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DSA C#programming~5 mins

Minimum Number of Platforms in DSA C# - Time & Space Complexity

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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 platforms changes as the number of trains increases.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


// Find minimum platforms needed for trains
int FindPlatforms(int[] arrivals, int[] departures) {
    Array.Sort(arrivals);
    Array.Sort(departures);
    int platformNeeded = 0, maxPlatforms = 0;
    int i = 0, j = 0;
    while (i < arrivals.Length && j < departures.Length) {
        if (arrivals[i] <= departures[j]) {
            platformNeeded++;
            i++;
            if (platformNeeded > maxPlatforms) maxPlatforms = platformNeeded;
        } else {
            platformNeeded--;
            j++;
        }
    }
    return maxPlatforms;
}
    

This code sorts arrival and departure times, then counts platforms needed by comparing times.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: One while loop that moves through both arrival and departure arrays.
  • How many times: Each element in arrivals and departures is visited once, so about 2n steps for n trains.
How Execution Grows With Input

As the number of trains grows, the sorting and the single loop both take more time.

Input Size (n)Approx. Operations
10About 10 log 10 + 20 steps
100About 100 log 100 + 200 steps
1000About 1000 log 1000 + 2000 steps

Pattern observation: Sorting grows faster than the loop, so sorting dominates the time.

Final Time Complexity

Time Complexity: O(n log n)

This means the time grows a bit faster than the number of trains because sorting takes more time as input grows.

Common Mistake

[X] Wrong: "The time is just O(n) because we only loop once through the arrays."

[OK] Correct: Sorting the arrays before looping takes more time, so the sorting step makes the total time O(n log n), not just O(n).

Interview Connect

Understanding how sorting affects the overall time helps you explain your solution clearly and shows you know how to analyze real problems efficiently.

Self-Check

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

Practice

(1/5)
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 output

    The output is the minimum number of platforms required to handle all trains 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: 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 speed or duration instead
2. Which of the following C# code snippets correctly declares arrays for arrival and departure times for the Minimum Number of Platforms problem?
easy
A. int arrival = [900, 940, 950]; int departure = [910, 1200, 1120];
B. int arrival[] = (900, 940, 950); int departure[] = (910, 1200, 1120);
C. int[] arrival = {900, 940, 950}; int[] departure = {910, 1200, 1120};
D. int[] arrival = (900, 940, 950); int[] departure = (910, 1200, 1120);

Solution

  1. Step 1: Recall C# array declaration syntax

    In C#, arrays are declared with type[], and values inside curly braces {}.
  2. Step 2: Check each option for correct syntax

    int[] arrival = {900, 940, 950}; int[] departure = {910, 1200, 1120}; uses int[] and curly braces {}, which is correct. Others use parentheses or wrong syntax.
  3. Final Answer:

    int[] arrival = {900, 940, 950}; int[] departure = {910, 1200, 1120}; -> Option C
  4. Quick Check:

    C# arrays use type[] and {} [OK]
Hint: Use int[] and curly braces {} for arrays in C# [OK]
Common Mistakes:
  • Using parentheses () instead of curly braces {}
  • Declaring arrays without [] after type
  • Using square brackets [] for initialization
3. What is the output of the following C# code snippet that calculates minimum platforms?
int[] arrival = {900, 940, 950, 1100, 1500, 1800};
int[] departure = {910, 1200, 1120, 1130, 1900, 2000};

int platforms = CalculateMinPlatforms(arrival, departure);
Console.WriteLine(platforms);

// Assume CalculateMinPlatforms correctly implements the algorithm.
medium
A. 5
B. 4
C. 2
D. 3

Solution

  1. Step 1: Sort arrival and departure times

    Arrival: 900, 940, 950, 1100, 1500, 1800 Departure: 910, 1120, 1130, 1200, 1900, 2000
  2. Step 2: Count overlapping trains to find max platforms

    At 950, trains at 900, 940, 950 overlap (3 trains). Later overlaps are fewer.
  3. Final Answer:

    3 -> Option D
  4. Quick Check:

    Max overlap = 3 platforms [OK]
Hint: Find max overlapping intervals for minimum platforms [OK]
Common Mistakes:
  • Not sorting times before counting overlaps
  • Counting total trains instead of overlaps
  • Mixing arrival and departure arrays
4. Identify the error in this C# code snippet for calculating minimum platforms:
int[] arrival = {900, 940, 950};
int[] departure = {910, 1200, 1120};

Array.Sort(arrival);
Array.Sort(departure);

int i = 0, j = 0, platforms = 0, maxPlatforms = 0;
while (i < arrival.Length && j < departure.Length) {
    if (arrival[i] <= departure[j]) {
        platforms++;
        i++;
    } else {
        platforms--;
        j++;
    }
    maxPlatforms = Math.Max(maxPlatforms, platforms);
}
Console.WriteLine(maxPlatforms);
medium
A. The comparison should be 'arrival[i] < departure[j]' instead of '<='
B. The loop condition should be 'i < arrival.Length || j < departure.Length'
C. The arrays should not be sorted before processing
D. The platforms variable should not be incremented

Solution

  1. Step 1: Analyze the comparison operator in the loop

    Using '<=' causes counting a train as arriving exactly when another departs, increasing platforms unnecessarily.
  2. Step 2: Correct comparison to '<' to avoid overlap at exact departure time

    Changing to '<' means a train arriving at the same time another departs does not increase platform count.
  3. Final Answer:

    The comparison should be 'arrival[i] < departure[j]' instead of '<=' -> Option A
  4. Quick Check:

    Use '<' to avoid counting simultaneous arrival and departure as overlap [OK]
Hint: Use '<' not '<=' when comparing arrival and departure [OK]
Common Mistakes:
  • Using '<=' causes overcounting platforms
  • Not sorting arrays before processing
  • Incorrect loop condition causing missed trains
5. Given 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 equals max during iteration and store them
B. Return only the maximum platform count without intervals
C. Sort only arrival times and ignore departure times
D. Count total trains without checking overlaps

Solution

  1. Step 1: Modify algorithm to track platform count changes

    While iterating, record start time when platforms reach max and end time when count drops.
  2. Step 2: Store intervals where platform count equals maximum

    Keep a list of intervals with max platforms to return after processing all times.
  3. Final Answer:

    Track intervals when platform count equals max during iteration and store them -> Option A
  4. Quick Check:

    Record max platform intervals during count updates [OK]
Hint: Save start/end times when platform count hits max [OK]
Common Mistakes:
  • Ignoring time intervals, returning only counts
  • Not updating intervals when platform count changes
  • Sorting only one array causing wrong intervals
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