Minimum Number of Platforms in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the arrival times slice.

DSA Go

sort.[1](arrival)

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

ASwap

BShuffle

CSort

DReverse

Attempts:

3 left

💡 Hint

Common Mistakes

Using Reverse instead of Sort causes incorrect order.

Using Shuffle or Swap does not sort the slice.

We use sort.Sort to sort the slice in ascending order, which is necessary to process arrival times correctly.

2fill in blank

medium

Complete the code to increment the platform count when a train arrives before the previous one departs.

DSA Go

if arrival[i] [1] departure[j] {

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

A<=

B>

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < reverses the logic.

Using == or <= does not correctly identify overlapping trains.

If the arrival time is less than the departure time, it means a new platform is needed.

3fill in blank

hard

Fix the error in the loop condition to avoid index out of range error.

DSA Go

for i, j := 0, 0; i < len(arrival) && j [1] len(departure); {

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

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= causes index out of range panic.

Using > or >= reverses the loop logic.

Both i and j must be strictly less than the length of their slices to avoid out of range errors.

4fill in blank

hard

Fill both blanks to correctly update platform counts and indices.

DSA Go

if arrival[i] [1] departure[j] {
    plat_needed++
    i[2]
} else {
    plat_needed--
    j++
}

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

A<

B++

C--

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < reverses the condition.

Using -- instead of ++ for i causes wrong index update.

When arrival is less than departure, we increase platform count and move to next arrival (i++).

5fill in blank

hard

Fill all three blanks to create a map of platform counts per time and filter times with more than one platform needed.

DSA Go

platforms := map[int]int{}
for i := 0; i < len(arrival); i++ {
    platforms[[1]]++
}
result := map[int]int{t: c for t, c in platforms if c [2] [3]

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

Aarrival[i]

B>

C1

Ddeparture[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using departure[i] instead of arrival[i] counts wrong times.

Using <= or < instead of > filters incorrectly.

We count platforms at arrival times and filter those with count greater than 1.

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