Why Minimum Number of Platforms in DSA Go?

The Big Idea

Discover how a simple sorting trick can save trains from waiting and chaos!

The Scenario

Imagine a busy train station where many trains arrive and depart at different times. You want to park all trains without any waiting, but you only have a few platforms. If you try to manage this by writing down each train's arrival and departure times on paper and checking overlaps manually, it becomes very confusing and slow.

The Problem

Manually checking each train's schedule to find overlapping times is slow and error-prone. You might miss some overlaps or waste platforms by not organizing the schedule well. This can cause trains to wait or block each other, leading to delays and frustration.

The Solution

The "Minimum Number of Platforms" concept helps you quickly find the smallest number of platforms needed so that no train has to wait. It uses a smart way to compare arrival and departure times, counting overlaps efficiently without checking every pair manually.

Before vs After

✗ Before

max_platforms := 0
for i := 0; i < n; i++ {
  count := 1
  for j := 0; j < n; j++ {
    if i != j && trains[i].arrival < trains[j].departure && trains[j].arrival < trains[i].departure {
      count++
    }
  }
  if count > max_platforms {
    max_platforms = count
  }
}

✓ After

sort.Ints(arrivals)
sort.Ints(departures)
platforms := 0
max_platforms := 0
i := 0
j := 0
for i < n && j < n {
  if arrivals[i] <= departures[j] {
    platforms++
    if platforms > max_platforms {
      max_platforms = platforms
    }
    i++
  } else {
    platforms--
    j++
  }
}

What It Enables

This concept enables efficient scheduling so that trains never wait, saving time and resources at busy stations.

Real Life Example

Railway stations use this method to decide how many platforms to build, ensuring smooth train arrivals and departures without delays.

Key Takeaways

Manual checking of train overlaps is slow and error-prone.

Sorting arrival and departure times helps count overlaps efficiently.

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 earliest train arrival time

B. The total number of trains arriving in a day

C. The longest time a train stays on a platform

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

Solution

1. Step 1: Understand the problem context

   The problem wants to find how many platforms are needed so that no train has to wait for another to leave.

2. Step 2: Identify the key requirement

   We must find the minimum number of platforms so that all trains can be accommodated without delay.

3. Final Answer:

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

4. Quick Check:

   Minimum platforms = no waiting trains [OK]

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

Common Mistakes:

• Confusing total trains with platforms needed
• Ignoring overlapping arrival and departure times
• Thinking about train durations instead of platform count

2. Which Go code snippet correctly declares slices for arrival and departure times for the Minimum Number of Platforms problem?

easy

A. arrivals := [900, 940, 950]; departures := [910, 1200, 1120]

B. arrivals := []int{900, 940, 950}; departures := []int{910, 1200, 1120}

C. arrivals := (900, 940, 950); departures := (910, 1200, 1120)

D. arrivals := {900, 940, 950}; departures := {910, 1200, 1120}

Solution

1. Step 1: Recall Go slice syntax

   In Go, slices are declared with []int{values}, not with parentheses or braces alone.

2. Step 2: Match correct syntax

   arrivals := []int{900, 940, 950}; departures := []int{910, 1200, 1120} uses []int{...} which is correct; others use invalid syntax.

3. Final Answer:

   arrivals := []int{900, 940, 950}; departures := []int{910, 1200, 1120} -> Option B

4. Quick Check:

   Go slices use []int{...} [OK]

Hint: Use []int{...} to declare slices in Go [OK]

Common Mistakes:

• Using parentheses () instead of braces {}
• Using braces {} without []int type
• Confusing arrays and slices syntax

3. What is the minimum number of platforms required for trains with arrival times [900, 940, 950, 1100, 1500, 1800] and departure times [910, 1200, 1120, 1130, 1900, 2000]?

medium

A. 3

B. 2

C. 4

D. 1

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 overlapping trains

   Between 940 and 1120, three trains overlap (940, 950, 1100 arrivals with departures after 940). So minimum platforms = 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Max overlap = 3 trains [OK]

Hint: Sort times and count overlaps to find max platforms [OK]

Common Mistakes:

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

4. Given this Go code snippet for Minimum Number of Platforms, what is the bug?

arrivals := []int{900, 940, 950}
departures := []int{910, 1200, 1120}
i, j, platforms := 0, 0, 0
for i < len(arrivals) && j < len(departures) {
  if arrivals[i] <= departures[j] {
    platforms++
    i++
  } else {
    platforms--
    j++
  }
}
fmt.Println(platforms)

medium

A. platforms should not be decremented when a train departs

B. The loop should run until i < len(arrivals) only

C. The comparison should be arrivals[i] < departures[j], not <=

D. platforms variable should be initialized to 1

Solution

1. Step 1: Understand platform counting logic

   When arrival time equals departure time, we should consider the train has left before the next arrives to avoid extra platform count.

2. Step 2: Fix comparison operator

   Using <= causes increment even when arrival equals departure, inflating platform count. Changing to < fixes this.

3. Final Answer:

   The comparison should be arrivals[i] < departures[j], not <= -> Option C

4. Quick Check:

   Use < to avoid counting simultaneous arrival/departure as overlap [OK]

Hint: Use strict less than (<) to avoid counting equal times as overlap [OK]

Common Mistakes:

• Using <= instead of < in time comparison
• Incorrect loop termination conditions
• Wrong initialization of platform count

5. You have arrival times [900, 940, 950, 1100, 1500, 1800] and departure times [910, 1200, 1120, 1130, 1900, 2000]. How would you modify the algorithm to also return the time intervals when the maximum number of platforms are needed?

hard

A. Track intervals when platform count reaches max during iteration

B. Sort only arrival times and print max arrival time intervals

C. Count total trains and divide by 2 to get intervals

D. Use a queue to store trains and print all arrival times

Solution

1. Step 1: Track platform count and max count during iteration

   While iterating sorted arrivals and departures, keep updating current platform count and record when it reaches a new maximum.

2. Step 2: Record time intervals for max platforms

   Store the start time when max platforms count is reached and end time when it decreases, to get intervals.

3. Final Answer:

   Track intervals when platform count reaches max during iteration -> Option A

4. Quick Check:

   Record max platform count intervals while counting platforms [OK]

Hint: Save start/end times when platform count hits max [OK]

Common Mistakes:

• Ignoring departure times when tracking intervals
• Only sorting arrivals without departures
• Using total trains count instead of overlaps