Minimum Number of Platforms in DSA Go - Time & Space Complexity

Time Complexity: Minimum Number of Platforms

O(n log n)

Understanding Time Complexity

We want to find out how the time needed to find the minimum number of train platforms grows as the number of trains increases.

How does the program's work change when we have more trains arriving and leaving?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func findPlatform(arrivals []int, departures []int) int {
    n := len(arrivals)
    sort.Ints(arrivals)
    sort.Ints(departures)

    i, j, platforms, maxPlatforms := 0, 0, 0, 0
    for i < n && j < n {
        if arrivals[i] <= departures[j] {
            platforms++
            if platforms > maxPlatforms {
                maxPlatforms = platforms
            }
            i++
        } else {
            platforms--
            j++
        }
    }
    return maxPlatforms
}
    

This code finds the minimum number of platforms needed so that no train waits, by sorting arrival and departure times and then comparing them.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the arrival and departure arrays, and then a single loop comparing elements.
• How many times: Sorting takes time depending on n (number of trains), and the loop runs at most 2n times because each pointer moves through the arrays once.

How Execution Grows With Input

As the number of trains increases, sorting takes more time, and the loop compares more times.

Input Size (n)	Approx. Operations
10	About 10 log 10 (sorting) + 20 (loop) = ~50 operations
100	About 100 log 100 + 200 = ~900 operations
1000	About 1000 log 1000 + 2000 = ~13000 operations

Pattern observation: The sorting dominates and grows faster than the loop, so total work grows a bit faster than just doubling when input doubles.

Final Time Complexity

Time Complexity: O(n log n)

This means the time needed grows a little faster than the number of trains, mainly because sorting takes more time as trains increase.

Common Mistake

[X] Wrong: "The loop alone makes this O(n) time, so the whole code is O(n)."

[OK] Correct: Sorting the arrays takes more time than the loop, so the sorting step controls the total time, making it O(n log n).

Interview Connect

Understanding how sorting and single-pass loops combine helps you explain your approach clearly and shows you know how to analyze real problems efficiently.

Self-Check

"What if the arrival and departure times were already sorted? How would the time complexity change?"

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