Minimum Number of Platforms in DSA C - 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 changes as the number of trains increases.

Specifically, we ask: How does the running time grow when we have more arrival and departure times?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void findPlatforms(int arr[], int dep[], int n) {
    sort(arr, arr + n);
    sort(dep, dep + n);
    int i = 0, j = 0, platforms = 0, maxPlatforms = 0;
    while (i < n && j < n) {
        if (arr[i] <= dep[j]) {
            platforms++;
            i++;
            if (platforms > maxPlatforms) maxPlatforms = platforms;
        } else {
            platforms--;
            j++;
        }
    }
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the arrival and departure arrays, and then scanning both arrays once.
• How many times: Sorting takes multiple comparisons (depends on n), scanning uses a single loop that runs at most 2n times.

How Execution Grows With Input

Sorting takes more time as the number of trains grows, and scanning grows linearly with the number of trains.

Input Size (n)	Approx. Operations
10	About 10 log 10 + 20 = 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 scanning, roughly n log n growth.

Final Time Complexity

Time Complexity: O(n log n)

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

Common Mistake

[X] Wrong: "The scanning loop alone makes this O(n) time."

[OK] Correct: Sorting the arrays first takes more time than scanning, so the total time is not just linear but includes sorting time.

Interview Connect

Understanding this time complexity helps you explain how sorting and scanning work together efficiently to solve real scheduling problems.

Self-Check

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

Practice

1. What does the problem "Minimum Number of Platforms" ask you to find given train arrival and departure times?

easy

A. The earliest train arrival time

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

C. The total number of trains arriving in a day

D. The maximum number of trains arriving at the same time

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum number of platforms required so that no train has to wait for another to leave.

2. Step 2: Clarify what is being counted

   We count how many trains overlap in time, which determines how many platforms are needed simultaneously.

3. Final Answer:

   The minimum number of platforms needed so no train waits -> Option B

4. Quick Check:

   Minimum platforms = max trains overlapping [OK]

Hint: Count overlapping trains to find minimum platforms needed [OK]

Common Mistakes:

• Confusing total trains with platforms needed
• Ignoring overlapping times
• Counting only arrivals or departures

2. Which of the following C code snippets correctly declares arrays for train arrival and departure times for 5 trains?

easy

A. int arrival[5] = {900, 940, 950, 1100, 1500}; int departure[5] = {910, 1200, 1120, 1130, 1900};

B. int arrival = {900, 940, 950, 1100, 1500}; int departure = {910, 1200, 1120, 1130, 1900};

C. int arrival[5] = (900, 940, 950, 1100, 1500); int departure[5] = (910, 1200, 1120, 1130, 1900);

D. int arrival[] = 900, 940, 950, 1100, 1500; int departure[] = 910, 1200, 1120, 1130, 1900;

Solution

1. Step 1: Check array declaration syntax in C

   Arrays must be declared with size or empty brackets and initialized with curly braces {}.

2. Step 2: Validate each option

   int arrival[5] = {900, 940, 950, 1100, 1500}; int departure[5] = {910, 1200, 1120, 1130, 1900}; uses correct syntax with size and braces. Others use incorrect syntax like parentheses or missing braces.

3. Final Answer:

   int arrival[5] = {900, 940, 950, 1100, 1500}; int departure[5] = {910, 1200, 1120, 1130, 1900}; -> Option A

4. Quick Check:

   Arrays use {} with size or empty brackets [OK]

Hint: Use curly braces {} to initialize arrays in C [OK]

Common Mistakes:

• Using parentheses () instead of braces {}
• Missing array size or empty brackets
• Assigning values without braces

3. Given arrival times {900, 940, 950, 1100, 1500} and departure times {910, 1200, 1120, 1130, 1900}, what is the minimum number of platforms needed? Consider the following code snippet:

int arrival[] = {900, 940, 950, 1100, 1500};
int departure[] = {910, 1200, 1120, 1130, 1900};
int n = 5;

// Assume arrays are sorted
int i = 1, j = 0, platforms = 1, result = 1;
while (i < n && j < n) {
  if (arrival[i] <= departure[j]) {
    platforms++;
    i++;
    if (platforms > result) result = platforms;
  } else {
    platforms--;
    j++;
  }
}
printf("%d", result);

medium

A. 4

B. 2

C. 1

D. 3

Solution

1. Step 1: Trace the platform count with given times

   Starts platforms=1 (900 arrival), i=1, j=0. 940 <= 910? No, so platforms--, j=1. Then 940 <= 1200: platforms++, i=2. 950 <= 1200: platforms++, i=3. 1100 <= 1200: platforms++, i=4 (max=3). Later departures reduce platforms.

2. Step 2: Find max platforms during iteration

   Maximum platforms reached is 3 (trains 940-1200, 950-1120, 1100-1130 overlap).

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Max overlapping trains = 3 [OK]

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

Common Mistakes:

• Not sorting arrival and departure arrays
• Incorrectly incrementing or decrementing platform count
• Confusing <= with < in comparisons

4. Identify the error in the following code snippet for calculating minimum platforms:

int arrival[] = {900, 940, 950, 1100, 1500};
int departure[] = {910, 1200, 1120, 1130, 1900};
int n = 5;
int i = 0, j = 0, platforms = 0, result = 0;
while (i < n && j < n) {
  if (arrival[i] < departure[j]) {
    platforms++;
    i++;
    if (platforms > result) result = platforms;
  } else {
    platforms--;
    j++;
  }
}
printf("%d", result);

medium

A. No error, code is correct

B. Initialization of platforms should start at 1

C. Loop condition should be i <= n && j <= n

D. Comparison should be arrival[i] <= departure[j]

Solution

1. Step 1: Verify initialization and first steps

   platforms=0, i=0, j=0 correct. 900 < 910: platforms=1, i=1, result=1.

2. Step 2: Check logic and trace

   940 < 910? No: platforms=0, j=1. Then 940 < 1200: platforms=1, i=2; 950<1200:2, i=3; 1100<1200:3, i=4 (max=3). Continues correctly. < handles equal times optimally.

3. Final Answer:

   No error, code is correct -> Option A

4. Quick Check:

   Code outputs 3, matching overlaps [OK]

Hint: plat=0, i=0,j=0, if arr[i] < dep[j] correctly minimizes platforms [OK]

Common Mistakes:

• Using arrival[i] <= departure[j] (overcounts on equal times)
• Initializing platforms=1 with i=0 (overcounts first arrival)
• Loop i <= n && j <= n (off-by-one)

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 handle trains arriving and departing at the same time (e.g., arrival[i] == departure[j]) to minimize platforms needed?

hard

A. Ignore trains with equal arrival and departure times

B. Treat arrival[i] == departure[j] as arrival first to allocate platform

C. Treat arrival[i] == departure[j] as departure first to free platform

D. Increase platform count by 2 when times are equal

Solution

1. Step 1: Understand impact of equal arrival and departure times

   If a train departs at the same time another arrives, the platform can be reused without adding a new one.

2. Step 2: Adjust comparison logic

   When arrival[i] == departure[j], count departure first to free platform before arrival increments platform count.

3. Final Answer:

   Treat arrival[i] == departure[j] as departure first to free platform -> Option C

4. Quick Check:

   Departure before arrival frees platform [OK]

Hint: Count departures before arrivals when times are equal [OK]

Common Mistakes:

• Counting arrival before departure causing extra platform
• Ignoring equal times leading to wrong count
• Adding platforms unnecessarily