runtime panic, so vector size must be at least n+1.

3. Final Answer:

   Vector size is too small causing index out of bounds -> Option A

4. Quick Check:

   Memo size < n+1 causes index error [OK]

Hint: Ensure memo vector size is n+1 for fib(n) [OK]

Common Mistakes:

• Not allocating enough space for memo
• Ignoring Rust's zero-based indexing
• Assuming unwrap() always safe

5. You want to find the minimum cost path in a grid from top-left to bottom-right, moving only right or down. Which property allows using dynamic programming to solve this efficiently?

hard

A. Greedy choice: always pick the cell with the smallest immediate cost.

B. Overlapping subproblems: each cell's cost is unique and never reused.

C. Optimal substructure: the minimum cost path to a cell depends on minimum cost paths to adjacent cells.

D. Divide and conquer: split the grid into independent parts solved separately.

Solution

1. Step 1: Understand minimum cost path problem

   The cost to reach a cell depends on the best path to its neighbors (top or left).

2. Step 2: Identify optimal substructure property

   Optimal substructure means the optimal solution to the whole problem contains optimal solutions to subproblems.

3. Final Answer:

   Optimal substructure: the minimum cost path to a cell depends on minimum cost paths to adjacent cells. -> Option C

4. Quick Check:

   Minimum cost path uses optimal substructure [OK]

Hint: Look for solutions built from optimal subproblems [OK]

Common Mistakes:

• Confusing overlapping subproblems with unique subproblems
• Assuming greedy always works here
• Thinking divide and conquer applies without overlap