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

Recall & Review

beginner

What is the main goal of the Minimum Number of Platforms problem?

To find the least number of train platforms required at a station so that no train has to wait for another to leave.

intermediate

Which data structures are commonly used to solve the Minimum Number of Platforms problem efficiently?

Two sorted arrays (one for arrival times and one for departure times) and pointers to traverse them.

intermediate

Explain the two-pointer technique used in the Minimum Number of Platforms problem.

Use one pointer to track arrivals and another for departures. Increment platform count when a train arrives before the earliest departure, else decrement when a train departs.

beginner

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 easier to count overlapping trains and platforms needed.

beginner

What does the maximum value of platforms needed during the traversal represent?

It represents the minimum number of platforms required so that no train waits.

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

A Calculate total travel time

B Count total trains

C Assign platforms randomly

D Sort arrival and departure times separately

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

When do we increase the platform count during traversal?

A When a train arrives before the earliest departure

B When no trains are at the station

C When trains arrive and depart at the same time

D When a train departs before the next arrival

We increase platform count when a train arrives before the earliest train departs, indicating overlap.

What does decrementing the platform count signify in the algorithm?

A A train has departed

B A train has arrived

C A platform is broken

D A train is delayed

Decrementing means a train has left, freeing a platform.

What is the time complexity of the efficient solution for Minimum Number of Platforms?

A O(n^2)

B O(n)

C O(n log n)

D O(log n)

Sorting takes O(n log n), and traversal is O(n), so overall O(n log n).

If arrival times are [9:00, 9:40, 9:50] and departure times are [9:10, 12:00, 11:20], what is the minimum number of platforms needed?

A 1

B 2

C 3

D 4

At 9:50, two trains overlap, so 2 platforms are needed.

Describe the step-by-step approach to find the minimum number of platforms needed for trains arriving and departing at different times.

Think about how to count overlapping trains using sorted times.

Explain why sorting arrival and departure times separately is crucial in solving the Minimum Number of Platforms problem.

Consider how events happen over time at the station.

Practice

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

easy

A. The earliest train arrival time

B. The minimum number of platforms needed so no train waits

C. The total number of trains arriving in a day

D. The maximum number of trains arriving at the same 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, which determines how many platforms are needed simultaneously.

3. Final Answer:

   The minimum number of platforms needed so no train waits -> Option B

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 only arrivals or departures

2. Which of the following C code snippets correctly declares arrays for train arrival and departure times for 5 trains?

easy

A. int arrival[5] = {900, 940, 950, 1100, 1500}; int departure[5] = {910, 1200, 1120, 1130, 1900};

B. int arrival = {900, 940, 950, 1100, 1500}; int departure = {910, 1200, 1120, 1130, 1900};

C. int arrival[5] = (900, 940, 950, 1100, 1500); int departure[5] = (910, 1200, 1120, 1130, 1900);

D. int arrival[] = 900, 940, 950, 1100, 1500; int departure[] = 910, 1200, 1120, 1130, 1900;

Solution

1. Step 1: Check array declaration syntax in C

   Arrays must be declared with size or empty brackets and initialized with curly braces {}.

2. Step 2: Validate each option

   int arrival[5] = {900, 940, 950, 1100, 1500}; int departure[5] = {910, 1200, 1120, 1130, 1900}; uses correct syntax with size and braces. Others use incorrect syntax like parentheses or missing braces.

3. Final Answer:

   int arrival[5] = {900, 940, 950, 1100, 1500}; int departure[5] = {910, 1200, 1120, 1130, 1900}; -> Option A

4. Quick Check:

   Arrays use {} with size or empty brackets [OK]

Hint: Use curly braces {} to initialize arrays in C [OK]

Common Mistakes:

• Using parentheses () instead of braces {}
• Missing array size or empty brackets
• Assigning values without braces

3. Given arrival times {900, 940, 950, 1100, 1500} and departure times {910, 1200, 1120, 1130, 1900}, what is the minimum number of platforms needed? Consider the following code snippet:

int arrival[] = {900, 940, 950, 1100, 1500};
int departure[] = {910, 1200, 1120, 1130, 1900};
int n = 5;

// Assume arrays are sorted
int i = 1, j = 0, platforms = 1, result = 1;
while (i < n && j < n) {
  if (arrival[i] <= departure[j]) {
    platforms++;
    i++;
    if (platforms > result) result = platforms;
  } else {
    platforms--;
    j++;
  }
}
printf("%d", result);

medium

A. 4

B. 2

C. 1

D. 3

Solution

1. Step 1: Trace the platform count with given times

   Starts platforms=1 (900 arrival), i=1, j=0. 940 <= 910? No, so platforms--, j=1. Then 940 <= 1200: platforms++, i=2. 950 <= 1200: platforms++, i=3. 1100 <= 1200: platforms++, i=4 (max=3). Later departures reduce platforms.

2. Step 2: Find max platforms during iteration

   Maximum platforms reached is 3 (trains 940-1200, 950-1120, 1100-1130 overlap).

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Max overlapping trains = 3 [OK]

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

Common Mistakes:

• Not sorting arrival and departure arrays
• Incorrectly incrementing or decrementing platform count
• Confusing <= with < in comparisons

4. Identify the error in the following code snippet for calculating minimum platforms:

int arrival[] = {900, 940, 950, 1100, 1500};
int departure[] = {910, 1200, 1120, 1130, 1900};
int n = 5;
int i = 0, j = 0, platforms = 0, result = 0;
while (i < n && j < n) {
  if (arrival[i] < departure[j]) {
    platforms++;
    i++;
    if (platforms > result) result = platforms;
  } else {
    platforms--;
    j++;
  }
}
printf("%d", result);

medium

A. No error, code is correct

B. Initialization of platforms should start at 1

C. Loop condition should be i <= n && j <= n

D. Comparison should be arrival[i] <= departure[j]

Solution

1. Step 1: Verify initialization and first steps

   platforms=0, i=0, j=0 correct. 900 < 910: platforms=1, i=1, result=1.

2. Step 2: Check logic and trace

   940 < 910? No: platforms=0, j=1. Then 940 < 1200: platforms=1, i=2; 950<1200:2, i=3; 1100<1200:3, i=4 (max=3). Continues correctly. < handles equal times optimally.

3. Final Answer:

   No error, code is correct -> Option A

4. Quick Check:

   Code outputs 3, matching overlaps [OK]

Hint: plat=0, i=0,j=0, if arr[i] < dep[j] correctly minimizes platforms [OK]

Common Mistakes:

• Using arrival[i] <= departure[j] (overcounts on equal times)
• Initializing platforms=1 with i=0 (overcounts first arrival)
• Loop i <= n && j <= n (off-by-one)

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 handle trains arriving and departing at the same time (e.g., arrival[i] == departure[j]) to minimize platforms needed?

hard

A. Ignore trains with equal arrival and departure times

B. Treat arrival[i] == departure[j] as arrival first to allocate platform

C. Treat arrival[i] == departure[j] as departure first to free platform

D. Increase platform count by 2 when times are equal

Solution

1. Step 1: Understand impact of equal arrival and departure times

   If a train departs at the same time another arrives, the platform can be reused without adding a new one.

2. Step 2: Adjust comparison logic

   When arrival[i] == departure[j], count departure first to free platform before arrival increments platform count.

3. Final Answer:

   Treat arrival[i] == departure[j] as departure first to free platform -> Option C

4. Quick Check:

   Departure before arrival frees platform [OK]

Hint: Count departures before arrivals when times are equal [OK]

Common Mistakes:

• Counting arrival before departure causing extra platform
• Ignoring equal times leading to wrong count
• Adding platforms unnecessarily