Minimum Number of Platforms in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to read the number of trains.

DSA C

int n;
scanf("%[1]", &n);

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

Ad

Bf

Cs

Dc

Attempts:

3 left

💡 Hint

Common Mistakes

Using %f or %c instead of %d for integer input.

Forgetting the ampersand (&) before the variable.

The format specifier %d is used to read an integer in C.

2fill in blank

medium

Complete the code to sort the arrival times array.

DSA C

qsort(arrival, n, sizeof(int), [1]);

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Acompare_ints

Bcmpfunc

CcompareInts

Dcompare

Attempts:

3 left

💡 Hint

Common Mistakes

Passing NULL or wrong function name to qsort.

Not defining the comparison function properly.

The comparison function cmpfunc is used with qsort to sort integers.

3fill in blank

hard

Fix the error in the loop condition to find minimum platforms.

DSA C

while (i < n && j [1] n) {
    if (arrival[i] <= departure[j]) {
        platforms++;
        i++;
    } else {
        platforms--;
        j++;
    }
    if (platforms > max_platforms)
        max_platforms = platforms;
}

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A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using greater than or equal operators causing infinite loops.

Using logical OR instead of AND.

The loop should continue while both i and j are less than n, so the condition is i < n && j < n.

4fill in blank

hard

Fill both blanks to correctly compare arrival and departure times.

DSA C

if (arrival[[1]] <= departure[[2]]) {
    platforms++;
    i++;
} else {
    platforms--;
    j++;
}

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Ai

Bj

Ck

Dm

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong indices causing incorrect comparisons.

Mixing up arrival and departure indices.

We compare arrival[i] with departure[j] to decide platform count changes.

5fill in blank

hard

Fill all three blanks to initialize variables correctly.

DSA C

int i = [1], j = [2], platforms = [3];

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A0

B1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting indices at 1 causing out-of-bound errors.

Initializing platforms to 1 instead of 0.

Indices i and j start at 0, and platforms start at 0.

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