Activity Selection Problem in DSA Java - 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.

// Assumes activities sorted by finish time
int selectActivities(int[] start, int[] finish) {
  int count = 1; // first activity always selected
  int lastFinish = finish[0];
  for (int i = 1; i < finish.length; i++) {
    if (start[i] >= lastFinish) {
      count++;
      lastFinish = finish[i];
    }
  }
  return count;
}
    

This code picks the maximum number of activities that don't overlap by checking each activity once.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop that goes through all activities once.
• How many times: Exactly once for each activity after the first.

How Execution Grows With Input

As the number of activities grows, the code 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 work grows directly with the number of activities.

Final Time Complexity

Time Complexity: O(n)

This means the time to select activities grows in a straight line with the number of activities.

Common Mistake

[X] Wrong: "The code checks every pair of activities, so it must be O(n²)."

[OK] Correct: The code only looks at each activity once in order, not comparing all pairs.

Interview Connect

Understanding this linear time solution shows you can find efficient answers by smart ordering and simple loops, a key skill in many problems.

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. 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