Greedy vs DP How to Know Which to Apply in DSA Rust - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the problem has optimal substructure property.

DSA Rust

fn has_optimal_substructure(problem: &Problem) -> bool {
    return problem.can_be_broken_down [1] smaller_problems();
}

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

Aand

Bwith

Cinto

Dor

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'and' or 'or' instead of 'into' changes the meaning.

Confusing prepositions that don't fit the phrase.

The optimal substructure means the problem can be broken into smaller subproblems.

2fill in blank

medium

Complete the code to decide if greedy approach is suitable by checking if choices are locally optimal.

DSA Rust

fn is_greedy_applicable(problem: &Problem) -> bool {
    return problem.has_greedy_choice_property [1] true;
}

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

A=

B==

C!=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' instead of '==' causes syntax errors.

Using '!=' or '<' changes the logic incorrectly.

We check if the greedy choice property is equal to true to confirm greedy suitability.

3fill in blank

hard

Fix the error in the code that tries to memoize results for dynamic programming.

DSA Rust

fn dp_solve(problem: &Problem, memo: &mut Vec<Option<i32>>, index: usize) -> i32 {
    if let Some(result) = memo[[1]] {
        return result;
    }
    // compute result
    let result = problem.compute(index);
    memo[index] = Some(result);
    result
}

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

Aindex

Bresult

Cmemo

Dproblem

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'result' or 'memo' as index causes errors.

Using 'problem' as index is invalid.

The memoization array is indexed by index to store and retrieve results.

4fill in blank

hard

Fill both blanks to create a function that chooses between greedy and DP based on problem properties.

DSA Rust

fn choose_strategy(problem: &Problem) -> &'static str {
    if problem.has_greedy_choice_property [1] true && problem.has_optimal_substructure [2] true {
        "Greedy"
    } else {
        "Dynamic Programming"
    }
}

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

A==

B!=

C&&

D||

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' or logical OR changes the logic.

Mixing operators causes wrong strategy choice.

Both properties must be equal to true to choose greedy; otherwise, use DP.

5fill in blank

hard

Fill all three blanks to implement a DP solution with memoization and recursion.

DSA Rust

fn dp_recursive(problem: &Problem, memo: &mut Vec<Option<i32>>, [1]: usize) -> i32 {
    if [2] >= problem.size() {
        return 0;
    }
    if let Some(result) = memo[[3]] {
        return result;
    }
    let include = problem.value([2]) + dp_recursive(problem, memo, [2] + 1);
    let exclude = dp_recursive(problem, memo, [2] + 1);
    let result = include.max(exclude);
    memo[[2]] = Some(result);
    result
}

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

Aindex

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names causes errors.

Using 'count' instead of 'index' breaks logic.

The variable index is used consistently to track the current position in recursion and memoization.

Practice

1. Which of the following best describes when to use a greedy algorithm instead of dynamic programming?

easy

A. When the problem has overlapping subproblems and no optimal substructure

B. When the problem has an optimal substructure and no overlapping subproblems

C. When the problem requires checking all possible solutions

D. When the problem is unsolvable

Solution

1. Step 1: Understand greedy algorithm conditions

   Greedy algorithms work when choosing the best option at each step leads to a global optimum, which means the problem has an optimal substructure and no overlapping subproblems.

2. Step 2: Compare with dynamic programming conditions

   Dynamic programming is used when the problem has overlapping subproblems and optimal substructure, requiring storing intermediate results.

3. Final Answer:

   When the problem has an optimal substructure and no overlapping subproblems -> Option B

4. Quick Check:

   Greedy = optimal substructure + no overlap [OK]

Hint: Use greedy if no overlapping subproblems exist [OK]

Common Mistakes:

• Confusing overlapping subproblems as a greedy condition
• Using DP when greedy suffices
• Assuming greedy always finds optimal solution
• Ignoring problem structure before choosing approach

2. Which Rust code snippet correctly declares a vector and uses it to implement a greedy approach for selecting intervals sorted by end time?

easy

A. let intervals = vec![(1,3), (2,5), (4,7)]; intervals.reverse();

B. let intervals = vec![(1,3), (2,5), (4,7)]; intervals.sort();

C. let mut intervals = vec![(1,3), (2,5), (4,7)]; intervals.sort_by(|a,b| a.0.cmp(&b.0));

D. let mut intervals = vec![(1,3), (2,5), (4,7)]; intervals.sort_by_key(|k| k.1);

Solution

1. Step 1: Identify correct vector declaration and sorting

   let mut intervals = vec![(1,3), (2,5), (4,7)]; intervals.sort_by_key(|k| k.1); declares a mutable vector and sorts it by the second element (end time), which is needed for greedy interval scheduling.

2. Step 2: Check other options for correctness

   let intervals = vec![(1,3), (2,5), (4,7)]; intervals.sort(); tries to sort an immutable vector and uses default sort (by first element), let mut intervals = vec![(1,3), (2,5), (4,7)]; intervals.sort_by(|a,b| a.0.cmp(&b.0)); sorts by start time, not end time, and let intervals = vec![(1,3), (2,5), (4,7)]; intervals.reverse(); reverses without sorting.

3. Final Answer:

   let mut intervals = vec![(1,3), (2,5), (4,7)]; intervals.sort_by_key(|k| k.1); -> Option D

4. Quick Check:

   Sort intervals by end time with mutable vector [OK]

Hint: Sort intervals by end time for greedy scheduling [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Using immutable vector when sorting is needed
• Not sorting intervals before greedy selection
• Using reverse instead of sort

3. What is the output of the following Rust code implementing a greedy approach to select maximum intervals?

let mut intervals = vec![(1,3), (2,5), (4,7)];
intervals.sort_by_key(|k| k.1);
let mut count = 0;
let mut end = 0;
for &(start, finish) in &intervals {
    if start >= end {
        count += 1;
        end = finish;
    }
}
println!("{}", count);

medium

A. 2

B. 3

C. 1

D. 0

Solution

1. Step 1: Sort intervals by end time

   Sorted intervals: (1,3), (2,5), (4,7)

2. Step 2: Iterate and count non-overlapping intervals

   First interval (1,3): start=1 >= end=0, count=1, end=3 Second interval (2,5): start=2 < end=3, skip Third interval (4,7): start=4 >= end=3, count=2, end=7

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Greedy interval count = 2 [OK]

Hint: Count intervals where start >= last end [OK]

Common Mistakes:

• Counting overlapping intervals
• Not updating end after selection
• Sorting incorrectly
• Confusing start and end values

4. The following Rust code attempts to solve the coin change problem using a greedy approach. What is the bug?

let coins = vec![1, 3, 4];
let mut amount = 6;
let mut count = 0;
for &coin in coins.iter().rev() {
    while amount >= coin {
        amount -= coin;
        count += 1;
    }
}
println!("{}", count);

medium

A. The amount variable is not mutable

B. The coins vector is not sorted ascending

C. Greedy fails because coin denominations do not guarantee optimal solution

D. The loop should not reverse the coins vector

Solution

1. Step 1: Understand greedy coin change limitation

   Greedy works only if coin denominations are canonical (like 1,5,10). Here, 1,3,4 can cause greedy to fail.

2. Step 2: Check code behavior

   Greedy picks 4 once (amount=2), then picks 1 twice (total 3 coins). But optimal is 2 coins (3+3).

3. Final Answer:

   Greedy fails because coin denominations do not guarantee optimal solution -> Option C

4. Quick Check:

   Greedy fails on non-canonical coins [OK]

Hint: Greedy fails if coin set is not canonical [OK]

Common Mistakes:

• Assuming greedy always works for coin change
• Ignoring coin order importance
• Not checking for optimal substructure
• Confusing mutability of variables

5. You want to find the minimum number of coins to make amount 6 using coins [1, 3, 4]. Which approach is best and why? A) Greedy picking largest coin first B) Brute force checking all combinations C) Dynamic programming storing minimum coins for each amount D) Sorting coins ascending and picking smallest first

hard

A. Dynamic programming storing minimum coins for each amount

B. Brute force checking all combinations

C. Greedy picking largest coin first

D. Sorting coins ascending and picking smallest first

Solution

1. Step 1: Analyze greedy approach limitations

   Greedy may pick 4 then 1s, resulting in 3 coins, but optimal is 2 coins (3+3).

2. Step 2: Understand dynamic programming advantage

   DP stores minimum coins needed for each amount, ensuring optimal solution even with tricky coin sets.

3. Step 3: Compare other options

   Brute force is slow; sorting ascending and picking smallest first is not optimal strategy.

4. Final Answer:

   Dynamic programming storing minimum coins for each amount -> Option A

5. Quick Check:

   DP handles overlapping subproblems for optimal coin count [OK]

Hint: Use DP for minimum coins with tricky denominations [OK]

Common Mistakes:

• Using greedy for non-canonical coin sets
• Ignoring performance of brute force
• Assuming sorting smallest first helps
• Not recognizing overlapping subproblems