Activity Selection Problem in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Activity Selection Problem?

To select the maximum number of activities that don't overlap in time from a given set of activities.

beginner

Which algorithmic approach is commonly used to solve the Activity Selection Problem?

A greedy algorithm that selects activities based on earliest finishing time first.

intermediate

Why do we sort activities by their finishing times in the Activity Selection Problem?

Sorting by finishing times helps pick the activity that leaves the most room for others, maximizing the total number of activities selected.

beginner

In the Activity Selection Problem, what condition must the next chosen activity satisfy?

Its start time must be greater than or equal to the finish time of the last selected activity.

intermediate

What is the time complexity of the greedy solution to the Activity Selection Problem after sorting?

O(n log n) due to sorting the activities by finish time; the selection process itself is O(n).

What is the first step in solving the Activity Selection Problem using a greedy approach?

A Sort activities by their finishing times

B Sort activities by their starting times

C Select the longest activity first

D Select activities randomly

Sorting activities by finishing times allows the greedy algorithm to pick the earliest finishing activity first.

After selecting an activity, what is the next step in the greedy algorithm for the Activity Selection Problem?

A Select the activity with the earliest start time

B Select the activity with the latest finishing time

C Select the activity with the longest duration

D Select the next activity that starts after the current one finishes

The algorithm picks the next activity whose start time is at or after the finish time of the last selected activity.

What does the Activity Selection Problem help to optimize?

A Minimum number of activities

B Maximum number of non-overlapping activities

C Maximum total duration of activities

D Minimum total duration of activities

The goal is to select the largest set of activities that do not overlap in time.

Which of these is NOT true about the greedy solution to the Activity Selection Problem?

A It selects activities based on earliest start time

B It requires sorting activities by finish time

C It always produces an optimal solution

D It runs in O(n log n) time due to sorting

The greedy solution selects activities based on earliest finish time, not start time.

If two activities have the same finishing time, which one should the greedy algorithm pick first?

A The one with the earlier start time

B The one with the later start time

C Either one, since they finish at the same time

D The one with the longer duration

Since they finish at the same time, either can be picked; the greedy choice property holds regardless, ensuring optimality.

Explain the steps of the greedy algorithm used to solve the Activity Selection Problem.

Think about how sorting helps and how you pick activities one by one.

Why does sorting activities by their finishing times guarantee an optimal solution in the Activity Selection Problem?

Consider how choosing the earliest finishing activity affects the rest.

Practice

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

easy

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

B. Find the activity with the longest duration

C. Sort activities by their start times

D. Calculate the total time of all activities

Solution

1. Step 1: Understand the problem goal

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

2. Step 2: Clarify what is NOT the goal

   Finding the longest activity or sorting alone does not solve the problem; the focus is on maximizing non-overlapping activities.

3. Final Answer:

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

4. Quick Check:

   Maximize non-overlapping activities = Select max activities [OK]

Hint: Focus on non-overlapping activities to maximize count [OK]

Common Mistakes:

• Thinking it's about longest activity duration
• Confusing sorting with selection
• Trying to maximize total time instead of count

2. Which of the following is the correct way to sort activities by their finish time in JavaScript?

easy

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

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

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

D. activities.sort((a, b) => a.duration - b.duration);

Solution

1. Step 1: Identify sorting key for Activity Selection

   We must sort activities by their finish time in ascending order to apply the greedy approach.

2. Step 2: Choose correct JavaScript sort syntax

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

3. Final Answer:

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

4. Quick Check:

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

Hint: Sort activities by finish time ascending for greedy selection [OK]

Common Mistakes:

• Sorting by start time instead of finish time
• Sorting finish time descending
• Sorting by duration which is irrelevant here

3. What will be the output of the following code snippet?

const activities = [
  {start: 1, finish: 4},
  {start: 3, finish: 5},
  {start: 0, finish: 6},
  {start: 5, finish: 7},
  {start: 8, finish: 9},
  {start: 5, finish: 9}
];

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

let count = 1;
let lastFinish = activities[0].finish;

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

medium

A. 6

B. 3

C. 5

D. 4

Solution

1. Step 1: Sort activities by finish time

   Sorted activities by finish time ascending: [{start:1, finish:4}, {start:3, finish:5}, {start:0, finish:6}, {start:5, finish:7}, {start:8, finish:9}, {start:5, finish:9}].

2. Step 2: Select activities greedily

   Start with first activity (finish=4). Next, select activities starting >=4: (start=5, finish=7), count=2; then (start=8, finish=9), count=3; then (start=9, finish=9) does not exist but next is (start=5, finish=9) which starts before 9, so skip. Total count is 4.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Greedy selection count = 4 [OK]

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

Common Mistakes:

• Not sorting activities before selection
• Counting overlapping activities
• Updating lastFinish incorrectly

4. Identify the bug in the following code for Activity Selection Problem:

function selectActivities(activities) {
  activities.sort((a, b) => a.start - b.start);
  let count = 1;
  let lastFinish = activities[0].finish;
  for (let i = 1; i < activities.length; i++) {
    if (activities[i].start > lastFinish) {
      count++;
      lastFinish = activities[i].finish;
    }
  }
  return count;
}

medium

A. Initialize count to 0 instead of 1

B. Comparison should be activities[i].start >= lastFinish

C. Sorting should be by finish time, not start time

D. No bug, code is correct

Solution

1. Step 1: Check sorting criteria

   The greedy approach requires sorting activities by their finish time, but the code sorts by start time, which is incorrect.

2. Step 2: Understand impact of wrong sorting

   Sorting by start time can cause wrong selection order, leading to suboptimal count of activities.

3. Final Answer:

   Sorting should be by finish time, not start time -> Option C

4. Quick Check:

   Sort by finish time for correct greedy selection [OK]

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

Common Mistakes:

• Sorting by start time instead of finish time
• Using > instead of >= in condition (less critical)
• Incorrect initialization of count

5. Given the activities with start and finish times:

const activities = [
  {start: 2, finish: 3},
  {start: 1, finish: 4},
  {start: 5, finish: 7},
  {start: 6, finish: 9},
  {start: 8, finish: 10},
  {start: 9, finish: 11}
];

What is the maximum number of non-overlapping activities that can be selected?

hard

A. 6

B. 4

C. 5

D. 3

Solution

1. Step 1: Sort activities by finish time

   Sorted activities by finish time ascending: [{start:2, finish:3}, {start:1, finish:4}, {start:5, finish:7}, {start:6, finish:9}, {start:8, finish:10}, {start:9, finish:11}].

2. Step 2: Select activities greedily

   Select (2,3), then (5,7), then (8,10), then (9,11) overlaps with (8,10) so skip; but (6,9) overlaps with (5,7) so skip. Total count is 4.

3. Step 3: Check for more selections

   Others overlap and cannot be selected without violating non-overlap. Maximum count is 4.

4. Final Answer:

   4 -> Option B

5. Quick Check:

   Max non-overlapping activities = 4 [OK]

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

Common Mistakes:

• Not sorting activities before selection
• Selecting overlapping activities
• Ignoring start time conditions