Minimum Number of Platforms in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the arrival times vector.

DSA Rust

arrivals.sort_by(|a, b| a.[1]b);

Drag options to blanks, or click blank then click option'

Apartial_cmp

Beq

Ccmp

Dne

Attempts:

3 left

💡 Hint

Common Mistakes

Using eq or ne which return booleans, not ordering.

Using partial_cmp which returns Option.

The method cmp is used to compare two values for sorting in Rust.

2fill in blank

medium

Complete the code to increment the platform count when a train arrives.

DSA Rust

if arrivals[i] <= departures[j] {
    platforms_needed[1];
    i += 1;
}

Drag options to blanks, or click blank then click option'

A-= 1

B+= 1

C= 1

D*= 1

Attempts:

3 left

💡 Hint

Common Mistakes

Decrementing platforms_needed instead of incrementing.

Assigning platforms_needed to 1 instead of incrementing.

When a train arrives before the earliest departure, we need one more platform, so increment by 1.

3fill in blank

hard

Fix the error in updating max_platforms when platforms_needed changes.

DSA Rust

if platforms_needed [1] max_platforms {
    max_platforms = platforms_needed;
}

Drag options to blanks, or click blank then click option'

A>=

B<=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using < or <= which would update max_platforms incorrectly.

Using >= which updates max_platforms even when equal.

We update max_platforms only if platforms_needed is greater than current max_platforms.

4fill in blank

hard

Fill both blanks to correctly handle train departure and decrement platform count.

DSA Rust

else {
    platforms_needed [1];
    j [2];
}

Drag options to blanks, or click blank then click option'

A-= 1

B+= 1

Attempts:

3 left

💡 Hint

Common Mistakes

Incrementing platforms_needed on departure.

Decrementing j instead of incrementing.

When a train departs, we reduce platforms_needed by 1 and move to next departure by incrementing j.

5fill in blank

hard

Fill all three blanks to complete the loop that calculates minimum platforms.

DSA Rust

while i < arrivals.len() && j < departures.len() {
    if arrivals[i] [1] departures[j] {
        platforms_needed [2];
        i += 1;
    } else {
        platforms_needed [3];
        j += 1;
    }
}

Drag options to blanks, or click blank then click option'

A<=

B+= 1

C-= 1

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using == instead of <= for comparison.

Mixing increment and decrement operators.

If arrival is less or equal to departure, increment platforms_needed; else decrement it.

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