Minimum Number of Platforms in DSA Typescript - Execution Trace

Concept Flow - Minimum Number of Platforms

Sort arrival times

↓

Sort departure times

↓

Initialize pointers i=0, j=0

↓

Compare arrival[i

↓

Increment platforms

↓

Move i or j pointer

↓

Update max platforms

↓

Repeat until all trains processed

Sort arrival and departure times, then use two pointers to track platforms needed by comparing arrivals and departures.

Execution Sample

DSA Typescript

const arrival = [900, 940, 950, 1100, 1500, 1800];
const departure = [910, 1200, 1120, 1130, 1900, 2000];
// Find minimum platforms needed

This code finds the minimum number of platforms needed so that no train waits.

Execution Table

Step	Operation	i (arrival index)	j (departure index)	Current Platforms	Max Platforms	Visual State
1	Compare arrival[0]=900 and departure[0]=910	0	0	1	1	Platforms: 1 (Train at 900 arrived)
2	arrival[1]=940 < departure[0]=910? No, departure[0] < arrival[1]	1	0	0	1	Platforms: 0 (Train at 900 departed)
3	Move j to 1	1	1	0	1	Platforms: 0
4	Compare arrival[1]=940 and departure[1]=1200	1	1	1	1	Platforms: 1 (Train at 940 arrived)
5	Compare arrival[2]=950 and departure[1]=1200	2	1	2	2	Platforms: 2 (Train at 950 arrived)
6	Compare arrival[3]=1100 and departure[1]=1200	3	1	3	3	Platforms: 3 (Train at 1100 arrived)
7	Compare arrival[4]=1500 and departure[1]=1200	4	1	2	3	Platforms: 2 (Train at 1200 departed)
8	Move j to 2	4	2	2	3	Platforms: 2
9	Compare arrival[4]=1500 and departure[2]=1120	4	2	1	3	Platforms: 1 (Train at 1120 departed)
10	Move j to 3	4	3	1	3	Platforms: 1
11	Compare arrival[4]=1500 and departure[3]=1130	4	3	0	3	Platforms: 0 (Train at 1130 departed)
12	Move j to 4	4	4	0	3	Platforms: 0
13	Compare arrival[4]=1500 and departure[4]=1900	4	4	1	3	Platforms: 1 (Train at 1500 arrived)
14	Compare arrival[5]=1800 and departure[4]=1900	5	4	2	3	Platforms: 2 (Train at 1800 arrived)
15	Compare arrival[6]=undefined and departure[4]=1900	6	4	2	3	No more arrivals, process departures
16	Move j to 5	6	5	1	3	Platforms: 1 (Train at 1900 departed)
17	Move j to 6	6	6	0	3	Platforms: 0 (Train at 2000 departed)
18	End	6	6	0	3	All trains processed

💡 All trains processed: i and j pointers reached end of arrival and departure arrays.

Variable Tracker

Variable	Start	After Step 1	After Step 5	After Step 10	After Step 15	Final
i (arrival index)	0	1	3	4	6	6
j (departure index)	0	0	1	3	4	6
Current Platforms	0	1	3	1	2	0
Max Platforms	0	1	3	3	3	3

Key Moments - 3 Insights

Why do we increment 'j' (departure pointer) when departure[j] < arrival[i]?

Because a train has left before the next train arrives, so one platform is freed. See execution_table steps 2, 7, 9.

Why do we update max platforms only when arrival[i] < departure[j]?

Because a new train arrives before any train leaves, increasing platform need. See execution_table steps 1, 5, 6.

What happens when all arrivals are processed but departures remain?

We keep moving the departure pointer to free platforms until all trains have left. See execution_table steps 15-17.

Visual Quiz - 3 Questions

Look at the execution_table at step 6, how many platforms are currently needed?

A 1

B 2

C 3

D 0

💡 Hint

Check the 'Current Platforms' column at step 6 in execution_table.

At which step does the number of platforms first decrease from 3 to 2?

A Step 7

B Step 5

C Step 10

D Step 2

💡 Hint

Look for when 'Current Platforms' changes from 3 to 2 in execution_table.

If arrival[2] was 1300 instead of 950, how would max platforms change?

A Max platforms would increase

B Max platforms would decrease

C Max platforms would stay the same

D Cannot determine

💡 Hint

Consider how later arrival affects overlapping trains and platform count in variable_tracker.

Concept Snapshot

Minimum Number of Platforms:
- Sort arrival and departure times separately.
- Use two pointers i, j for arrivals and departures.
- Increment platforms if arrival[i] < departure[j].
- Decrement platforms if departure[j] <= arrival[i].
- Track max platforms needed during traversal.

Full Transcript

To find the minimum number of platforms needed for trains, first sort the arrival and departure times. Then use two pointers to compare the next arrival and departure. If a train arrives before the earliest departure, increase platform count. If a train departs before the next arrival, decrease platform count. Keep track of the maximum platforms needed at any time. Continue until all trains are processed. This ensures no train waits for a platform.

Practice

1. What does the "Minimum Number of Platforms" problem ask you to find given train arrival and departure times?

easy

A. The total time trains spend on the platform

B. The total number of trains arriving in a day

C. The maximum speed of trains on the platform

D. The least number of platforms needed so no train waits

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum number of platforms required so that no train has to wait for another to leave.

2. Step 2: Clarify what is being counted

   We count overlapping train schedules to find the peak number of trains at the station simultaneously.

3. Final Answer:

   The least number of platforms needed so no train waits -> Option D

4. Quick Check:

   Minimum platforms = max trains overlapping [OK]

Hint: Count overlapping trains to find minimum platforms needed [OK]

Common Mistakes:

• Confusing total trains with platforms needed
• Ignoring overlapping times
• Counting total arrival or departure times only

2. Which of the following TypeScript code snippets correctly sorts an array of train arrival times stored as numbers?

easy

A. arrivals.sort();

B. arrivals.sort((a, b) => a - b);

C. arrivals.sort((a, b) => b - a);

D. arrivals.sort((a, b) => a + b);

Solution

1. Step 1: Understand sorting numbers in TypeScript

   Default sort() treats elements as strings, so numeric sort needs a compare function.

2. Step 2: Identify correct compare function

   The function (a, b) => a - b sorts numbers in ascending order correctly.

3. Final Answer:

   arrivals.sort((a, b) => a - b); -> Option B

4. Quick Check:

   Numeric ascending sort uses (a, b) => a - b [OK]

Hint: Use (a, b) => a - b to sort numbers ascending [OK]

Common Mistakes:

• Using sort() without compare function for numbers
• Using (a, b) => a + b which is invalid
• Sorting descending when ascending needed

3. Given arrival times [900, 940, 950, 1100, 1500, 1800] and departure times [910, 1200, 1120, 1130, 1900, 2000], what is the minimum number of platforms needed?

medium

A. 2

B. 3

C. 4

D. 5

Solution

1. Step 1: Sort arrival and departure times

   Arrivals: 900, 940, 950, 1100, 1500, 1800; Departures: 910, 1120, 1130, 1200, 1900, 2000.

2. Step 2: Count platforms needed by comparing arrivals and departures

   At 900 arrive (1 platform), 910 depart (0), 940 arrive (1), 950 arrive (2), 1100 arrive (3), 1120 depart (2), 1130 depart (1), 1200 depart (0), 1500 arrive (1), 1800 arrive (2), 1900 depart (1), 2000 depart (0). Maximum platforms needed is 4.

3. Final Answer:

   4 -> Option C

4. Quick Check:

   Max overlapping trains = 4 [OK]

Hint: Track arrivals and departures to find max overlap [OK]

Common Mistakes:

• Not sorting times before processing
• Counting total trains instead of overlaps
• Mixing arrival and departure times

4. The following TypeScript code tries to find the minimum number of platforms but has a bug. What is the bug?

function findPlatforms(arrivals: number[], departures: number[]): number {
  arrivals.sort((a, b) => a - b);
  departures.sort((a, b) => a - b);
  let platformNeeded = 1, maxPlatforms = 1;
  let i = 1, j = 0;
  while (i < arrivals.length && j < departures.length) {
    if (arrivals[i] <= departures[j]) {
      platformNeeded++;
      i++;
    } else {
      platformNeeded--;
      j++;
    }
    if (platformNeeded > maxPlatforms) maxPlatforms = platformNeeded;
  }
  return maxPlatforms;
}

medium

A. Start i from 0 instead of 1 to include first arrival

B. Use < instead of <= in comparison to avoid counting overlaps wrongly

C. Initialize platformNeeded to 0 instead of 1

D. Sort arrivals and departures in descending order

Solution

1. Step 1: Analyze loop initialization

   Variable i starts at 1, skipping the first arrival time, which causes missing counting the first train.

2. Step 2: Correct initialization for accurate counting

   Starting i at 0 ensures all arrivals are considered, giving correct platform count.

3. Final Answer:

   Start i from 0 instead of 1 to include first arrival -> Option A

4. Quick Check:

   Include all arrivals by starting i=0 [OK]

Hint: Start pointers at 0 to include all trains [OK]

Common Mistakes:

• Skipping first arrival by starting i at 1
• Changing comparison operator incorrectly
• Initializing platformNeeded wrongly

5. You have trains arriving and departing as follows: Arrivals: [1000, 1010, 1025, 1030, 1040] Departures: [1015, 1020, 1035, 1045, 1050] If you want to minimize the number of platforms by rescheduling trains without changing their arrival order, which train's departure time should you delay to reduce platform count?

hard

A. Delay the departure of train arriving at 1010

B. Delay the departure of train arriving at 1000

C. Delay the departure of train arriving at 1040

D. Delay the departure of train arriving at 1025

Solution

1. Step 1: Identify overlapping trains causing max platforms

   Trains at 1010 and 1025 overlap because 1010 departs at 1020 and 1025 arrives before that.

2. Step 2: Delay departure of train at 1010 to after 1025 departs

   Delaying 1010's departure reduces overlap with 1025, lowering platform need.

3. Final Answer:

   Delay the departure of train arriving at 1010 -> Option A

4. Quick Check:

   Delay overlapping train's departure to reduce platforms [OK]

Hint: Delay departure of earliest overlapping train to reduce platforms [OK]

Common Mistakes:

• Delaying wrong train's departure
• Changing arrival order which is fixed
• Ignoring overlap timing details