Minimum Number of Platforms in DSA C - Build from Scratch

Minimum Number of Platforms

📖 Scenario: Imagine a small train station where trains arrive and depart at different times. We want to find out the minimum number of platforms needed so that no train has to wait.

🎯 Goal: You will write a program to find the minimum number of platforms required for the trains based on their arrival and departure times.

📋 What You'll Learn

Create two arrays called arrivals and departures with exact train times

Create an integer variable called n for the number of trains

Write a function called minPlatforms that takes arrivals, departures, and n as input and returns the minimum number of platforms needed

Print the result using printf

💡 Why This Matters

🌍 Real World

Train stations and airports use this logic to manage platforms and gates efficiently to avoid delays.

💼 Career

Understanding scheduling and resource allocation algorithms is important for software engineers working on transportation, logistics, and event management systems.

1

Create train arrival and departure arrays

Create two integer arrays called arrivals and departures with these exact values: arrivals = {900, 940, 950, 1100, 1500, 1800} and departures = {910, 1200, 1120, 1130, 1900, 2000}. Also create an integer variable called n and set it to 6.

DSA C

# Create arrays arrivals and departures with given times and integer n = 6
// Your code here

Hint

Use int arrivals[] = {...}; and int departures[] = {...}; to create the arrays. Set n to 6.

2

Declare the minPlatforms function

Declare a function called minPlatforms that takes three parameters: two integer arrays arrivals and departures, and an integer n. The function should return an integer.

DSA C

#include 

// Declare minPlatforms function here
// Your code here

int main() {
    int arrivals[] = {900, 940, 950, 1100, 1500, 1800};
    int departures[] = {910, 1200, 1120, 1130, 1900, 2000};
    int n = 6;
    return 0;
}

Hint

Write the function declaration with the correct parameters and return type.

3

Implement the minPlatforms function logic

Implement the minPlatforms function to calculate the minimum number of platforms needed. Use sorting for arrivals and departures, then use two pointers to count platforms needed. Return the maximum platforms required.

DSA C

#include 

int minPlatforms(int arrivals[], int departures[], int n) {
    // Your code here
}

int main() {
    int arrivals[] = {900, 940, 950, 1100, 1500, 1800};
    int departures[] = {910, 1200, 1120, 1130, 1900, 2000};
    int n = 6;
    return 0;
}

Hint

Sort both arrays. Use two indexes i and j to traverse arrivals and departures. Increase platform count if arrival is before or equal to departure, else decrease. Track max platforms.

4

Print the minimum number of platforms

Call the minPlatforms function with arrivals, departures, and n. Store the result in an integer variable called result. Then print Minimum number of platforms needed: followed by result using printf.

DSA C

#include 

void sort(int arr[], int n) {
    for (int i = 0; i  arr[j + 1]) {
                int temp = arr[j];
                arr[j] = arr[j + 1];
                arr[j + 1] = temp;
            }
        }
    }
}

int minPlatforms(int arrivals[], int departures[], int n) {
    sort(arrivals, n);
    sort(departures, n);

    int platform_needed = 1, max_platforms = 1;
    int i = 1, j = 0;

    while (i  max_platforms) {
                max_platforms = platform_needed;
            }
            i++;
        } else {
            platform_needed--;
            j++;
        }
    }
    return max_platforms;
}

int main() {
    int arrivals[] = {900, 940, 950, 1100, 1500, 1800};
    int departures[] = {910, 1200, 1120, 1130, 1900, 2000};
    int n = 6;
    // Call minPlatforms and print the result
    // Your code here
    return 0;
}

Hint

Call minPlatforms(arrivals, departures, n), store in result, then print with printf.

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