Minimum Number of Platforms in DSA C++ - Build from Scratch

Minimum Number of Platforms

📖 Scenario: You are managing a small train station. You want to find out the minimum number of platforms needed so that no train has to wait.You have the arrival and departure times of trains for the day.

🎯 Goal: Build a program that calculates the minimum number of platforms required at the station so that no train waits.

📋 What You'll Learn

Create two arrays called arrival and departure with exact train times.

Create an integer variable called n for the number of trains.

Write a function called findMinimumPlatforms that takes arrival, departure, and n as parameters and returns the minimum number of platforms needed.

Print the minimum number of platforms after calling the function.

💡 Why This Matters

🌍 Real World

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

💼 Career

This problem teaches sorting and two-pointer techniques useful in scheduling, resource allocation, and interval problems in software engineering.

1

Create train arrival and departure arrays

Create two integer arrays called arrival and departure with these exact values:
arrival = {900, 940, 950, 1100, 1500, 1800}
departure = {910, 1200, 1120, 1130, 1900, 2000}

DSA C++

# Create the arrival and departure arrays
# Your code here

Hint

Use integer arrays with the exact times given for arrival and departure.

2

Create variable for number of trains

Create an integer variable called n and set it to the number of trains, which is 6.

DSA C++

int arrival[] = {900, 940, 950, 1100, 1500, 1800};
int departure[] = {910, 1200, 1120, 1130, 1900, 2000};

// Create the variable n for number of trains
# Your code here

Hint

Count the number of trains and assign it to n.

3

Write function to find minimum platforms

Write a function called findMinimumPlatforms that takes arrival, departure, and n as parameters and returns an integer. Inside the function:
- Sort the arrival and departure arrays.
- Use two pointers i and j starting at 0.
- Use variables platforms_needed and max_platforms initialized to 0.
- While i < n and j < n, if arrival[i] <= departure[j], increment platforms_needed and i, update max_platforms if needed; else decrement platforms_needed and increment j.
- Return max_platforms.

DSA C++

int arrival[] = {900, 940, 950, 1100, 1500, 1800};
int departure[] = {910, 1200, 1120, 1130, 1900, 2000};
int n = 6;

// Write the function findMinimumPlatforms below
# Your code here

Hint

Sort both arrays and use two pointers to count platforms needed.

4

Print the minimum number of platforms

Call the function findMinimumPlatforms with arrival, departure, and n and print the returned value using std::cout.

DSA C++

#include 
#include 

int arrival[] = {900, 940, 950, 1100, 1500, 1800};
int departure[] = {910, 1200, 1120, 1130, 1900, 2000};
int n = 6;

int findMinimumPlatforms(int arrival[], int departure[], int n) {
    std::sort(arrival, arrival + n);
    std::sort(departure, departure + n);

    int platforms_needed = 0, max_platforms = 0;
    int i = 0, j = 0;

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

int main() {
    // Print the minimum number of platforms
    std::cout << findMinimumPlatforms(arrival, departure, n) << std::endl;
    return 0;
}

Hint

Call the function and print the result using std::cout.

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