Overlapping Subproblems and Optimal Substructure in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the nth Fibonacci number using recursion.

DSA Rust

fn fib(n: u32) -> u32 {
    if n <= 1 {
        return n;
    }
    fib(n - 1) + fib([1])
}

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An - 5

Bn - 3

Cn - 4

Dn - 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using n - 3 or other values instead of n - 2.

Forgetting to return the sum of two recursive calls.

The Fibonacci number is the sum of the two previous numbers: fib(n-1) + fib(n-2).

2fill in blank

medium

Complete the code to memoize Fibonacci results using a vector.

DSA Rust

fn fib_memo(n: usize, memo: &mut Vec<Option<u32>>) -> u32 {
    if let Some(value) = memo[n] {
        return value;
    }
    let result = if n <= 1 {
        n as u32
    } else {
        fib_memo(n - 1, memo) + fib_memo([1], memo)
    };
    memo[n] = Some(result);
    result
}

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An - 2

Bn - 3

Cn - 4

Dn - 5

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect indices like n - 3.

Not storing the computed result in the memo vector.

Memoization stores results of fib(n-1) and fib(n-2). The second call must be fib(n-2).

3fill in blank

hard

Fix the error in the code to correctly compute the minimum cost path in a grid using dynamic programming.

DSA Rust

fn min_cost_path(cost: &Vec<Vec<u32>>, m: usize, n: usize) -> u32 {
    if m == 0 && n == 0 {
        return cost[0][0];
    }
    if m == 0 {
        return min_cost_path(cost, 0, n - 1) + cost[0][n];
    }
    if n == 0 {
        return min_cost_path(cost, m - 1, 0) + cost[m][0];
    }
    let left = min_cost_path(cost, m, n - 1) + cost[m][n];
    let up = min_cost_path(cost, m - 1, n) + cost[m][n];
    if left < [1] {
        left
    } else {
        [2]
    }
}

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Aup

Bleft

Ccost[m][n]

Dm + n

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing left with wrong variable.

Returning left in both branches.

We compare left and up costs and return the smaller one. So if left < up, return left else return up.

4fill in blank

hard

Fill both blanks to create a dynamic programming table for the coin change problem.

DSA Rust

fn coin_change(coins: &Vec<u32>, amount: usize) -> usize {
    let mut dp = vec![amount + 1; amount + 1];
    dp[0] = 0;
    for i in 1..=amount {
        for &coin in coins.iter() {
            if coin as usize <= i {
                dp[i] = dp[i].[1] dp[i - coin as usize] + 1;
            }
        }
    }
    if dp[amount] > amount [2] 0 else dp[amount]
}

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Amin

Bmax

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using max instead of min.

Using wrong comparison operator in the if condition.

We update dp[i] with the minimum between current dp[i] and dp[i - coin] + 1. If dp[amount] > amount, return 0 meaning no solution.

5fill in blank

hard

Fill all three blanks to implement a memoized recursive solution for the longest common subsequence (LCS).

DSA Rust

fn lcs(s1: &str, s2: &str, m: usize, n: usize, memo: &mut Vec<Vec<Option<usize>>>) -> usize {
    if m == 0 || n == 0 {
        return 0;
    }
    if let Some(val) = memo[m - 1][n - 1] {
        return val;
    }
    let result = if s1.chars().nth(m - 1) == s2.chars().nth(n - 1) {
        1 + lcs(s1, s2, [1], [2], memo)
    } else {
        std::cmp::max(lcs(s1, s2, m - 1, n, memo), lcs(s1, s2, m, [3], memo))
    };
    memo[m - 1][n - 1] = Some(result);
    result
}

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Am - 1

Bn - 1

Dm - 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong indices in recursive calls.

Not memoizing results properly.

When characters match, move both indices back by 1. When not, compare lcs(m-1, n) and lcs(m, n-1).

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