Why Minimum Number of Platforms in DSA Typescript?

The Big Idea

Discover how to solve a real-world scheduling puzzle with simple sorting and counting!

The Scenario

Imagine a busy train station where many trains arrive and depart at different times. You want to know how many platforms are needed so that no train has to wait for a platform to be free.

The Problem

Trying to count platforms manually by checking each train's arrival and departure times is slow and confusing. It's easy to make mistakes, especially when many trains overlap in time.

The Solution

The "Minimum Number of Platforms" method helps us quickly find the smallest number of platforms needed by sorting arrival and departure times and counting overlaps efficiently.

Before vs After

✗ Before

let platforms = 0;
for (let i = 0; i < trains.length; i++) {
  for (let j = 0; j < trains.length; j++) {
    if (trains[i].arrival < trains[j].departure && trains[j].arrival < trains[i].departure) {
      platforms++;
    }
  }
}

✓ After

let arrivals = trains.map(t => t.arrival).sort((a, b) => a - b);
let departures = trains.map(t => t.departure).sort((a, b) => a - b);
let platformNeeded = 0, maxPlatforms = 0;
let i = 0, j = 0;
while (i < arrivals.length && j < departures.length) {
  if (arrivals[i] < departures[j]) {
    platformNeeded++;
    maxPlatforms = Math.max(maxPlatforms, platformNeeded);
    i++;
  } else {
    platformNeeded--;
    j++;
  }
}

What It Enables

This concept lets you efficiently plan resources to avoid delays and overcrowding in busy schedules.

Real Life Example

Railway stations use this to decide how many platforms to build so trains can arrive and leave smoothly without waiting.

Key Takeaways

Manual checking of overlapping trains is slow and error-prone.

Sorting arrivals and departures helps count overlaps quickly.

Minimum platforms needed equals the maximum number of overlapping trains.

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