Greedy vs DP How to Know Which to Apply in DSA Rust - Build Both Approaches

Greedy vs DP: How to Know Which to Apply

📖 Scenario: Imagine you are organizing a small event and need to select tasks to complete within a limited time. Some tasks give more value than others. You want to maximize the total value of tasks done.This project will help you understand when to use a simple greedy approach and when to use a dynamic programming (DP) approach by comparing both methods on the same problem.

🎯 Goal: You will build two Rust programs: one using a greedy method and one using dynamic programming to select tasks. You will see how each method works and when each is suitable.

📋 What You'll Learn

Create a list of tasks with exact durations and values

Create a time limit variable

Implement a greedy algorithm to select tasks based on value per duration

Implement a dynamic programming algorithm to select tasks for maximum total value within time limit

Print the selected tasks and total value for both methods

💡 Why This Matters

🌍 Real World

Choosing tasks or projects to maximize value within limited time or resources is common in event planning, project management, and resource allocation.

💼 Career

Understanding when to use greedy algorithms versus dynamic programming helps software engineers optimize solutions for scheduling, budgeting, and decision-making problems.

1

Create the list of tasks

Create a vector called tasks with these exact tuples representing (duration, value): (1, 10), (2, 15), (3, 40), (2, 30), (1, 20).

DSA Rust

let tasks = vec![(1, 10), (2, 15), (3, 40), (2, 30), (1, 20)];
// Your code here

Hint

Use vec![(duration, value), ...] to create the list.

2

Set the total available time

Create a variable called time_limit and set it to 5.

DSA Rust

fn main() {
    let tasks = vec![(1, 10), (2, 15), (3, 40), (2, 30), (1, 20)];
    let time_limit = 5;
    // Your code here
}

Hint

Just create a variable with the exact name and value.

3

Implement the greedy selection

Write code to select tasks greedily by sorting tasks by value per duration in descending order, then pick tasks while total duration is within time_limit. Store selected tasks in greedy_selected and total value in greedy_value.

DSA Rust

fn main() {
    let tasks = vec![(1, 10), (2, 15), (3, 40), (2, 30), (1, 20)];
    let time_limit = 5;

    // Your code here
}

Hint

Sort by value/duration ratio, then pick tasks until time limit is reached.

4

Implement dynamic programming and print results

Write code to use dynamic programming to find the maximum total value of tasks within time_limit. Store the result in dp_value. Then print the selected tasks and total values for both greedy and DP methods exactly as shown: println!("Greedy selected tasks: {:?}, total value: {}", greedy_selected, greedy_value); and println!("DP maximum value: {}", dp_value);

DSA Rust

fn main() {
    let tasks = vec![(1, 10), (2, 15), (3, 40), (2, 30), (1, 20)];
    let time_limit = 5;

    let mut tasks_sorted = tasks.clone();
    tasks_sorted.sort_by(|a, b| (b.1 as f64 / b.0 as f64).partial_cmp(&(a.1 as f64 / a.0 as f64)).unwrap());

    let mut total_time = 0;
    let mut greedy_value = 0;
    let mut greedy_selected = Vec::new();

    for &(duration, value) in &tasks_sorted {
        if total_time + duration <= time_limit {
            greedy_selected.push((duration, value));
            total_time += duration;
            greedy_value += value;
        }
    }

    // Your code here
}

Hint

Use a 2D vector for DP. Then print results exactly as shown.

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