Why Minimum Number of Platforms in DSA Java?

The Big Idea

What if you could instantly know the exact number of platforms needed to keep every train on time?

The Scenario

Imagine you are managing a busy train station with many trains arriving and departing at different times. You need to decide how many platforms to build so that no train has to wait.

Without a clear method, you try to track each train's arrival and departure manually on paper.

The Problem

Manually checking each train's schedule to find overlapping times is slow and confusing. You might miss overlaps or count platforms incorrectly, causing delays or wasted space.

As the number of trains grows, this manual method becomes impossible to manage accurately.

The Solution

The Minimum Number of Platforms concept uses sorting and simple counting to find the exact number of platforms needed. It checks overlaps efficiently by comparing arrival and departure times.

This method quickly tells you the minimum platforms required so trains never wait, saving time and space.

Before vs After

✗ Before

for each train:
  for each other train:
    if times overlap:
      increment platform count

✓ After

sort arrivals and departures
while arrivals and departures remain:
  if next arrival < next departure:
    increment platform count
  else:
    decrement platform count

What It Enables

This concept enables efficient scheduling and resource planning in busy transport hubs and similar scenarios.

Real Life Example

Railway stations use this to build just enough platforms so trains arrive and leave smoothly without waiting or overcrowding.

Key Takeaways

Manual checking of overlapping train times is slow and error-prone.

Sorting arrivals and departures helps count overlaps efficiently.

Minimum Number of Platforms finds the least platforms needed to avoid train delays.

Practice

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

easy

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

B. The total number of trains arriving in a day

C. The longest time a train stays on a platform

D. The earliest train arrival time

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 how many trains overlap in time on the station to find the minimum platforms needed.

3. Final Answer:

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

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 earliest arrival only

2. Which Java data structure is best suited to efficiently find the minimum number of platforms needed given sorted arrival and departure arrays?

easy

A. PriorityQueue (Min-Heap)

B. Stack

C. LinkedList

D. HashMap

Solution

1. Step 1: Identify the need for quick minimum departure tracking

   We need to know the earliest train departure to free a platform, so a min-heap helps track minimum departure time efficiently.

2. Step 2: Match data structure to problem

   PriorityQueue in Java is a min-heap, perfect for this use case to manage platforms dynamically.

3. Final Answer:

   PriorityQueue (Min-Heap) -> Option A

4. Quick Check:

   Min-heap tracks earliest departure [OK]

Hint: Use min-heap to track earliest departure times [OK]

Common Mistakes:

• Using HashMap which doesn't maintain order
• Using Stack which is LIFO, not suitable
• Using LinkedList which is inefficient for min queries

3. What is the output of the following Java code snippet that calculates minimum platforms?

int[] arrivals = {900, 940, 950, 1100, 1500, 1800};
int[] departures = {910, 1200, 1120, 1130, 1900, 2000};
System.out.println(minPlatforms(arrivals, departures));

// Assume minPlatforms returns the minimum platforms needed

medium

A. 2

B. 4

C. 3

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 overlapping trains

   At 900-910: 1 train At 940-1120: 2 trains overlap At 950-1130: 3 trains overlap (max) After 1130, trains start leaving, so max platforms needed = 3

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Max overlap = 3 platforms [OK]

Hint: Find max overlapping intervals after sorting [OK]

Common Mistakes:

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

4. Identify the bug in this Java method to find minimum platforms:

public static int minPlatforms(int[] arr, int[] dep) {
    Arrays.sort(arr);
    Arrays.sort(dep);
    int i = 0, j = 0, platforms = 0, maxPlatforms = 0;
    while (i < arr.length && j < dep.length) {
        if (arr[i] <= dep[j]) {
            platforms++;
            i++;
        } else {
            platforms--;
            j++;
        }
        maxPlatforms = Math.max(maxPlatforms, platforms);
    }
    return maxPlatforms;
}

medium

A. The loop should run while i < arr.length only

B. The condition should be arr[i] < dep[j] instead of <= to avoid counting simultaneous arrival and departure

C. The platforms variable should not be decremented

D. Arrays.sort should be called only on arrivals

Solution

1. Step 1: Analyze the comparison condition

   The condition arr[i] <= dep[j] counts a train arriving exactly when another departs as overlapping, increasing platforms unnecessarily.

2. Step 2: Correct the condition to avoid overcounting

   Changing to arr[i] < dep[j] treats simultaneous arrival and departure as no overlap, which is correct for platform counting.

3. Final Answer:

   The condition should be arr[i] < dep[j] instead of <= to avoid counting simultaneous arrival and departure -> Option B

4. Quick Check:

   Use strict less than for arrival vs departure [OK]

Hint: Use < not <= to avoid counting simultaneous arrival and departure [OK]

Common Mistakes:

• Using <= causes overcounting platforms
• Not sorting both arrays
• Incorrect loop condition

5. Given arrival times [1000, 1010, 1025, 1030, 1040] and departure times [1020, 1035, 1045, 1050, 1100], what is the minimum number of platforms needed if trains arriving exactly when another departs do NOT require a new platform?

hard

A. 4

B. 5

C. 2

D. 3

Solution

1. Step 1: Sort and analyze arrival and departure times

   Arrivals: 1000, 1010, 1025, 1030, 1040 Departures: 1020, 1035, 1045, 1050, 1100

2. Step 2: Count overlapping trains considering no extra platform if arrival equals departure

   1000-1020: 1 train 1010-1035: overlaps with first, total 2 1025-1045: first gone at 1020, overlaps with second, total 2 1030-1050: overlaps with second (till 1035) and third (till 1045), total 3 1040-1100: second gone at 1035, overlaps with third, fourth, total 3 (max)

3. Step 3: Adjust for arrival equals departure rule

   No arrivals equal departures here, so max platforms = 3

4. Final Answer:

   3 -> Option D

5. Quick Check:

   Max overlap = 3 platforms [OK]

Hint: Count overlaps, exclude equal arrival and departure as overlap [OK]

Common Mistakes:

• Counting equal arrival and departure as overlap
• Not sorting times
• Miscounting overlaps