Minimum Number of Platforms in DSA Typescript - Time & Space Complexity

Time Complexity: Minimum Number of Platforms

O(n log n)

Understanding Time Complexity

We want to find out how the time needed to calculate the minimum number of train platforms changes as the number of trains increases.

Specifically, how does the code handle more arrival and departure times?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function findMinPlatforms(arrivals: number[], departures: number[]): number {
  arrivals.sort((a, b) => a - b);
  departures.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++;
      i++;
      if (platformNeeded > maxPlatforms) maxPlatforms = platformNeeded;
    } else {
      platformNeeded--;
      j++;
    }
  }
  return maxPlatforms;
}
    

This code finds the minimum number of platforms needed so that no train waits, by sorting arrival and departure times and comparing them.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the arrival and departure arrays.
• How many times: Each sort runs once on n elements, and the while loop runs at most 2n times.

How Execution Grows With Input

As the number of trains (n) grows, sorting takes more time, and the loop compares arrivals and departures more times.

Input Size (n)	Approx. Operations
10	About 10 log 10 + 20 = 50 operations
100	About 100 log 100 + 200 = 1200 operations
1000	About 1000 log 1000 + 2000 = 13000 operations

Pattern observation: The operations grow a bit faster than the input size because of sorting, but the main loop grows linearly.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a little faster than the number of trains because sorting takes more time than just checking each train.

Common Mistake

[X] Wrong: "The while loop alone makes this O(n²) because it compares arrivals and departures repeatedly."

[OK] Correct: The loop actually moves forward through both arrays without restarting, so it runs at most 2n times, which is linear, not quadratic.

Interview Connect

Understanding how sorting and linear scanning combine helps you explain your approach clearly and shows you can analyze real problems efficiently.

Self-Check

"What if the arrival and departure arrays were already sorted? How would the time complexity change?"

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