Activity Selection Problem in DSA Typescript - Build from Scratch

Activity Selection Problem

📖 Scenario: You are organizing a conference with many talks. Each talk has a start time and an end time. You want to attend the maximum number of talks without any overlaps.

🎯 Goal: Build a program that selects the maximum number of non-overlapping activities (talks) you can attend.

📋 What You'll Learn

Create an array of activities with start and end times

Sort activities by their end times

Select the maximum number of activities that do not overlap

Print the selected activities in order

💡 Why This Matters

🌍 Real World

Scheduling talks, meetings, or tasks to maximize usage of time without conflicts.

💼 Career

Useful for roles in project management, event planning, and software development involving scheduling algorithms.

1

Create the list of activities

Create an array called activities with these exact objects: { start: 1, end: 4 }, { start: 3, end: 5 }, { start: 0, end: 6 }, { start: 5, end: 7 }, { start: 8, end: 9 }, { start: 5, end: 9 }

DSA Typescript

// Your code here

Hint

Use an array of objects with start and end properties.

2

Sort activities by their end time

Create a new array called sortedActivities by sorting activities in ascending order of their end property using activities.slice().sort((a, b) => a.end - b.end)

DSA Typescript

const activities = [
  { start: 1, end: 4 },
  { start: 3, end: 5 },
  { start: 0, end: 6 },
  { start: 5, end: 7 },
  { start: 8, end: 9 },
  { start: 5, end: 9 }
];
// Your code here

Hint

Use slice() to copy the array before sorting to keep original intact.

3

Select maximum number of non-overlapping activities

Create an array called selectedActivities and add the first activity from sortedActivities. Then use a for loop with variable i from 1 to sortedActivities.length. Inside the loop, if sortedActivities[i].start is greater or equal to the end of the last selected activity, add sortedActivities[i] to selectedActivities

DSA Typescript

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

const sortedActivities = activities.slice().sort((a, b) => a.end - b.end);
// Your code here

Hint

Start with the first activity and add others only if they start after the last selected activity ends.

4

Print the selected activities

Use console.log to print the selectedActivities array

DSA Typescript

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

const sortedActivities = activities.slice().sort((a, b) => a.end - b.end);

const selectedActivities = [sortedActivities[0]];

for (let i = 1; i = selectedActivities[selectedActivities.length - 1].end) {
    selectedActivities.push(sortedActivities[i]);
  }
}
// Your code here

Hint

Print the selectedActivities array to see the chosen talks.

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