Minimum Number of Platforms in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to sort the arrival times array.

DSA C++

std::sort(arr, arr + [1]);

Drag options to blanks, or click blank then click option'

An

Barr

Cplatforms

Ddep

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable for array length.

Sorting the departure array instead of arrival.

We sort the arrival times array from index 0 to n to process trains in order of arrival.

2fill in blank

medium

Complete the code to initialize the platform count to 1.

DSA C++

int platform_needed = [1];

Drag options to blanks, or click blank then click option'

A0

B1

Cn

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing platform count to zero.

Using negative numbers.

At least one platform is needed initially for the first train.

3fill in blank

hard

Fix the error in the while loop condition to compare arrival and departure times correctly.

DSA C++

while (i < n && j < n && arr[i] [1] dep[j]) {

Drag options to blanks, or click blank then click option'

A>

B>=

C<=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using greater than or equal instead of less than.

Confusing arrival and departure arrays.

We check if the next train arrives before the earliest departure to decide platform allocation.

4fill in blank

hard

Fill both blanks to update platform count and move pointers correctly inside the loop.

DSA C++

if (arr[i] [1] dep[j]) {
    platform_needed++;
    i[2];
} else {
    j++;
    platform_needed--;
}

Drag options to blanks, or click blank then click option'

A<

B++

C--

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operator.

Decrementing i instead of incrementing.

If arrival is less than departure, increase platform count and move arrival pointer forward.

5fill in blank

hard

Fill all three blanks to track the maximum platforms needed during the process.

DSA C++

int max_platforms = 1;

while (i < n && j < n) {
    if (arr[i] [1] dep[j]) {
        platform_needed++;
        max_platforms = std::max(max_platforms, [2]);
        i[3];
    } else {
        j++;
        platform_needed--;
    }
}

Drag options to blanks, or click blank then click option'

A<

Bplatform_needed

C++

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Not updating max_platforms correctly.

Using wrong operators for comparison or increment.

We update max_platforms with the current platform_needed and increment i when arrival is less than departure.

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