Activity Selection Problem in DSA Javascript - Time & Space Complexity

Time Complexity: Activity Selection Problem

O(n)

Understanding Time Complexity

We want to understand how the time needed to pick activities grows as the number of activities increases.

How does the program's work change when we have more activities to choose from?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function activitySelection(start, finish) {
  let n = start.length;
  let selected = [0];
  let last = 0;
  for (let i = 1; i < n; i++) {
    if (start[i] >= finish[last]) {
      selected.push(i);
      last = i;
    }
  }
  return selected;
}
    

This code selects the maximum number of activities that don't overlap, assuming activities are sorted by finish time.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop that checks each activity once.
• How many times: The loop runs once for each activity after the first, so n-1 times.

How Execution Grows With Input

As the number of activities grows, the program checks each one once to decide if it fits.

Input Size (n)	Approx. Operations
10	About 9 checks
100	About 99 checks
1000	About 999 checks

Pattern observation: The number of operations grows roughly in a straight line with the number of activities.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows directly in proportion to the number of activities.

Common Mistake

[X] Wrong: "The loop runs multiple times for each activity because of nested checks."

[OK] Correct: The code only has one simple loop that goes through activities once, so it does not multiply work inside the loop.

Interview Connect

Understanding this helps you explain how greedy choices can lead to efficient solutions, a key skill in problem solving.

Self-Check

"What if the activities were not sorted by finish time? How would the time complexity change?"

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