Activity Selection Problem in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort activities by their finish time.

DSA Rust

activities.sort_by_key(|activity| activity.[1]);

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Aduration

Bfinish

Cstart

Dpriority

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by start time instead of finish time

Sorting by duration or priority which is not required

We sort activities by their finish time to select the maximum number of non-overlapping activities.

2fill in blank

medium

Complete the code to initialize the first selected activity index.

DSA Rust

let mut last_selected = [1];

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A1

Bactivities.len() - 1

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from last index

Starting from 1 instead of 0

Using -1 which is invalid index

The first selected activity is at index 0 after sorting by finish time.

3fill in blank

hard

Fix the error in the condition to check if the current activity can be selected.

DSA Rust

if activities[i].start >= activities[[1]].finish {

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Alast_selected

Bi

C0

Dactivities.len() - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with current index i

Using 0 or last index incorrectly

We compare the start time of the current activity with the finish time of the last selected activity.

4fill in blank

hard

Fill both blanks to add the current activity to the selected list and update the last selected index.

DSA Rust

selected.push(activities[[1]].clone());
last_selected = [2];

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Ai

Blast_selected

C0

Dactivities.len() - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using last_selected instead of i for push

Not updating last_selected correctly

We add the current activity at index i and update last_selected to i.

5fill in blank

hard

Fill all three blanks to complete the activity selection loop.

DSA Rust

for [1] in 1..activities.len() {
    if activities[[2]].start >= activities[last_selected].finish {
        selected.push(activities[[3]].clone());
        last_selected = [3];
    }
}

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

Ai

B0

Dlast_selected

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong loop variable

Mixing indices in condition and push

The loop variable is i, used to check and select activities starting from index 1.

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