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 grows as the number of trains increases.

Specifically, how the operations scale when checking train arrival and departure times.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Given arrival and departure times of trains
// Find minimum platforms needed
int findPlatform(int arr[], int dep[], int n) {
    sort(arr, arr + n);
    sort(dep, dep + n);
    int platform_needed = 1, result = 1;
    int i = 1, j = 0;
    while (i < n && j < n) {
        if (arr[i] <= dep[j]) {
            platform_needed++;
            i++;
        } else {
            platform_needed--;
            j++;
        }
        if (platform_needed > result)
            result = platform_needed;
    }
    return result;
}
    

This code sorts arrival and departure times, then uses two pointers to count platforms needed as trains arrive and leave.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the arrival and departure arrays, and then traversing them with two pointers.
• How many times: Sorting takes about n log n steps; the while loop runs at most 2n times because each pointer moves through the arrays once.

How Execution Grows With Input

As the number of trains (n) grows, sorting takes more time but grows slower than the number of trains squared.

Input Size (n)	Approx. Operations
10	About 10 * log 10 ≈ 33 + 20 (traversal) = 53
100	About 100 * log 100 ≈ 664 + 200 (traversal) = 864
1000	About 1000 * log 1000 ≈ 9966 + 2000 (traversal) = 11966

Pattern observation: The sorting dominates and grows a bit faster than linearly, but the traversal grows linearly.

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 of sorting.

Common Mistake

[X] Wrong: "The while loop alone makes this O(n²) because it has nested conditions."

[OK] Correct: The two pointers each move forward only once through the arrays, so the loop runs at most 2n times, not n squared.

Interview Connect

Understanding this time complexity helps you explain how sorting and linear scanning work together efficiently in real problems.

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 primarily ask you to find?

easy

A. The total number of trains passing through the station

B. The maximum number of trains arriving at the station

C. The least number of train platforms needed so no train waits

D. The shortest time a train stays on a platform

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum platforms so that no train waits due to platform unavailability.

2. Step 2: Clarify what is being counted

   We count platforms needed to handle overlapping train schedules without delay.

3. Final Answer:

   The least number of train platforms needed so no train waits -> Option C

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 distance

2. Which of the following C++ code snippets correctly declares vectors to store arrival and departure times for the Minimum Number of Platforms problem?

easy

A. vector<int> arrival; vector<int> departure;

B. vector arrival, departure int;

C. std::vector arrival(int), departure(int);

D. std::vector<int> arrival, departure;

Solution

1. Step 1: Check correct vector syntax in C++

   Vectors must be declared with type inside angle brackets and variable names, e.g., vector<int> arrival;

2. Step 2: Identify valid declarations

   vector<int> arrival; vector<int> departure; uses correct syntax for two separate vectors: vector<int> arrival; vector<int> departure;

3. Final Answer:

   vector<int> arrival; vector<int> departure; -> Option A

4. Quick Check:

   Correct vector syntax = vector<int> arrival; vector<int> departure; [OK]

Hint: Use vector<int> name; for integer vectors in C++ [OK]

Common Mistakes:

• Missing angle brackets <> for vector type
• Incorrect order of type and variable name
• Using wrong syntax like vector arrival(int)

3. Consider the following C++ code snippet for Minimum Number of Platforms problem:

std::vector<int> arrival = {900, 940, 950, 1100, 1500, 1800};
std::vector<int> departure = {910, 1200, 1120, 1130, 1900, 2000};

int platforms = 1, result = 1;
int i = 1, j = 0;

while (i < arrival.size() && j < departure.size()) {
  if (arrival[i] <= departure[j]) {
    platforms++;
    i++;
  } else {
    platforms--;
    j++;
  }
  if (platforms > result) result = platforms;
}

std::cout << result << std::endl;

What is the output of this code?

medium

A. 2

B. 3

C. 4

D. 1

Solution

1. Step 1: Trace arrival and departure comparisons

   Compare arrival[i] and departure[j]: - i=1(940) <= j=0(910)? No, so platforms-- (1->0), j=1 - i=1(940) <= j=1(1200)? Yes, platforms++ (0->1), i=2 - i=2(950) <= j=1(1200)? Yes, platforms++ (1->2), i=3 - i=3(1100) <= j=1(1200)? Yes, platforms++ (2->3), i=4 - i=4(1500) <= j=1(1200)? No, platforms-- (3->2), j=2 - i=4(1500) <= j=2(1120)? No, platforms-- (2->1), j=3 - i=4(1500) <= j=3(1130)? No, platforms-- (1->0), j=4 - i=4(1500) <= j=4(1900)? Yes, platforms++ (0->1), i=5 - i=5(1800) <= j=4(1900)? Yes, platforms++ (1->2), i=6 (end)

2. Step 2: Find maximum platforms during process

   Maximum platforms reached is 3 during the steps.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Max overlapping trains = 3 [OK]

Hint: Count overlaps by comparing sorted arrivals and departures [OK]

Common Mistakes:

• Not updating platform count correctly
• Mixing up arrival and departure indices
• Assuming max platforms equals total trains

4. The following code snippet is intended to find the minimum number of platforms needed. Identify the bug:

std::vector<int> arrival = {900, 940, 950, 1100};
std::vector<int> departure = {910, 1120, 1130, 1200};

int platforms = 1, result = 1;
int i = 1, j = 0;

while (i < arrival.size() && j < departure.size()) {
  if (arrival[i] < departure[j]) {
    platforms++;
    i++;
  } else {
    platforms--;
    j++;
  }
  if (platforms > result) result = platforms;
}

std::cout << result << std::endl;

medium

A. The comparison should be <= instead of < in if condition

B. The initial value of platforms should be 0, not 1

C. The vectors arrival and departure are not sorted

D. The loop condition should use || instead of &&

Solution

1. Step 1: Analyze the if condition for arrival and departure

   The condition uses arrival[i] < departure[j], but trains arriving exactly when another departs need a platform, so <= is correct.

2. Step 2: Understand impact of using < instead of <=

   Using < misses cases where arrival equals departure, causing undercount of platforms needed.

3. Final Answer:

   The comparison should be <= instead of < in if condition -> Option A

4. Quick Check:

   Use <= to count overlapping trains correctly [OK]

Hint: Use <= to include trains arriving exactly at departure time [OK]

Common Mistakes:

• Using < instead of <= causes wrong platform count
• Not sorting arrival and departure arrays
• Incorrect loop condition with && vs ||

5. You have the following train schedules:

Arrival: {1000, 1010, 1025, 1030, 1040}
Departure: {1015, 1020, 1035, 1045, 1050}

What is the minimum number of platforms required so that no train waits? Choose the correct approach and answer.

hard

A. Sort only arrival times and count unique values; answer is 4

B. Count total trains; answer is 5

C. Use a hash map to count trains per time; answer is 2

D. Sort both arrays, then use two pointers to count overlaps; answer is 2

Solution

1. Step 1: Sort arrival and departure times

   Both arrays are already sorted: arrival = {1000,1010,1025,1030,1040}, departure = {1015,1020,1035,1045,1050}

2. Step 2: Use two pointers to count overlapping trains

   Compare arrival[i] and departure[j]: - 1010 <= 1015: platforms=2 - 1025 <= 1015? No, platforms=1 - 1025 <= 1020? No, platforms=0 - 1025 <= 1035? Yes, platforms=1 - 1030 <= 1035? Yes, platforms=2 - 1040 <= 1035? No, platforms=1 - 1040 <= 1045? Yes, platforms=2 Maximum platforms needed = 2 during overlaps.

3. Final Answer:

   Sort both arrays, then use two pointers to count overlaps; answer is 2 -> Option D

4. Quick Check:

   Two pointer overlap count = 2 platforms [OK]

Hint: Sort arrivals and departures, count overlaps with two pointers [OK]

Common Mistakes:

• Counting total trains instead of overlaps
• Not sorting both arrays
• Using incorrect data structures like hash maps unnecessarily