Activity Selection Problem in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort activities by their finish time.

DSA Typescript

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

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

Aduration

Bfinish

Cstart

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by start time instead of finish time.

Using a wrong property name.

Sorting activities by their finish time helps us select the maximum number of non-overlapping activities.

2fill in blank

medium

Complete the code to select the first activity after sorting.

DSA Typescript

selected.push(activities[[1]]);

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

Aactivities.length - 1

B1

Cactivities.length

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting selection from index 1 instead of 0.

Using the last index instead of the first.

The first activity after sorting by finish time is at index 0 and is always selected first.

3fill in blank

hard

Fix the error in the condition to check if the next activity can be selected.

DSA Typescript

if (activities[i].start >= [1]) {

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

Aactivities[i-1].finish

Bselected[selected.length]

Cselected[selected.length - 1].finish

Dactivities[0].start

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with the wrong activity's finish time.

Using an out-of-range index.

We compare the start time of the current activity with the finish time of the last selected activity.

4fill in blank

hard

Fill both blanks to complete the loop that selects compatible activities.

DSA Typescript

for (let i = [1]; i < activities.length; i++) {
  if (activities[i].start >= [2]) {
    selected.push(activities[i]);
  }
}

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

A1

B0

Cselected[selected.length - 1].finish

Dactivities[0].start

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop from 0 causing duplicate selection.

Comparing with wrong finish time.

Start loop from 1 because first activity is already selected. Compare start time with last selected activity's finish time.

5fill in blank

hard

Fill all three blanks to create a function that returns the maximum set of compatible activities.

DSA Typescript

function selectActivities(activities: {start: number, finish: number}[]): {start: number, finish: number}[] {
  activities.sort((a, b) => a.[1] - b.[2]);
  const selected = [activities[0]];
  for (let i = [3]; i < activities.length; i++) {
    if (activities[i].start >= selected[selected.length - 1].finish) {
      selected.push(activities[i]);
    }
  }
  return selected;
}

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

Afinish

Bstart

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by start time instead of finish time.

Starting loop from 0 causing duplicate selection.

Sort by finish time (both a and b). Start loop from 1 since first activity is pre-selected.

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 activities that do not overlap and maximize the count.

2. Step 2: Identify the correct objective

   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 A

4. Quick Check:

   Goal = Max non-overlapping activities [OK]

Hint: Focus on maximizing count without time conflicts [OK]

Common Mistakes:

• Confusing longest duration with maximum count
• Sorting by start time only
• Summing all activity durations

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

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) => b.start - a.start);

Solution

1. Step 1: Identify sorting criteria for Activity Selection

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

2. Step 2: Match sorting code to criteria

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

3. Final Answer:

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

4. Quick Check:

   Sort by finish time ascending [OK]

Hint: Sort activities by finish time ascending [OK]

Common Mistakes:

• Sorting by start time instead of finish time
• Sorting finish time descending
• Using incorrect comparison operators

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

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);

const selected = [];
let lastFinish = 0;
for (const activity of activities) {
  if (activity.start >= lastFinish) {
    selected.push(activity);
    lastFinish = activity.finish;
  }
}

console.log(selected);

medium

A. [{start:0, finish:6}, {start:8, finish:9}]

B. [{start:1, finish:4}, {start:3, finish:5}, {start:5, finish:7}]

C. [{start:1, finish:4}, {start:5, finish:7}, {start:8, finish:9}]

D. [{start:3, finish:5}, {start:5, finish:9}]

Solution

1. Step 1: Sort activities by finish time

   Sorted list: [{1,4}, {3,5}, {0,6}, {5,7}, {8,9}, {5,9}]

2. Step 2: Select activities greedily

   Select {1,4} (start 1 >= 0), then {5,7} (start 5 >= 4), then {8,9} (start 8 >= 7). Others overlap.

3. Final Answer:

   [{start:1, finish:4}, {start:5, finish:7}, {start:8, finish:9}] -> Option C

4. Quick Check:

   Greedy selection = chosen activities [OK]

Hint: Pick earliest finishing activities first [OK]

Common Mistakes:

• Selecting overlapping activities
• Not sorting by finish time
• Updating lastFinish incorrectly

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

const activities = [
  {start: 1, finish: 3},
  {start: 2, finish: 5},
  {start: 4, finish: 7}
];

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

const selected = [];
let lastFinish = 0;
for (const activity of activities) {
  if (activity.start >= lastFinish) {
    selected.push(activity);
    lastFinish = activity.finish;
  }
}

console.log(selected);

medium

A. Not initializing selected array

B. Sorting by start time instead of finish time

C. Using 'let' instead of 'const' for lastFinish

D. Incorrect comparison operator in if condition

Solution

1. Step 1: Check sorting criteria

   The code sorts activities by start time, but the greedy algorithm requires sorting by finish time.

2. Step 2: Identify impact of sorting error

   Sorting by start time can cause incorrect selection of overlapping activities, breaking the algorithm.

3. Final Answer:

   Sorting by start time instead of finish time -> Option B

4. Quick Check:

   Sort by finish time needed [OK]

Hint: Always sort by finish time for this problem [OK]

Common Mistakes:

• Sorting by start time
• Confusing variable declarations
• Wrong if condition operators

5. Given the activities with start and finish times:

const activities = [
  {start: 0, finish: 3},
  {start: 1, finish: 4},
  {start: 3, finish: 5},
  {start: 4, finish: 7},
  {start: 5, finish: 9},
  {start: 8, finish: 10}
];

How many activities can be selected using the Activity Selection Problem's greedy approach, and which are they?

hard

A. 4 activities: {0,3}, {3,5}, {5,9}, {8,10}

B. 4 activities: {0,3}, {3,5}, {4,7}, {8,10}

C. 5 activities: {0,3}, {1,4}, {3,5}, {4,7}, {8,10}

D. 3 activities: {0,3}, {3,5}, {5,9}

Solution

1. Step 1: Sort activities by finish time

   Sorted: [{0,3}, {1,4}, {3,5}, {4,7}, {5,9}, {8,10}]

2. Step 2: Select greedily

   {0,3} (last=3); skip {1,4}; {3,5} (last=5); {4,7} overlaps with {3,5} so skip; {5,9} (last=9); {8,10} overlaps with {5,9} so skip. Actually, {8,10} starts before 9, so overlaps. So selected are {0,3}, {3,5}, {5,9} only 3 activities.

3. Final Answer:

   3 activities: {0,3}, {3,5}, {5,9} -> Option D

4. Quick Check:

   Greedy picks non-overlapping max activities [OK]

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

Common Mistakes:

• Selecting overlapping activities
• Miscounting due to off-by-one errors
• Ignoring sorting step