Activity Selection Problem in DSA Rust - Build from Scratch

Activity Selection Problem

📖 Scenario: You are organizing a conference and have a list of activities with their start and end times. You want to select the maximum number of activities that don't overlap so attendees can join as many as possible.

🎯 Goal: Build a Rust program that selects the maximum number of non-overlapping activities from a given list using the Activity Selection Problem approach.

📋 What You'll Learn

Create a vector of tuples called activities with exact start and end times

Create a variable called selected_activities to store chosen activities

Sort the activities by their end times

Use a for loop with variables start and end to select activities

Print the selected_activities vector at the end

💡 Why This Matters

🌍 Real World

Scheduling meetings, conferences, or tasks where time conflicts must be avoided to maximize participation.

💼 Career

Understanding greedy algorithms like the Activity Selection Problem is useful for software engineers working on scheduling, resource allocation, and optimization problems.

1

Create the list of activities

Create a vector called activities with these exact tuples representing start and end times: (1, 4), (3, 5), (0, 6), (5, 7), (8, 9), (5, 9).

DSA Rust

# Create the vector activities with given tuples
# Your code here

Hint

Use let mut activities = vec![(1, 4), (3, 5), (0, 6), (5, 7), (8, 9), (5, 9)]; to create the vector.

2

Prepare the selected activities list

Create a mutable vector called selected_activities to store the chosen activities. Initialize it as empty.

DSA Rust

fn main() {
    let mut activities = vec![(1, 4), (3, 5), (0, 6), (5, 7), (8, 9), (5, 9)];
    // Create an empty vector selected_activities
    // Your code here
}

Hint

Use let mut selected_activities: Vec<(i32, i32)> = Vec::new(); to create an empty vector.

3

Sort and select activities

Sort the activities vector by the end time (second element of the tuple). Then, use a for loop with variables start and end to iterate over activities. Select activities that start after or when the last selected activity ended. Use a variable last_end initialized to 0 to track the end time of the last selected activity. Append chosen activities to selected_activities.

DSA Rust

fn main() {
    let mut activities = vec![(1, 4), (3, 5), (0, 6), (5, 7), (8, 9), (5, 9)];
    let mut selected_activities: Vec = Vec::new();
    // Sort activities by end time
    // Your code here
    // Select activities using a for loop
    // Your code here
}

Hint

Use activities.sort_by_key(|k| k.1); to sort by end time. Then loop with for (start, end) in activities and select activities where start >= last_end.

4

Print the selected activities

Print the selected_activities vector to show the chosen activities.

DSA Rust

fn main() {
    let mut activities = vec![(1, 4), (3, 5), (0, 6), (5, 7), (8, 9), (5, 9)];
    let mut selected_activities: Vec = Vec::new();
    activities.sort_by_key(|k| k.1);
    let mut last_end = 0;
    for (start, end) in activities {
        if start >= last_end {
            selected_activities.push((start, end));
            last_end = end;
        }
    }
    // Print the selected_activities vector
    // Your code here
}

Hint

Use println!("{:?}", selected_activities); to print the vector.

Practice

1. What is the main goal of the Activity Selection Problem?

easy

A. Calculate the total duration of all activities

B. Select the maximum number of activities that don't overlap in time

C. Sort activities by their start time

D. Find the activity with the longest duration

Solution

1. Step 1: Understand the problem objective

   The Activity Selection Problem aims to choose the largest set of activities that do not overlap in their time intervals.

2. Step 2: Identify the correct goal

   Among the options, only selecting the maximum number of non-overlapping activities matches the problem's goal.

3. Final Answer:

   Select the maximum number of activities that don't overlap in time -> Option B

4. Quick Check:

   Goal = Max non-overlapping activities [OK]

Hint: Focus on maximizing non-overlapping activities [OK]

Common Mistakes:

• Confusing longest duration with maximum count
• Sorting by start time only
• Summing durations instead of counting activities

2. Which Rust code snippet correctly sorts a vector of activities by their finish time? Assume each activity is a tuple (start: i32, finish: i32).

easy

A. activities.sort_by_key(|a| a.1);

B. activities.sort_by_key(|a| a.0);

C. activities.sort_unstable_by(|a, b| b.1.cmp(&a.1));

D. activities.sort_by(|a, b| a.0.cmp(&b.0));

Solution

1. Step 1: Identify sorting key for finish time

   Each activity is a tuple (start, finish). The finish time is at index 1.

2. Step 2: Choose correct sorting method

   Sorting by finish time ascending uses sort_by_key(|a| a.1). activities.sort_by_key(|a| a.1); matches this.

3. Final Answer:

   activities.sort_by_key(|a| a.1); -> Option A

4. Quick Check:

   Sort by finish time = a.1 [OK]

Hint: Sort activities by finish time using .sort_by_key(|a| a.1) [OK]

Common Mistakes:

• Sorting by start time instead of finish time
• Sorting in descending order incorrectly
• Using unstable sort unnecessarily

3. What is the output of the following Rust code that implements the activity selection logic?

let mut activities = vec![(1, 4), (3, 5), (0, 6), (5, 7), (8, 9), (5, 9)];
activities.sort_by_key(|a| a.1);
let mut selected = vec![activities[0]];
let mut last_finish = activities[0].1;
for &activity in &activities[1..] {
    if activity.0 >= last_finish {
        selected.push(activity);
        last_finish = activity.1;
    }
}
println!("{:?}", selected);

medium

A. [(1, 4), (3, 5), (5, 7), (8, 9)]

B. [(0, 6), (5, 7), (8, 9)]

C. [(1, 4), (5, 7), (8, 9)]

D. [(1, 4), (5, 9)]

Solution

1. Step 1: Sort activities by finish time

   Sorted activities: [(1,4), (3,5), (0,6), (5,7), (8,9), (5,9)]

2. Step 2: Select activities greedily

   Start with (1,4). Next, (3,5) starts before 4, skip. (0,6) starts before 4, skip. (5,7) starts at 5 ≥ 4, select. (8,9) starts at 8 ≥ 7, select. (5,9) starts at 5 < 7, skip.

3. Final Answer:

   [(1, 4), (5, 7), (8, 9)] -> Option C

4. Quick Check:

   Greedy selection = [(1,4),(5,7),(8,9)] [OK]

Hint: Pick earliest finishing activities that start after last finish [OK]

Common Mistakes:

• Selecting overlapping activities
• Not sorting by finish time first
• Including activities starting before last finish

4. Identify the bug in this Rust code snippet for the Activity Selection Problem:

let mut activities = vec![(1, 3), (2, 5), (4, 7)];
activities.sort_by_key(|a| a.0);
let mut selected = vec![activities[0]];
let mut last_finish = activities[0].1;
for &activity in &activities[1..] {
    if activity.0 > last_finish {
        selected.push(activity);
        last_finish = activity.1;
    }
}
println!("{:?}", selected);

medium

A. Using > instead of >= in the if condition

B. Not pushing the first activity into selected

C. Not initializing last_finish correctly

D. Sorting by start time instead of finish time

Solution

1. Step 1: Check sorting criteria

   The code sorts activities by start time (a.0), but the greedy approach requires sorting by finish time.

2. Step 2: Understand impact of sorting

   Sorting by start time breaks the greedy logic, causing incorrect selection of activities.

3. Final Answer:

   Sorting by start time instead of finish time -> Option D

4. Quick Check:

   Sort by finish time needed, not start time [OK]

Hint: Always sort activities by finish time, not start time [OK]

Common Mistakes:

• Sorting by start time
• Using strict > instead of >= (less critical)
• Incorrect initialization of last_finish

5. Given a list of activities with start and finish times, how would you modify the standard Activity Selection Problem to select activities that maximize total duration instead of count? Which approach is correct?

hard

A. Use dynamic programming to consider all subsets and maximize total duration

B. Sort activities by start time and greedily pick those with longest duration

C. Sort by finish time and pick activities with shortest duration first

D. Select activities randomly and check total duration

Solution

1. Step 1: Understand difference in optimization goal

   Maximizing total duration is different from maximizing count; greedy approach fails here.

2. Step 2: Choose correct algorithmic approach

   Dynamic programming can consider all subsets to find the maximum total duration without overlaps.

3. Final Answer:

   Use dynamic programming to consider all subsets and maximize total duration -> Option A

4. Quick Check:

   Max total duration needs DP, not greedy [OK]

Hint: Max duration needs DP, not greedy selection [OK]

Common Mistakes:

• Trying greedy for max duration
• Sorting by start time only
• Random selection without logic