Minimum Number of Platforms in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the problem statement of the Minimum Number of Platforms?

Given arrival and departure times of trains, find the minimum number of platforms needed so that no train waits.

beginner

What is the main approach to solve the Minimum Number of Platforms problem?

Sort arrival and departure times separately, then use two pointers to count how many trains are at the station at the same time.

intermediate

Why do we sort arrival and departure times separately in the Minimum Number of Platforms problem?

Sorting helps to process events in chronological order, making it easy to track how many trains overlap at any time.

beginner

What does the variable 'platform_needed' represent in the algorithm?

'platform_needed' tracks the current number of trains at the station, increasing on arrival and decreasing on departure.

intermediate

How do you update the maximum number of platforms needed during the algorithm?

Keep updating the maximum platforms needed whenever 'platform_needed' increases beyond the current maximum.

What is the first step in solving the Minimum Number of Platforms problem?

A Sort arrival and departure times separately

B Count total trains

C Calculate average waiting time

D Merge arrival and departure times

Sorting arrival and departure times separately allows us to process events in order.

If a train arrives at 10:00 and another departs at 10:00, do they overlap?

A Yes, they overlap

B Only if arrival is before departure

C No, they do not overlap

D Depends on platform count

If arrival time equals departure time, the train leaving frees the platform before the next arrives.

What data structure is commonly used to keep track of platforms needed?

A Two pointers with sorted arrays

B Stack

C Queue

D Hash map

Two pointers iterate over sorted arrival and departure arrays to count overlapping trains.

What does increasing 'platform_needed' indicate?

A No trains at station

B A train has departed

C Platform is freed

D A train has arrived

'platform_needed' increases when a train arrives and occupies a platform.

What is the time complexity of the Minimum Number of Platforms algorithm?

A O(n^2)

B O(n log n)

C O(n)

D O(log n)

Sorting takes O(n log n), and the two-pointer traversal is O(n), so total is O(n log n).

Explain the step-by-step process to find the minimum number of platforms needed for trains.

Think about how to count overlapping trains using sorted times.

Why is it important to sort arrival and departure times separately in this problem?

Sorting helps to simulate the timeline of train events.

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