Activity Selection Problem in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort activities by their finish time.

DSA Java

Arrays.sort(activities, (a, b) -> a.[1] - b.finish);

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

Astart

Bduration

Cfinish

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by start time instead of finish time.

Using wrong field names.

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

2fill in blank

medium

Complete the code to select the first activity after sorting.

DSA Java

int count = 1;
int lastFinish = activities[0].[1];

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

Aduration

Bfinish

Cstart

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using start time instead of finish time.

Not initializing count to 1.

We select the first activity and store its finish time to compare with next activities.

3fill in blank

hard

Fix the error in the condition to select the next compatible activity.

DSA Java

for (int i = 1; i < activities.length; i++) {
    if (activities[i].[1] >= lastFinish) {
        count++;
        lastFinish = activities[i].finish;
    }
}

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

Aindex

Bfinish

Cduration

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing finish time instead of start time.

Using wrong comparison operator.

We check if the start time of the current activity is after or equal to the last selected activity's finish time.

4fill in blank

hard

Fill both blanks to create a method that returns the maximum number of activities.

DSA Java

public static int maxActivities(Activity[] activities) {
    Arrays.sort(activities, (a, b) -> a.[1] - b.[2]);
    int count = 1;
    int lastFinish = activities[0].finish;
    for (int i = 1; i < activities.length; i++) {
        if (activities[i].start >= lastFinish) {
            count++;
            lastFinish = activities[i].finish;
        }
    }
    return count;
}

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

Afinish

Bstart

Cduration

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using different fields for sorting.

Sorting by start time instead of finish time.

Both blanks must be 'finish' to sort activities by their finish time.

5fill in blank

hard

Fill all three blanks to create a loop that counts compatible activities.

DSA Java

int count = 1;
int lastFinish = activities[0].finish;
for (int i = [1]; i < activities.length; i[2]) {
    if (activities[i].[3] >= lastFinish) {
        count++;
        lastFinish = activities[i].finish;
    }
}

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

A1

B++

Cstart

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop at 0 instead of 1.

Using wrong increment operator.

Comparing finish time instead of start time.

Start loop at 1, increment i by ++, and compare start time with lastFinish.

Practice

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

easy

A. Calculate the total time of all activities

B. Find the activity with the longest duration

C. Sort activities by their start time

D. Select the maximum number of activities that don't overlap

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 time.

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 -> Option D

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 all activity durations

2. Which of the following Java code snippets correctly sorts activities by their finish time for the Activity Selection Problem?

easy

A. Arrays.sort(activities, (a, b) -> a.finish - b.finish);

B. Arrays.sort(activities, (a, b) -> a.start - b.start);

C. Arrays.sort(activities, (a, b) -> b.finish - a.finish);

D. Arrays.sort(activities, (a, b) -> b.start - a.start);

Solution

1. Step 1: Identify sorting criteria for Activity Selection

   Activities must be sorted by their finish time in ascending order to apply the greedy approach.

2. Step 2: Match code to sorting by finish time ascending

   Arrays.sort(activities, (a, b) -> a.finish - b.finish); sorts by finish time ascending (a.finish - b.finish), which is correct.

3. Final Answer:

   Arrays.sort(activities, (a, b) -> a.finish - b.finish); -> Option A

4. Quick Check:

   Sort by finish time ascending = Arrays.sort(activities, (a, b) -> a.finish - b.finish); [OK]

Hint: Sort activities by finish time ascending [OK]

Common Mistakes:

• Sorting by start time instead of finish time
• Sorting in descending order
• Using incorrect comparator syntax

3. What is the output of the following Java code snippet for the Activity Selection Problem?

class Activity {
  int start, finish;
  Activity(int s, int f) { start = s; finish = f; }
}

Activity[] activities = {
  new Activity(1, 4),
  new Activity(3, 5),
  new Activity(0, 6),
  new Activity(5, 7),
  new Activity(8, 9),
  new Activity(5, 9)
};

Arrays.sort(activities, (a, b) -> a.finish - b.finish);

int count = 1;
int lastFinish = activities[0].finish;
for (int i = 1; i < activities.length; i++) {
  if (activities[i].start >= lastFinish) {
    count++;
    lastFinish = activities[i].finish;
  }
}
System.out.println(count);

medium

A. 5

B. 3

C. 4

D. 6

Solution

1. Step 1: Sort activities by finish time ascending

   The activities in sorted order (stable sort preserves order of ties): (1,4), (3,5), (0,6), (5,7), (8,9), (5,9)

2. Step 2: Trace the greedy selection

   Pick (1,4), count=1, lastFinish=4
   Skip (3,5) since 3 < 4
   Skip (0,6) since 0 < 4
   Pick (5,7) since 5 >= 4, count=2, lastFinish=7
   Pick (8,9) since 8 >= 7, count=3, lastFinish=9
   Skip (5,9) since 5 < 9

3. Step 3: Check if any other activities can be selected

   All activities considered, total count = 4 (including the first selected activity)

4. Final Answer:

   4 -> Option C

5. Quick Check:

   Greedy selection count = 4 [OK]

Hint: Sort by finish time, pick next starting after last finish [OK]

Common Mistakes:

• Not sorting activities before selection
• Selecting overlapping activities
• Incorrectly updating last finish time

4. Identify the error in the following Java code snippet for the Activity Selection Problem:

Arrays.sort(activities, (a, b) -> a.finish + b.finish);

int count = 1;
int lastFinish = activities[0].finish;
for (int i = 1; i < activities.length; i++) {
  if (activities[i].start >= lastFinish) {
    count++;
    lastFinish = activities[i].finish;
  }
}
System.out.println(count);

medium

A. Loop starts from index 0 instead of 1

B. Comparator uses addition instead of subtraction for sorting

C. Condition should check start > lastFinish, not >= lastFinish

D. Count variable is not initialized

Solution

1. Step 1: Analyze the comparator in Arrays.sort

   The comparator uses a.finish + b.finish, which does not correctly compare finish times. It should use subtraction (a.finish - b.finish) to sort ascending.

2. Step 2: Understand impact of incorrect comparator

   Using addition returns a positive number regardless of order, causing incorrect sorting and wrong activity selection.

3. Final Answer:

   Comparator uses addition instead of subtraction for sorting -> Option B

4. Quick Check:

   Comparator must subtract finish times [OK]

Hint: Use subtraction in comparator, not addition [OK]

Common Mistakes:

• Using addition in comparator instead of subtraction
• Starting loop at wrong index
• Misunderstanding condition operator

5. Given the following activities with start and finish times:

int[] start = {1, 3, 0, 5, 8, 5};
int[] finish = {2, 4, 6, 7, 9, 9};

Which set of activities selected by the greedy algorithm (selecting earliest finish time first) is correct?

hard

A. [Activity(1,2), Activity(3,4), Activity(5,7), Activity(8,9)]

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

C. [Activity(1,2), Activity(5,9)]

D. [Activity(3,4), Activity(5,7), Activity(5,9)]

Solution

1. Step 1: Sort activities by finish time ascending

   Activities sorted by finish time:
   (1,2), (3,4), (0,6), (5,7), (8,9), (5,9)

2. Step 2: Select activities greedily

   Pick (1,2) first.
   Next, (3,4) start=3 >=2 select.
   (0,6) start=0 <4 skip.
   (5,7) start=5 >=4 select.
   (8,9) start=8 >=7 select.
   (5,9) start=5 <7 skip.

3. Step 3: Final selected set

   [Activity(1,2), Activity(3,4), Activity(5,7), Activity(8,9)]

4. Final Answer:

   [Activity(1,2), Activity(3,4), Activity(5,7), Activity(8,9)] -> Option A

5. Quick Check:

   Greedy selection = [Activity(1,2), Activity(3,4), Activity(5,7), Activity(8,9)] [OK]

Hint: Pick activities with earliest finish time, skip overlapping [OK]

Common Mistakes:

• Choosing activities with longer duration
• Selecting overlapping activities
• Ignoring sorting by finish time