DP vs Recursion vs Greedy Choosing the Right Tool in DSA Rust - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate Fibonacci numbers using simple recursion.

DSA Rust

fn fib(n: u32) -> u32 {
    if n <= 1 {
        return n;
    }
    fib(n - 1) [1] fib(n - 2)
}

Drag options to blanks, or click blank then click option'

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Forgetting to return the sum.

The Fibonacci sequence is defined as fib(n) = fib(n-1) + fib(n-2). So, the correct operator is '+'.

2fill in blank

medium

Complete the code to use memoization in Fibonacci calculation to avoid repeated work.

DSA Rust

fn fib_memo(n: usize, memo: &mut Vec<Option<u32>>) -> u32 {
    if let Some(val) = memo[n] {
        return val;
    }
    let result = if n <= 1 {
        n as u32
    } else {
        fib_memo(n - 1, memo) [1] fib_memo(n - 2, memo)
    };
    memo[n] = Some(result);
    result
}

Drag options to blanks, or click blank then click option'

A/

B-

C*

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong operator causing incorrect results.

Not storing the computed result in memo.

Memoization stores results of fib(n-1) and fib(n-2) and adds them to get fib(n). The operator is '+'.

3fill in blank

hard

Fix the error in the greedy coin change function to return the minimum coins for amount.

DSA Rust

fn min_coins(coins: &[u32], amount: u32) -> u32 {
    let mut remaining = amount;
    let mut count = 0;
    for &coin in coins.iter() {
        count += remaining / coin;
        remaining = remaining [1] coin * (remaining / coin);
    }
    if remaining == 0 { count } else { u32::MAX }
}

Drag options to blanks, or click blank then click option'

A*

B+

C-

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using addition or multiplication instead of subtraction.

Not updating remaining amount correctly.

To reduce the remaining amount, subtract the total value of coins used: coin * (remaining / coin).

4fill in blank

hard

Fill both blanks to create a DP table for the 0/1 knapsack problem.

DSA Rust

fn knapsack(weights: &[usize], values: &[usize], capacity: usize) -> usize {
    let n = weights.len();
    let mut dp = vec![vec![0; capacity + 1]; n + 1];
    for i in 1..=n {
        for w in 0..=capacity {
            dp[i][w] = if weights[i - 1] <= w {
                dp[i - 1][w].[1] values[i - 1] [2] dp[i - 1][w - weights[i - 1]]
            } else {
                dp[i - 1][w]
            };
        }
    }
    dp[n][capacity]
}

Drag options to blanks, or click blank then click option'

Amax(

B+

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Not using max to compare options.

The DP relation chooses the maximum between not taking and taking the item. Taking adds the item's value to the best value for remaining capacity.

5fill in blank

hard

Fill all three blanks to create a recursive function with memoization for climbing stairs problem.

DSA Rust

fn climb_stairs(n: usize, memo: &mut Vec<Option<u32>>) -> u32 {
    if n <= 1 {
        return 1;
    }
    if let Some(val) = memo[n] {
        return val;
    }
    let result = climb_stairs(n [1], memo) [2] climb_stairs(n [3], memo);
    memo[n] = Some(result);
    result
}

Drag options to blanks, or click blank then click option'

A-1

B+

C-2

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication or wrong indices.

Not adding the two recursive calls.

The number of ways to climb stairs is sum of ways to climb n-1 and n-2 steps. So, use n-1 and n-2 with addition.

Practice

1. Which approach is best suited when a problem has overlapping subproblems and optimal substructure?

easy

A. Dynamic Programming

B. Greedy Algorithm

C. Simple Recursion without memoization

D. Brute Force

Solution

1. Step 1: Understand problem characteristics

   Overlapping subproblems means the same smaller problems repeat. Optimal substructure means the best solution can be built from best solutions of smaller problems.

2. Step 2: Match approach to characteristics

   Dynamic Programming stores results of subproblems to avoid recomputation and uses optimal substructure to build the solution efficiently.

3. Final Answer:

   Dynamic Programming -> Option A

4. Quick Check:

   Overlapping subproblems + optimal substructure = Dynamic Programming [OK]

Hint: Use DP when subproblems repeat and build optimal solutions [OK]

Common Mistakes:

• Choosing greedy when subproblems overlap
• Using simple recursion without memoization causing slow code
• Confusing optimal substructure with greedy choice property

2. Which of the following Rust function signatures correctly represents a recursive function that returns an integer?

easy

A. fn factorial(n: u32) -> u32 { let mut result = 1; for i in 1..=n { result *= i; } result }

B. fn factorial(n: u32) { if n == 0 { 1 } else { n * factorial(n - 1) } }

C. fn factorial(n: u32) -> String { if n == 0 { "1".to_string() } else { (n * factorial(n - 1)).to_string() } }

D. fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } }

Solution

1. Step 1: Check function return type and recursion

   fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } } returns u32 and calls itself recursively correctly. fn factorial(n: u32) { if n == 0 { 1 } else { n * factorial(n - 1) } } lacks return type, so it returns unit ().

2. Step 2: Validate correctness of recursion and return

   fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } } correctly returns the factorial value recursively. fn factorial(n: u32) -> u32 { let mut result = 1; for i in 1..=n { result *= i; } result } is iterative, not recursive. fn factorial(n: u32) -> String { if n == 0 { "1".to_string() } else { (n * factorial(n - 1)).to_string() } } returns String, not integer.

3. Final Answer:

   fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } } -> Option D

4. Quick Check:

   Recursive function with correct return type = fn factorial(n: u32) -> u32 { if n == 0 { 1 } else { n * factorial(n - 1) } } [OK]

Hint: Recursive functions must specify return type matching recursive calls [OK]

Common Mistakes:

• Omitting return type causes compilation errors
• Confusing iterative and recursive implementations
• Returning wrong data types in recursion

3. What is the output of the following Rust code using recursion to compute Fibonacci numbers?

fn fib(n: u32) -> u32 {
    if n <= 1 {
        n
    } else {
        fib(n - 1) + fib(n - 2)
    }
}

fn main() {
    println!("{}", fib(5));
}

medium

A. 13

B. 5

C. 3

D. 8

Solution

1. Step 1: Understand Fibonacci recursion

   fib(5) = fib(4) + fib(3). Calculate fib(4) and fib(3) similarly until base cases fib(1)=1 and fib(0)=0.

2. Step 2: Calculate fib(5) step-by-step

   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:

   Fibonacci of 5 = 5 [OK]

Hint: Trace base cases and add results up to n [OK]

Common Mistakes:

• Confusing fib(5) with fib(6)
• Miscounting base cases
• Assuming fib(0) = 1 instead of 0

4. The following Rust code attempts to implement a greedy algorithm for coin change but has a bug. What is the bug?

fn coin_change(coins: &[u32], amount: u32) -> u32 {
    let mut remaining = amount;
    let mut count = 0;
    for &coin in coins {
        count += remaining / coin;
        remaining %= coin;
    }
    if remaining == 0 { count } else { 0 }
}

fn main() {
    let coins = [1, 2, 5];
    println!("{}", coin_change(&coins, 11));
}

medium

A. The loop should break when remaining is zero

B. The function returns 0 instead of -1 when change is impossible

C. Coins are not sorted in descending order, so greedy fails

D. The function uses integer division which causes errors

Solution

1. Step 1: Check coin order

   Greedy coin change requires coins sorted in descending order to pick largest first. Here coins=[1, 2, 5] are ascending, so it picks smallest first.

2. Step 2: Trace execution

   For amount=11: coin=1, count=11, remaining=0; subsequent coins add 0. Returns 11, but optimal is 3 (5+5+1). Greedy suboptimal.

3. Final Answer:

   Coins are not sorted in descending order, so greedy fails -> Option C

4. Quick Check:

   Greedy needs descending coin order [OK]

Hint: Always sort coins descending for greedy coin change [OK]

Common Mistakes:

• Assuming greedy always works for any coin set
• Ignoring coin order importance
• Confusing return values for impossible change

5. You want to find the minimum number of coins needed to make a certain amount using coins of denominations [1, 3, 4]. Which approach is best and why?

hard

A. Use dynamic programming to store minimum coins needed for smaller amounts and build up

B. Use recursion without memoization to explore all combinations

C. Use a greedy algorithm because picking the largest coin first always gives minimum coins

D. Use simple iteration over coins without storing intermediate results

Solution

1. Step 1: Analyze greedy suitability

   Greedy fails here because picking largest coin first (4) may not lead to minimum coins for all amounts (e.g., amount=6).

2. Step 2: Consider recursion and DP

   Recursion without memoization repeats calculations and is inefficient. Dynamic programming stores results for smaller amounts, avoiding recomputation and ensuring minimum coins.

3. Final Answer:

   Use dynamic programming to store minimum coins needed for smaller amounts and build up -> Option A

4. Quick Check:

   Minimum coins with overlapping subproblems = Dynamic Programming [OK]

Hint: DP stores sub-results to find minimum coins efficiently [OK]

Common Mistakes:

• Assuming greedy always works for coin change
• Using recursion without memoization causing slow code
• Ignoring overlapping subproblems in coin change