Minimum Number of Platforms in DSA Rust - 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 key idea to solve the Minimum Number of Platforms problem efficiently?

Sort arrival and departure times separately, then use two pointers to track overlapping trains to find the maximum platforms needed at any time.

intermediate

Why do we sort arrival and departure times separately in this problem?

Sorting separately helps us process events in chronological order and easily count how many trains are at the station simultaneously.

intermediate

In the two-pointer approach, what does it mean when arrival[i] <= departure[j]?

It means a train arrives before the earliest departure, so we need an extra platform and move the arrival pointer forward.

beginner

What data structure is commonly used to solve the Minimum Number of Platforms problem in O(n log n) time?

Arrays for arrival and departure times, sorted separately, and two pointers to traverse them.

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 Use a queue to store trains

D Calculate total waiting time

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

If arrival[i] > departure[j], what should we do in the two-pointer approach?

A Stop the algorithm

B Increase platform count

C Move arrival pointer forward

D Move departure pointer forward

If arrival is after departure, a platform is freed, so move departure pointer forward.

What does the maximum number of platforms needed represent?

A Maximum trains at station at the same time

B Total trains arriving

C Total trains departing

D Minimum waiting time

It represents the peak number of trains present simultaneously.

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

A O(n)

B O(n^2)

C O(n log n)

D O(log n)

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

Which of these is NOT required to solve the Minimum Number of Platforms problem?

A Arrival times array

B Graph data structure

C Departure times array

D Sorting both arrays

Graph is not needed; arrays and sorting suffice.

Explain the two-pointer approach to find the minimum number of platforms needed for trains.

Think about processing train arrival and departure events in order.

Describe why sorting arrival and departure times separately helps solve the Minimum Number of Platforms problem.

Consider how to know when trains overlap in time.

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 at the station

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

C. The maximum speed of any train

D. The total number of trains arriving in a day

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 B

4. Quick Check:

   Minimum platforms = no waiting trains [OK]

Hint: Focus on overlapping train times to find platform count [OK]

Common Mistakes:

• Confusing total trains with platforms needed
• Ignoring overlapping arrival and departure times
• Thinking about train speed or travel time

2. Which Rust data structure is best suited to keep track of train departure times when solving the Minimum Number of Platforms problem?

easy

A. BinaryHeap (priority queue)

B. VecDeque (double-ended queue)

C. HashMap (key-value store)

D. LinkedList

Solution

1. Step 1: Identify the need for ordering

   We need to quickly access the earliest departure time to compare with arrivals.

2. Step 2: Choose a data structure that supports this

   A BinaryHeap (priority queue) efficiently gives the smallest or largest element, perfect for earliest departure.

3. Final Answer:

   BinaryHeap (priority queue) -> Option A

4. Quick Check:

   Priority queue = earliest departure access [OK]

Hint: Use priority queue to track earliest departure times [OK]

Common Mistakes:

• Using HashMap which doesn't keep order
• Using LinkedList which is inefficient for min extraction
• Using VecDeque which is FIFO, not sorted

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

let arrivals = vec![900, 940, 950, 1100, 1500, 1800];
let departures = vec![910, 1200, 1120, 1130, 1900, 2000];

// After sorting both vectors
// Calculate minimum platforms needed
// Output the result

// Assume correct implementation of algorithm
println!("{}", min_platforms(arrivals, departures));

medium

A. 3

B. 2

C. 4

D. 1

Solution

1. Step 1: Sort arrival and departure times

   Arrivals sorted: [900, 940, 950, 1100, 1500, 1800] Departures sorted: [910, 1120, 1130, 1200, 1900, 2000]

2. Step 2: Count overlapping trains

   At 1100 arrival, trains at station are 3 (940-1200, 950-1120, 1100-1130 overlap). So minimum platforms = 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Max overlap = 3 platforms [OK]

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

Common Mistakes:

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

4. Identify the bug in this Rust function snippet for minimum platforms calculation:

fn min_platforms(arrivals: Vec<i32>, departures: Vec<i32>) -> i32 {
    let mut arr = arrivals.clone();
    let mut dep = departures.clone();
    arr.sort();
    dep.sort();
    let mut platform_needed = 1;
    let mut max_platforms = 1;
    let mut i = 1;
    let mut j = 0;
    while i < arr.len() && j < dep.len() {
        if arr[i] <= dep[j] {
            platform_needed += 1;
            i += 1;
        } else {
            platform_needed -= 1;
            j += 1;
        }
        if platform_needed > max_platforms {
            max_platforms = platform_needed;
        }
    }
    max_platforms
}

medium

A. Using clone() causes runtime error

B. Sorting departures is unnecessary

C. Starting i from 1 skips first arrival comparison

D. The loop condition should be OR instead of AND

Solution

1. Step 1: Check loop initialization

   Variable i starts at 1, skipping the first arrival time comparison which can cause incorrect platform count.

2. Step 2: Understand impact of skipping first arrival

   Missing the first arrival means initial platform count may be wrong, leading to incorrect max platforms.

3. Final Answer:

   Starting i from 1 skips first arrival comparison -> Option C

4. Quick Check:

   Indexing from 1 skips first train [OK]

Hint: Start indices at 0 to include all trains [OK]

Common Mistakes:

• Ignoring index start positions
• Thinking sorting is unnecessary
• Confusing clone() with error cause

5. Given arrival times [100, 140, 150, 200, 215, 400] and departure times [110, 300, 220, 230, 315, 600], what is the minimum number of platforms needed? Explain why this number is required.

hard

A. 5

B. 3

C. 2

D. 4

Solution

1. Step 1: Sort arrival and departure times

   Arrivals sorted: [100, 140, 150, 200, 215, 400] Departures sorted: [110, 220, 230, 300, 315, 600]

2. Step 2: Count overlapping trains step-by-step

   At 215 arrival, trains at station: 140-300, 150-220, 200-230, 215-315 (4) overlap. Maximum overlap is 4, so 4 platforms needed.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Max overlap = 4 platforms [OK]

Hint: Find max overlapping intervals for platform count [OK]

Common Mistakes:

• Not sorting times before counting
• Miscounting overlaps at arrival times
• Assuming platforms equal total trains