Activity Selection Problem in DSA C - Time & Space Complexity

Time Complexity: Activity Selection Problem

O(n)

Understanding Time Complexity

We want to understand how the time needed to select 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.

// Activities sorted by finish time
int selectActivities(int start[], int finish[], int n) {
    int count = 1; // first activity always selected
    int last = 0;
    for (int i = 1; i < n; i++) {
        if (start[i] >= finish[last]) {
            count++;
            last = i;
        }
    }
    return count;
}
    

This code picks 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 (n - 1) times, where n is the number of activities.

How Execution Grows With Input

As the number of activities grows, the program checks each one once in order.

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 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 loops once through the list, comparing each activity to the last selected one, not all pairs.

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. Calculate the total time of all activities

B. Find the activity with the longest duration

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

D. Sort activities by their start time

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: 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 C

4. Quick Check:

   Goal = Max non-overlapping activities [OK]

Hint: Choose activities that finish earliest to fit more [OK]

Common Mistakes:

• Confusing longest duration with maximum count
• Sorting by start time only
• Trying to sum all activity times

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

easy

A. qsort(activities, n, sizeof(Activity), compareStart);

B. qsort(activities, n, sizeof(Activity), compareFinish);

C. sort(activities, n, sizeof(Activity), compareDuration);

D. qsort(activities, n, sizeof(Activity), compareName);

Solution

1. Step 1: Identify sorting criteria

   For the Activity Selection Problem, activities must be sorted by their finish time to select the maximum number of non-overlapping activities.

2. Step 2: Match sorting function

   Only qsort with compareFinish sorts by finish time correctly.

3. Final Answer:

   qsort(activities, n, sizeof(Activity), compareFinish); -> Option B

4. Quick Check:

   Sort by finish time = qsort + compareFinish [OK]

Hint: Sort activities by finish time using qsort and compareFinish [OK]

Common Mistakes:

• Sorting by start time instead of finish time
• Using undefined compare functions
• Using wrong sorting function name

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

typedef struct {
  int start;
  int finish;
} Activity;

Activity activities[] = {{1, 4}, {3, 5}, {0, 6}, {5, 7}, {8, 9}, {5, 9}};
int n = 6;

// Assume activities are sorted by finish time
int count = 1;
int last_finish = activities[0].finish;
for (int i = 1; i < n; i++) {
  if (activities[i].start >= last_finish) {
    count++;
    last_finish = activities[i].finish;
  }
}
printf("%d\n", count);

medium

A. 6

B. 3

C. 5

D. 4

Solution

1. Step 1: Trace the selection process

   Activities sorted by finish time: (1,4), (3,5), (0,6), (5,7), (8,9), (5,9). Start with count=1, last_finish=4.
   Check each activity:
   - (3,5): start 3 < 4, skip
   - (0,6): start 0 < 4, skip
   - (5,7): start 5 >= 4, select, count=2, last_finish=7
   - (8,9): start 8 >= 7, select, count=3, last_finish=9
   - (5,9): start 5 < 9, skip

2. Step 2: Final count of selected activities

   Total selected activities = 4

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Selected count = 4 [OK]

Hint: Count activities starting after last selected finish time [OK]

Common Mistakes:

• Not sorting activities by finish time
• Selecting overlapping activities
• Incorrectly updating last_finish

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

typedef struct {
  int start;
  int finish;
} Activity;

Activity activities[] = {{1, 3}, {2, 5}, {4, 7}};
int n = 3;

// Activities are sorted by finish time
int count = 0;
int last_finish = 0;
for (int i = 0; i < n; i++) {
  if (activities[i].start > last_finish) {
    count++;
    last_finish = activities[i].finish;
  }
}
printf("%d\n", count);

medium

A. The condition should be activities[i].start >= last_finish

B. The loop should start from i = 1, not 0

C. last_finish should be initialized to activities[0].finish

D. The activities array is not sorted

Solution

1. Step 1: Analyze the selection condition

   The condition uses '>' which excludes activities starting exactly at last_finish time. It should allow activities starting at or after last_finish, so '>=' is correct.

2. Step 2: Check other parts

   Starting loop at 0 and initializing last_finish to 0 is fine if condition is corrected. The activities are sorted as stated.

3. Final Answer:

   The condition should be activities[i].start >= last_finish -> Option A

4. Quick Check:

   Use >= to include activities starting at last finish [OK]

Hint: Use >= to allow activities starting exactly at last finish [OK]

Common Mistakes:

• Using > excludes valid activities
• Starting loop from 1 unnecessarily
• Wrong initialization of last_finish

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

Activities = [(1, 2), (3, 4), (0, 6), (5, 7), (8, 9), (5, 9)]

Which set of activities represents the maximum number of non-overlapping activities selected by the greedy algorithm?

hard

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

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

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

D. [(3, 4), (5, 7), (8, 9)]

Solution

1. Step 1: Sort activities by finish time

   Sorted activities: (1,2), (3,4), (0,6), (5,7), (8,9), (5,9)

2. Step 2: Select activities greedily

   Select (1,2) first.
   Next, (3,4) starts after 2, select.
   (0,6) overlaps, skip.
   (5,7) starts after 4, select.
   (8,9) starts after 7, select.
   (5,9) overlaps, skip.
   Selected set: [(1,2), (3,4), (5,7), (8,9)]

3. Final Answer:

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

4. Quick Check:

   Greedy selects max non-overlapping activities [OK]

Hint: Pick earliest finishing activities to maximize count [OK]

Common Mistakes:

• Choosing overlapping activities
• Not sorting by finish time
• Selecting longer activities first